Quantum physics

Class 12 Physics, unit 21, book pages 124–140

At the atomic scale light and matter behave in ways classical physics cannot explain. This unit shows that electromagnetic radiation comes in packets called photons, each with energy and momentum , and uses this idea to explain the photoelectric effect, the Compton effect, pair production and pair annihilation. It then shows that particles such as electrons behave as waves (), which makes the electron microscope possible, introduces Heisenberg's uncertainty principle, and explains line spectra with discrete energy levels ().

Learning outcomes (19)
  • State that electromagnetic radiation has a particulate nature.
  • Explain and apply the photonic model of light to solve problems (use ), and use the electronvolt (eV) as a unit of energy.
  • Explain that a photon has momentum given by , and connect this with the idea that light can exert a force.
  • Describe that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation.
  • Describe and use the terms threshold frequency and threshold wavelength.
  • Explain photoelectric emission in terms of photon energy and work function energy.
  • State and apply .
  • Explain why the maximum kinetic energy of photoelectrons is independent of intensity, whereas the photoelectric current is proportional to intensity.
  • Juxtapose the evidence for light as a wave and as a particle: the photoelectric effect provides evidence for a particulate nature of electromagnetic radiation, while interference and diffraction provide evidence for a wave nature.
  • Analyse qualitatively the evidence provided by electron diffraction for the wave nature of particles.
  • Explain and apply the de Broglie wavelength to solve problems (use ).
  • State that there are discrete electron energy levels in isolated atoms (for example atomic hydrogen).
  • Explain the appearance and formation of emission and absorption line spectra.
  • Use to solve problems.
  • Describe the Compton effect qualitatively.
  • Explain the phenomena of pair production and pair annihilation.
  • Explain how electron microscopes achieve very high resolution.
  • State and explain Heisenberg's uncertainty principle qualitatively.
  • Use the uncertainty principle to explain why empirical measurements must necessarily have uncertainty in them.

What quantum physics is about

Newton's mechanics and Maxwell's theory of electricity and magnetism (together called classical physics) describe everyday objects very well. Around 1900, experiments on atoms and light gave results that classical physics could not explain at all: the radiation given out by hot bodies, the photoelectric effect and the line spectra of gases.

Quantum physics is the branch of physics that studies the behaviour of matter and energy at the atomic and subatomic level. At this scale the rules of classical physics are replaced by new ones that at first look strange. This unit builds four of those ideas.

The big ideas of Unit 21
Quantum ideaWhat it saysEvidence in this unit
Light comes in packetsLight energy is carried by photons, each of energy and momentum .Photoelectric effect, Compton effect, pair production
Matter behaves as wavesA moving particle has a wavelength .Electron diffraction, electron microscope
Limits to measurementPosition and momentum cannot both be known exactly.Heisenberg's uncertainty principle
Atoms have fixed energiesElectrons in atoms occupy only certain energy levels.Emission and absorption line spectra

21.1Electromagnetic waves and the spectrum

In 1864 the British physicist James Clerk Maxwell predicted that an accelerated electric charge produces linked electric and magnetic disturbances that travel through space. A changing electric field produces a magnetic field, and a changing magnetic field produces an electric field, so the two fields keep each other going. This travelling pair of fields is an electromagnetic (EM) wave.

Electromagnetic wave
A wave made of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of travel. It is produced by accelerated charges and travels through vacuum at .
An electromagnetic wave. The electric field (vertical) and the magnetic field (drawn in perspective, into and out of the page) oscillate in step, at right angles to each other and to the direction of travel .

Properties of electromagnetic waves

  • The electric and magnetic fields are perpendicular to each other and to the direction of propagation, so EM waves are transverse waves.
  • The direction of propagation is the direction of the vector product .
  • The two fields oscillate in phase and with the same frequency.
  • All EM waves travel through vacuum with the same speed . They need no medium, which is why sunlight reaches Earth through empty space. They can also travel through some materials, more slowly.
  • EM waves carry energy and momentum from one point to another.

Frequency , wavelength and speed are related by the wave equation:

: speed of light (), : frequency in hertz (Hz), : wavelength in metres.

The electromagnetic spectrum

Visible light from the glowing filament of a bulb is one example of an EM wave. Radio waves, microwaves, infrared, ultraviolet, X-rays and gamma rays are the same kind of wave. They differ only in wavelength (and so in frequency). Arranged in order of wavelength they form the electromagnetic spectrum.

The electromagnetic spectrum with typical wavelengths. The visible band is a very narrow part, from about (red) to (violet).
RegionTypical wavelengthTypical source or use
Radio wavesRadio and TV broadcasting
MicrowavesRadar, mobile phones, microwave ovens
InfraredHeat radiation, TV remote controls
Visible light to Light seen by the eye
UltravioletSun, sterilising equipment
X-raysMedical imaging, crystal studies
Gamma raysRadioactive nuclei, cancer treatment

Inside the visible band the colours in order of decreasing wavelength are red, orange, yellow, green, blue, indigo and violet. Red light has the longest wavelength and the lowest frequency; violet has the shortest wavelength and the highest frequency.

Frequencies of red and violet light
Find the frequencies of red light () and violet light ().
  1. Use for red light, with in metres.
  2. Repeat for violet light.

21.1Planck's quantum theory and the photonOften examined

Classical physics treated light energy as a smooth, continuous flow. In 1900 the German physicist Max Planck studied the radiation emitted by hot bodies (from their atoms and molecules). He found that the measured results at all wavelengths could be explained only if energy is emitted or absorbed in small, fixed packets. His idea is called Planck's quantum theory, and it gave birth to quantum physics.

Postulates of Planck's quantum theory

  1. Atoms and molecules emit or absorb energy only in discrete amounts (not continuously). The energy is , where is a whole number called the quantum number.
  2. The smallest amount of energy that can be emitted or absorbed as electromagnetic radiation is called a quantum (plural quanta). For radiation of frequency one quantum has energy .
  3. The energies of atoms and molecules are therefore quantized. The allowed energy states are called quantum states. An atom emits or absorbs energy only when it jumps from one quantum state to another.

(quantum number), : frequency of the radiation, : Planck's constant.

Planck's constant (h)
The constant that links the energy of a quantum to its frequency, (usually rounded to ). Its unit is joule second ().
Quantum
The smallest packet of energy that can be emitted or absorbed as electromagnetic radiation, .
Photon
A quantum of electromagnetic radiation (light). It travels at speed , has zero rest mass, energy and momentum .

Energy of a photon

In 1905 Einstein went further: light itself travels as a stream of these packets, now called photons. The energy of one photon depends only on the frequency of the light. Using :

Photon energy is directly proportional to frequency and inversely proportional to wavelength.

The electronvolt

Photon energies in joules are tiny numbers such as . Atomic and nuclear physics therefore uses a much smaller unit of energy, the electronvolt.

Electronvolt (eV)
The energy gained (or lost) by an electron when it moves through a potential difference of one volt. .

Derivation Relation between the electronvolt and the joule

Aim. Express in joules, the SI unit of energy.

  1. When a charge moves through a potential difference , the work done on it (and the kinetic energy it gains) is the charge times the p.d.
  2. For one electron the charge is . By definition of the electronvolt, the p.d. is .
  3. A coulomb times a volt is a joule, because .
  4. Larger multiples follow from the prefixes.
Result

The electronvolt is a very small unit, the right size for single electrons, atoms and photons. To change eV into joules multiply by ; to change joules into eV divide by .

Example 21.1
What is the frequency of a photon whose energy is ?
Given
  • photon energy
To find
  • frequency
  1. Change the energy into joules, because is in .
  2. Use and make the subject.
Assignment 21.1
What will be the photon energy for a wavelength of angstroms, if the energy of a photon corresponding to a wavelength of angstroms is ?
  1. Photon energy is inversely proportional to wavelength (), so write a ratio. and cancel.
  2. Substitute , and . The units of wavelength cancel, so there is no need to convert angstroms.

21.3Momentum of a photonOften examined

A photon has zero rest mass: a photon cannot be at rest, it always moves at speed . Even so it carries momentum. Maxwell's theory already showed that an electromagnetic wave carrying energy also carries momentum . In the photon picture every photon carries its own share of this momentum.

Derivation Momentum of a photon

Aim. Show that a photon of energy has momentum .

Assumptions.
  • The photon moves at the speed of light and has zero rest mass.
  • Its energy is (Planck).
  1. By Einstein's mass-energy relation, energy is equivalent to a mass . A photon moving at speed therefore behaves as if it had this mass (it is not a rest mass).
  2. Momentum is mass times velocity, and the photon's velocity is .
  3. Put in the photon energy .
  4. Use , so .
Result

A photon's momentum grows with its frequency: a short-wavelength X-ray photon carries far more momentum than a photon of visible light. Rewriting as shows that the energy of a photon is directly proportional to its momentum.

Light can exert a force

When light falls on a surface, the photons are absorbed or reflected, so their momentum changes. By Newton's second law, a rate of change of momentum is a force. So a beam of light pushes on any surface it hits. The push per unit area is called radiation pressure.

Derivation Force exerted by a beam of light

Aim. Find the force on a surface when a light beam of power falls on it at right angles.

  1. Each photon has energy , so the number of photons arriving per second is the power divided by the energy of one photon.
  2. If the surface absorbs the light, each photon gives up all its momentum .
  3. Force equals the momentum delivered per second: number of photons per second times momentum per photon.
  4. If the surface reflects the light straight back, each photon's momentum reverses from to , a change of , so the force doubles.
Result

Because is so large the force is tiny: a beam absorbed by a surface pushes with only . It is still real and measurable.

Force from a laser beam
A laser beam of wavelength is completely absorbed by a black surface. Find the number of photons striking the surface per second and the force on it.
  1. Energy of one photon.
  2. Photons per second = power / energy per photon.
  3. Force for complete absorption.

21.2Photoelectric effect: experiment and observationsOften examined

The word photoelectric joins two words: photo (light) and electric (electron). The photoelectric effect is an interaction between light (photons) and the electrons of a metal.

Photoelectric effect
The emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it.
Photoelectrons
The electrons emitted from a metal surface because of light falling on it.
Photoelectric current
The current produced when the photoelectrons are collected and flow round a circuit.

Experimental set-up

Set-up to study the photoelectric effect. Light falls on the emitter plate E inside an evacuated tube; the photoelectrons travel to the collector C and the ammeter A records the current. The battery sets the potential difference V and can be reversed.
  • An evacuated glass tube contains two metal plates. The emitter E (the metal surface under study) is joined to the negative terminal of a battery, and the collector C (also called the anode or detector) is kept at a positive potential.
  • The tube is evacuated so that the emitted electrons do not collide with air molecules and the metal surface stays clean.
  • When the tube is kept in the dark the ammeter reads zero: there is no current in the circuit.
  • When light of suitable frequency falls on E, the ammeter shows a current at once. Electrons are emitted from the negative plate, cross the gap and are collected by the positive plate, completing the circuit.

Effect of applied voltage on photoelectric current

Keep the frequency of the light fixed and plot the photoelectric current I against the potential V of the collector, for three different intensities. The graph shows these features:

Photoelectric current against collector potential
Light of one frequency at three intensities. Brighter light gives a larger saturation current, but every curve reaches zero at the same stopping potential -V_0.
  • For positive V the current rises and then becomes constant. This maximum value is the saturation current: every electron emitted per second is now collected by the anode.
  • The saturation current is larger for brighter light. Brighter light of the same frequency ejects more electrons per second, but it does not give each electron more energy.
  • A small current flows even at V=0, because the electrons leave the metal with some kinetic energy.
  • When V is made negative (battery reversed), the collector repels the electrons. Only electrons with kinetic energy greater than e|V| reach it, so the current falls.
  • At a certain negative potential -V_0 the current becomes zero: even the fastest electrons are turned back. V_0 is the stopping potential. It is the same for all three intensities.
Stopping potential (V_0)
The minimum negative (retarding) potential of the collector, relative to the emitter, at which the photoelectric current becomes zero. It measures the maximum kinetic energy of the photoelectrons.
Saturation current
The largest photoelectric current, reached when all the photoelectrons emitted per second are collected.

Derivation Maximum kinetic energy and stopping potential

Aim. Show that K.E_{\max}=eV_0.

Assumptions.
  • e=1.6\times10^{-19}\,\mathrm{C} is the charge of an electron.
  • The electrons move in a vacuum, so no energy is lost in collisions.
  1. The fastest photoelectron leaves the emitter with kinetic energy K.E_{\max}=\tfrac12mv_{\max}^{2} and moves towards the negative collector.
    K.E_{\max}=\tfrac12mv_{\max}^{2}
  2. Moving against a retarding p.d. V_0, the electron (charge e) does work eV_0 against the electric force. This work equals the electric potential energy it gains.
    \text{work done against field}=eV_0
  3. At the stopping potential this fastest electron just comes to rest at the collector, so all its kinetic energy has turned into potential energy (conservation of energy).
    \tfrac12mv_{\max}^{2}=eV_0
ResultK.E_{\max}=eV_0\qquad(21.2)

Measuring the stopping potential with a voltmeter gives the maximum kinetic energy directly. A stopping potential of 2\,\mathrm{V} means K.E_{\max}=2\,\mathrm{eV}=3.2\times10^{-19}\,\mathrm{J}.

Effect of intensity of incident radiation on photoelectric current

Keep the frequency and the potential difference fixed and change only the intensity. The photoelectric current increases linearly with intensity. The current is proportional to the number of photoelectrons emitted per second, so:

Photoelectric current against intensity
S is the intensity of the light. Frequency and voltage fixed. Doubling the intensity doubles the current, so the number of photoelectrons per second is proportional to the intensity.
Maximum kinetic energy against intensity
S is the intensity of the light, frequency fixed. Brighter light does not give faster electrons: K.E_{\max} stays the same.

Effect of frequency of incident radiation on the kinetic energy of photoelectrons

Now keep the intensity fixed and change the frequency. The maximum kinetic energy of the photoelectrons depends on the frequency. Even very dim light of high frequency releases electrons with large kinetic energy. For every metal there is a threshold frequency f_0: below it no electrons are emitted, however intense the light is. Classical physics has no explanation for this frequency dependence.

Maximum kinetic energy against frequency
No electrons are emitted below the threshold frequency f_0. Above f_0 the graph is a straight line of slope h; extended back (dashed) it cuts the energy axis at -\Phi. The line is drawn to scale for sodium (\Phi=2.3\,\mathrm{eV}, f_0=5.6\times10^{14}\,\mathrm{Hz}).
  • Below f_0: no emission at all, even with very bright light.
  • Above f_0: K.E_{\max} increases linearly with f. The graph is a straight line whose slope is Planck's constant h for every metal.
  • Extending the line backwards (dashed) meets the energy axis at -\Phi, where \Phi is the work function of the metal.
Current against voltage for two frequencies
Same intensity, two frequencies. The higher frequency needs a larger stopping potential (V_{02}>V_{01}), because its electrons leave with more kinetic energy.
Threshold frequency (f_0)
The lowest frequency of electromagnetic radiation that can produce the photoelectric effect in a given metal.
Threshold wavelength (\lambda_0)
The longest wavelength of radiation that can eject electrons from a given metal, \lambda_0=c/f_0. No photoemission occurs if the wavelength is longer than this.
Work function (\Phi)
The minimum energy needed to remove an electron from the surface of a metal (the book also calls it the threshold energy). It is measured in joules or electronvolts, and \Phi=hf_0=hc/\lambda_0.
Approximate work functions. A small work function means a long threshold wavelength.
MetalWork function \PhiThreshold wavelength \lambda_0=hc/\PhiLight needed
Potassium2.1\,\mathrm{eV}590\,\mathrm{nm}visible (yellow or shorter)
Sodium2.3\,\mathrm{eV}540\,\mathrm{nm}visible (green or shorter)
Zinc4.3\,\mathrm{eV}290\,\mathrm{nm}ultraviolet
Copper4.7\,\mathrm{eV}265\,\mathrm{nm}ultraviolet

Why classical wave theory fails

Aspects of the photoelectric effect that classical physics cannot explain
ObservationClassical wave theory predictsExperiment shows
Threshold frequencyLight of any frequency should eject electrons if it is bright enough, because energy keeps building up.No electrons below f_0, however bright the light.
Kinetic energy and intensityBrighter light (larger amplitude) should give faster electrons.K.E_{\max} does not depend on intensity.
Kinetic energy and frequencyFrequency should not matter.K.E_{\max} increases linearly with frequency.
Time lagWith dim light an electron needs time to collect enough energy (seconds or even minutes).Emission starts at once (within about 10^{-9}\,\mathrm{s}), even in very dim light.

21.2Photon theory of the photoelectric effectOften examined

In 1905 Albert Einstein reinterpreted Planck's theory: light itself exists as discrete packets of energy called photons, each of energy E=hf. Photoelectric emission is the result of an interaction between one photon of the incident radiation and one electron in the metal. The photon gives all its energy to the electron and disappears.

A part of this energy (the work function \Phi) is used by the electron to break away from the metal. The rest appears as the kinetic energy of the electron. Electrons from just below the surface lose a little extra energy on the way out, so only electrons from the surface leave with the maximum kinetic energy.

Derivation Einstein's photoelectric equation

Aim. Relate the photon energy hf to the work function \Phi and the maximum kinetic energy of the photoelectrons.

Assumptions.
  • Light is a stream of photons, each of energy hf.
  • Each photon is absorbed completely by a single electron (one photon, one electron).
  • The minimum energy needed to remove an electron from the surface is the work function \Phi.
  1. One photon of frequency f brings energy hf and gives all of it to one electron.
    E=hf
  2. Energy is conserved: the photon energy pays the work function and the remainder becomes kinetic energy. For an electron at the surface the kinetic energy is the maximum.
    hf=\Phi+K.E_{\max}\qquad(21.3)
  3. At the threshold frequency f_0 the electron just escapes, with zero kinetic energy. Put K.E_{\max}=0 and f=f_0.
    hf_0=\Phi
  4. Substitute \Phi=hf_0 back into (21.3).
    K.E_{\max}=hf-hf_0=h(f-f_0)
  5. Using K.E_{\max}=\tfrac12mv_{\max}^{2}=eV_0 gives other useful forms.
    hf=\Phi+\tfrac12mv_{\max}^{2}=\Phi+eV_0
  6. Divide by e to see how the stopping potential depends on frequency.
    V_0=\dfrac{h}{e}\,f-\dfrac{\Phi}{e}
Resulthf=\Phi+K.E_{\max},\qquad K.E_{\max}=h(f-f_0)=eV_0

The equation is a straight line: K.E_{\max} (or V_0) against f has slope h (or h/e), cuts the frequency axis at f_0 and, extended back, cuts the energy axis at -\Phi (or -\Phi/e).

The equation gives three cases:

  • If hf<\Phi: there is no photoelectric effect.
  • If hf=\Phi: the photoelectric effect just occurs, but the ejected electron has zero kinetic energy.
  • If hf>\Phi: electrons are emitted with kinetic energy up to K.E_{\max}=hf-\Phi.

How the photon theory explains the observations

ObservationExplanation with photons
Threshold frequencyA photon with hf<\Phi cannot free an electron. An electron cannot add up the energy of two photons, so a brighter beam (more photons) still gives no emission.
K.E_{\max} independent of intensityIntensity is the number of photons per second. Each photon still has the same energy hf, so K.E_{\max}=hf-\Phi does not change.
Current proportional to intensityMore photons per second free more electrons per second, so the current increases in proportion.
K.E_{\max} increases with frequencyK.E_{\max}=hf-\Phi is a straight line in f with slope h.
No time lagThe energy arrives in one packet, so an electron can leave as soon as it absorbs a photon.

Stopping potential against frequency

Since V_0=\dfrac{h}{e}f-\dfrac{\Phi}{e}, a graph of stopping potential against frequency is a straight line. It lets us find both the work function and Planck's constant.

Stopping potential against frequency
Two metals give parallel lines: the slope h/e is the same for every metal. The frequency intercept is the threshold frequency f_0 and the extended line cuts the V_0 axis at -\Phi/e.
  • Slope =h/e, so h=e\times\text{slope}. The slope is the same for every metal.
  • Frequency intercept =f_0, so \Phi=hf_0.
  • Intercept on the stopping potential axis (line extended back to f=0) =-\Phi/e, so \Phi=e\times|\text{intercept}|.

Uses of the photoelectric effect

  • Photocells (photoelectric cells) convert light energy into electric current. They switch on street lights at dusk, open automatic doors, operate burglar alarms and count objects on a conveyor belt.
  • Photoconductive devices, such as light dependent resistors, whose resistance falls when light releases charge carriers.
  • Solar cells, which convert sunlight directly into electrical energy.
  • Light meters in cameras and the image sensors of digital cameras.
Example 21.2
A metal whose work function is 4.2\,\mathrm{eV} is irradiated by radiation whose wavelength is 2000\,\text{\AA}. Find the maximum kinetic energy of the emitted electron.
Given
  • \Phi = 4.2\,\mathrm{eV}=4.2\times1.6\times10^{-19}=6.72\times10^{-19}\,\mathrm{J} work function
  • \lambda = 2000\,\text{\AA}=2000\times10^{-10}\,\mathrm{m}=2\times10^{-7}\,\mathrm{m}
To find
  • K.E_{\max}
  1. Einstein's photoelectric equation, with f=c/\lambda.
    K.E_{\max}=hf-\Phi=\dfrac{hc}{\lambda}-\Phi
  2. Energy of one photon.
    \dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{2\times10^{-7}}=9.945\times10^{-19}\,\mathrm{J}
  3. Subtract the work function.
    K.E_{\max}=9.945\times10^{-19}-6.72\times10^{-19}=3.23\times10^{-19}\,\mathrm{J}
  4. Change into electronvolts.
    K.E_{\max}=\dfrac{3.23\times10^{-19}}{1.6\times10^{-19}}=2.02\,\mathrm{eV}
K.E_{\max}=3.23\times10^{-19}\,\mathrm{J}=2.02\,\mathrm{eV}
Assignment 21.2
Radiation of wavelength 3000\,\text{\AA} falls on a photoelectric surface for which the work function is 1.6\,\mathrm{eV}. What is the stopping potential for the emitted electrons?
  1. Photon energy in joules.
    E=\dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3000\times10^{-10}}=6.63\times10^{-19}\,\mathrm{J}
  2. Change into electronvolts so it can be compared with \Phi directly.
    E=\dfrac{6.63\times10^{-19}}{1.6\times10^{-19}}=4.14\,\mathrm{eV}
  3. Einstein's equation gives the maximum kinetic energy.
    K.E_{\max}=E-\Phi=4.14-1.6=2.54\,\mathrm{eV}
  4. K.E_{\max}=eV_0, so the stopping potential in volts equals K.E_{\max} in electronvolts.
    V_0=\dfrac{K.E_{\max}}{e}=\dfrac{2.54\,\mathrm{eV}}{e}=2.54\,\mathrm{V}
V_0\approx2.5\,\mathrm{V}
Threshold frequency and wavelength of sodium
The work function of sodium is 2.3\,\mathrm{eV}. Find its threshold frequency and threshold wavelength. Will red light of wavelength 700\,\mathrm{nm} eject electrons from sodium?
  1. At threshold the whole photon energy is used as work function.
    f_0=\dfrac{\Phi}{h}=\dfrac{2.3\times1.6\times10^{-19}}{6.63\times10^{-34}}=5.6\times10^{14}\,\mathrm{Hz}
  2. Threshold wavelength from \lambda_0=c/f_0.
    \lambda_0=\dfrac{3\times10^{8}}{5.55\times10^{14}}=5.4\times10^{-7}\,\mathrm{m}=540\,\mathrm{nm}
  3. Compare: 700\,\mathrm{nm} is longer than \lambda_0, so each red photon has less energy than \Phi.
    700\,\mathrm{nm}>540\,\mathrm{nm}\ \Rightarrow\ \text{no emission}
f_0=5.6\times10^{14}\,\mathrm{Hz},\quad \lambda_0=540\,\mathrm{nm};\ \text{red light cannot eject electrons}

21.3Compton's effectOften examined

In 1923 the American physicist Arthur H. Compton directed a beam of X-rays of wavelength \lambda at a block of graphite. He found that the scattered X-rays had a slightly longer wavelength \lambda' than the incident X-rays. The increase in wavelength (the loss of energy) depended on the angle through which the X-rays were scattered.

Compton effect
The scattering of X-ray (or gamma-ray) photons by electrons, in which the photons transfer energy and momentum to the electrons, so the wavelength of the scattered radiation increases.
Compton shift (\Delta\lambda)
The change in wavelength between the scattered and the incident X-ray, \Delta\lambda=\lambda'-\lambda.

Classical wave theory cannot explain this. A wave makes an electron oscillate at the wave's own frequency, so the electron should re-radiate at the same frequency and wavelength. Compton explained the result by treating the X-ray as a stream of photons.

Compton scattering. An incident photon (\lambda, f) collides with an electron at rest. The photon is scattered through angle \theta with a longer wavelength \lambda', and the electron recoils at angle \phi with kinetic energy K.E and momentum p_e.

Explanation with photons

Compton assumed that if a photon behaves like a particle, its collision with an electron is like a collision between two billiard balls: both energy and momentum must be conserved. The incident photon (energy hf, momentum h/\lambda) hits a free electron at rest and gives part of its energy and momentum to it. The electron recoils. The scattered photon is left with less energy, so its frequency is lower and its wavelength is longer.

Conservation of energy:

hf=K.E+hf'\qquad(21.4)

hf and hf': energies of the incident and scattered photons, K.E: kinetic energy given to the recoiling electron.

Conservation of momentum (a vector equation): the momentum \vec p of the incident photon equals the vector sum of the momenta of the scattered photon \vec p\,' and the recoil electron \vec p_e. The momentum of each photon is p=E/c=h/\lambda (Eq. 21.6).

\vec p=\vec p\,'+\vec p_e\qquad(21.5)

Take the incident direction as the x-axis. Resolving momenta into components:

\text{along }x:\quad \dfrac{h}{\lambda}=p_e\cos\phi+\dfrac{h}{\lambda'}\cos\theta\qquad(21.7)
\text{along }y:\quad 0=-p_e\sin\phi+\dfrac{h}{\lambda'}\sin\theta\qquad(21.8)

From these equations Compton derived the shift in wavelength:

\Delta\lambda=\lambda'-\lambda=\dfrac{h}{m_0c}\left(1-\cos\theta\right)\qquad(21.9)

m_0: rest mass of the electron, c: speed of light, \theta: scattering angle of the photon.

Derivation Compton shift formula (full working)

Aim. Derive \lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta) from conservation of energy and momentum.

Assumptions.
  • The electron is free and at rest before the collision (outer electrons in graphite are very loosely bound compared with the X-ray photon's energy).
  • Energy and momentum are both conserved in the collision.
  1. Write photon momenta as p=h/\lambda and p'=h/\lambda', so photon energies are pc and p'c. Energy conservation, with E_e the total energy of the electron after the collision, is:
    pc+m_0c^{2}=p'c+E_e\ \Rightarrow\ E_e=(p-p')c+m_0c^{2}
  2. Momentum conservation gives \vec p_e=\vec p-\vec p\,'. Squaring (or combining 21.7 and 21.8), the angle between \vec p and \vec p\,' is \theta:
    p_e^{2}=p^{2}+p'^{2}-2pp'\cos\theta
  3. The recoil electron can be fast, so use the relativistic energy-momentum relation for it.
    E_e^{2}=p_e^{2}c^{2}+m_0^{2}c^{4}
  4. Square the energy equation of step 1 and set it equal to the relation of step 3, using step 2 for p_e^{2}.
    (p-p')^{2}c^{2}+2(p-p')m_0c^{3}+m_0^{2}c^{4}=(p^{2}+p'^{2}-2pp'\cos\theta)c^{2}+m_0^{2}c^{4}
  5. Expand (p-p')^{2}=p^{2}-2pp'+p'^{2} and cancel the terms that appear on both sides, then divide by 2c^{2}.
    (p-p')\,m_0c=pp'(1-\cos\theta)
  6. Divide both sides by pp'm_0c. Note \dfrac{p-p'}{pp'}=\dfrac{1}{p'}-\dfrac{1}{p}.
    \dfrac{1}{p'}-\dfrac{1}{p}=\dfrac{1-\cos\theta}{m_0c}
  7. Put back 1/p'=\lambda'/h and 1/p=\lambda/h, then multiply by h.
    \lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta)
Result\Delta\lambda=\dfrac{h}{m_0c}\left(1-\cos\theta\right)

The shift depends only on the scattering angle \theta. It does not depend on the incident wavelength or on the scattering material. The book quotes this result without working; the steps are given here for completeness.

Compton wavelength (\lambda_c)
The constant \lambda_c=\dfrac{h}{m_0c}=2.43\times10^{-12}\,\mathrm{m}=0.00243\,\mathrm{nm}. It has the dimension of a wavelength and equals the Compton shift for scattering at 90^\circ.
\lambda_c=\dfrac{h}{m_0c}=\dfrac{6.63\times10^{-34}}{(9.11\times10^{-31})(3\times10^{8})}=2.43\times10^{-12}\,\mathrm{m}
Compton shift against scattering angle
The shift is zero for \theta=0, equals the Compton wavelength \lambda_c at 90^\circ and is largest, 2\lambda_c=4.86\,\mathrm{pm}, for a photon bounced straight back (180^\circ).

Features of the Compton shift

  • The shift depends only on the scattering angle \theta: \Delta\lambda=0 at \theta=0^\circ, \Delta\lambda=\lambda_c at 90^\circ and \Delta\lambda=2\lambda_c=4.86\times10^{-12}\,\mathrm{m} (the maximum) at 180^\circ.
  • It does not depend on the wavelength of the incident photon or on the nature of the scattering material.
  • Since (1-\cos\theta) is never negative, \lambda'\ge\lambda: the scattered photon never gains energy from an electron at rest. (The book says (1-\cos\theta) is always positive; it is zero only at \theta=0, where there is no scattering.)

Physical significance of the Compton effect

  • It is strong evidence for the particle nature of electromagnetic radiation: an X-ray photon collides with an electron like a particle.
  • It confirms that a photon carries momentum p=h/\lambda as well as energy E=hf.
  • It shows that the laws of conservation of energy and momentum hold in each individual photon-electron collision.
  • Together with the photoelectric effect it describes how photons interact with electrons: in the photoelectric effect the photon is completely absorbed; in the Compton effect it is scattered with less energy.
Example 21.3
An X-ray of 0.20\,\mathrm{nm} is incident on a target. What is the Compton shift that can be expected at 45^\circ?
Given
  • \lambda = 0.20\,\mathrm{nm}
  • \theta = 45^\circ
  • m_0 = 9.11\times10^{-31}\,\mathrm{kg}
To find
  • \Delta\lambda=\lambda'-\lambda
  1. Compton shift formula.
    \lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta)
  2. Substitute. The incident wavelength is not needed, because the shift does not depend on it.
    \lambda'-\lambda=\dfrac{6.63\times10^{-34}}{(9.11\times10^{-31})(3\times10^{8})}(1-\cos45^\circ)=2.43\times10^{-12}\times0.293
  3. Evaluate.
    \lambda'-\lambda=7.11\times10^{-13}\,\mathrm{m}
\Delta\lambda=7.11\times10^{-13}\,\mathrm{m}\ (0.000711\,\mathrm{nm})
Assignment 21.3
An X-ray of 71\,\mathrm{pm} is incident on a target. What is the shift that can be expected at 60^\circ? Find the wavelength of the scattered X-ray.
  1. Compton shift at \theta=60^\circ, where \cos60^\circ=0.5.
    \Delta\lambda=\dfrac{h}{m_0c}(1-\cos60^\circ)=2.43\times10^{-12}\times0.5=1.21\times10^{-12}\,\mathrm{m}=1.21\,\mathrm{pm}
  2. The scattered wavelength is the incident wavelength plus the shift.
    \lambda'=\lambda+\Delta\lambda=71+1.21=72.2\,\mathrm{pm}
\Delta\lambda=1.21\,\mathrm{pm},\qquad \lambda'=72.2\,\mathrm{pm}

21.4Pair productionOften examined

Every particle has an antiparticle with the same mass but opposite charge. The antiparticle of the electron is the positron (e^+): mass m_0=9.11\times10^{-31}\,\mathrm{kg}, charge +e.

Pair production
The creation of a particle and its antiparticle from energy. When a high-energy photon passes near a heavy nucleus, it disappears and an electron-positron pair is created. The photon is totally absorbed.
Pair production. A gamma-ray photon of energy hf\ge1.02\,\mathrm{MeV} passing close to a heavy nucleus changes into an electron and a positron.

Pair production is direct evidence that energy can be converted into mass, as Einstein's relation E=mc^{2} says. Three quantities are conserved in the process: charge, momentum and energy.

Conservation laws in pair production

Conservation of charge. The photon has zero charge. The electron (-e) and positron (+e) together also have zero charge. So a photon can never produce a single electron, a single positron, two electrons or two positrons: charge would not be conserved.

Conservation of momentum. The photon's momentum hf/c must equal the total momentum of the products. Here m v_{e^+} and m v_{e^-} are the momenta of the positron and electron (vector sum), and the nucleus takes a small recoil:

\dfrac{hf}{c}=mv_{e^+}+mv_{e^-}\qquad(21.11)

Conservation of energy. The photon energy becomes the total energy of the electron and positron:

E_{\text{photon}}=E_{\text{electron}}+E_{\text{positron}}\qquad(21.12\,\text{a})

The total energy of a particle is its kinetic energy plus its rest mass energy m_0c^{2}, so

hf=(K.E_{\text{electron}}+m_0c^{2})+(K.E_{\text{positron}}+m_0c^{2})\qquad(21.12\,\text{b})

Derivation Minimum photon energy for pair production

Aim. Find the threshold (minimum) energy a photon needs to create an electron-positron pair.

  1. Rearrange Eq. (21.12 b): the photon energy provides two rest mass energies plus the kinetic energies.
    hf=2m_0c^{2}+K.E_{\text{electron}}+K.E_{\text{positron}}
  2. The photon energy is least when the pair is created with (almost) zero kinetic energy.
    hf_{\min}=2m_0c^{2}
  3. Rest mass energy of one electron (or positron).
    m_0c^{2}=(9.11\times10^{-31})(3\times10^{8})^{2}=8.2\times10^{-14}\,\mathrm{J}=0.511\,\mathrm{MeV}
  4. Double it for the pair.
    hf_{\min}=2\times0.511\,\mathrm{MeV}=1.02\,\mathrm{MeV}=1.64\times10^{-13}\,\mathrm{J}
  5. The largest wavelength that can do this follows from E=hc/\lambda.
    \lambda_{\max}=\dfrac{hc}{hf_{\min}}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.64\times10^{-13}}=1.21\times10^{-12}\,\mathrm{m}
Resulthf_{\min}=2m_0c^{2}=1.02\,\mathrm{MeV}

Only gamma-ray photons (wavelength about 10^{-12}\,\mathrm{m} or less) can produce a pair. If the photon energy is more than 1.02\,\mathrm{MeV}, the extra energy appears as kinetic energy of the electron and positron.

Conditions for pair production

  • The photon energy must be at least 2m_0c^{2}=1.02\,\mathrm{MeV} (a gamma-ray photon).
  • A heavy nucleus (or another particle) must be nearby to take up momentum.
  • Charge, energy and momentum must all be conserved, so a particle and its antiparticle (electron and positron) are always created together.
Energy shared by the pair
A gamma-ray photon of energy 3.0\,\mathrm{MeV} produces an electron-positron pair near a nucleus. How much kinetic energy do the two particles share? Find the kinetic energy of each if it is shared equally.
  1. Energy balance: the photon supplies both rest mass energies and the kinetic energies.
    K.E_{\text{total}}=hf-2m_0c^{2}=3.0-1.02=1.98\,\mathrm{MeV}
  2. Equal sharing.
    K.E_{\text{each}}=\dfrac{1.98}{2}=0.99\,\mathrm{MeV}
K.E_{\text{total}}=1.98\,\mathrm{MeV},\qquad K.E_{\text{each}}=0.99\,\mathrm{MeV}

21.5Pair annihilationOften examined

Pair annihilation is the reverse of pair production. In pair production energy (a photon) is converted into mass (an electron-positron pair). In annihilation mass (an electron-positron pair) is converted into energy (photons).

Pair annihilation
The process in which an electron and a positron combine, both disappear, and their mass is converted into two gamma-ray photons.
Pair annihilation. A positron and an electron with almost zero total momentum meet and disappear; two gamma-ray photons of 0.511\,\mathrm{MeV} each fly off in opposite directions.

Why two photons and not one

Just before annihilation the electron and positron are nearly at rest, so their total momentum is (almost) zero. A single photon always moves at speed c and carries momentum hf/c, which is never zero. One photon therefore cannot conserve momentum. Two photons of equal energy moving in opposite directions have equal and opposite momenta, so their total momentum is zero, just as before. Charge is also conserved: (+e)+(-e)=0, and photons carry no charge.

Derivation Energy of each annihilation photon

Aim. Find the energy and wavelength of each photon produced when an electron and a positron at rest annihilate.

  1. The total energy before annihilation is the rest mass energy of two particles (kinetic energy is negligible).
    E_{\text{before}}=2m_0c^{2}
  2. Two photons share this energy equally (they have equal momenta, so equal energies).
    2hf=2m_0c^{2}\ \Rightarrow\ hf=m_0c^{2}
  3. Substitute the rest mass energy of the electron.
    hf=0.511\,\mathrm{MeV}=8.2\times10^{-14}\,\mathrm{J}
  4. Wavelength of each photon.
    \lambda=\dfrac{hc}{m_0c^{2}}=\dfrac{h}{m_0c}=2.43\times10^{-12}\,\mathrm{m}
  5. If the particles are moving, their kinetic energies are also given to the photons.
    2m_0c^{2}+K.E_{\text{electron}}+K.E_{\text{positron}}=\text{energy of two gamma photons}\qquad(21.13)
Resulthf=m_0c^{2}=0.511\,\mathrm{MeV}\ \text{for each photon}

The photons are gamma rays. Their wavelength happens to equal the Compton wavelength h/m_0c. Annihilation shows mass being turned completely into energy, E=mc^{2}.

Pair production and pair annihilation compared
FeaturePair productionPair annihilation
ConversionEnergy into massMass into energy
BeforeOne gamma photon (hf\ge1.02\,\mathrm{MeV})An electron and a positron
AfterAn electron and a positronTwo gamma photons (0.511\,\mathrm{MeV} each when the pair is at rest)
MomentumA heavy nucleus takes the extra momentumPhotons move in opposite directions
Energy equationhf=2m_0c^{2}+K.E_{e^-}+K.E_{e^+}2m_0c^{2}+K.E_{e^-}+K.E_{e^+}=\text{energy of two photons}

Another example of annihilation is proton-antiproton annihilation. Besides confirming the photon model of electromagnetic radiation, pair production and pair annihilation clearly illustrate Einstein's idea of mass and rest energy.

21.6Wave-particle duality and de Broglie wavesOften examined

Interference, diffraction and polarization are the basic properties of all electromagnetic waves. Light shows all three, so light can be considered a wave. But the photoelectric effect can be explained only if light is made of small energy packets (particles) called photons. So light displays a dual nature: it has particle characteristics as well as wave characteristics.

Wave-particle duality
The property of light (and of matter) of behaving as a wave in some experiments and as a stream of particles in others.
Light as a wave and as a particle
Evidence for the wave natureEvidence for the particle nature
Interference (Young's double slit experiment)Photoelectric effect
Diffraction (single slit, diffraction grating)Compton effect (photon momentum)
Polarization (light is a transverse wave)Pair production and pair annihilation
Speed c, wavelength \lambda and frequency f with c=f\lambdaLine spectra (atoms emit photons of definite energy)

The two pictures are joined by two equations. The particle properties (energy E and momentum p) are fixed by the wave properties (frequency f and wavelength \lambda):

E=hf,\qquad p=\dfrac{h}{\lambda}

de Broglie's hypothesis

In 1924 the French physicist Louis de Broglie argued that nature should be symmetric. If light, which shows pure wave properties, can also have particle-like properties, then particles such as electrons can also behave as waves. Particles of matter obey a wave equation just as photons do. Compton's work showed that a photon of wavelength \lambda has momentum p=h/\lambda; de Broglie proposed the same relation for every moving particle.

Derivation de Broglie wavelength

Aim. Find the wavelength associated with a particle of mass m moving with speed v.

  1. For a photon, the momentum and wavelength are related by Compton's result.
    p=\dfrac{h}{\lambda}
  2. de Broglie assumed the same relation holds for a material particle, so its wavelength is Planck's constant divided by its momentum.
    \lambda=\dfrac{h}{p}\qquad(21.14)
  3. For a particle of mass m and speed v (much less than c) the momentum is p=mv.
    \lambda=\dfrac{h}{mv}
Result\lambda=\dfrac{h}{p}=\dfrac{h}{mv}

Every moving particle has a wavelength. The greater its momentum, the shorter its wavelength. Heavy everyday objects have wavelengths far too small to observe, but electrons have wavelengths comparable to the spacing of atoms, so their wave nature can be seen.

de Broglie wavelength
The wavelength of the matter wave associated with a moving particle, \lambda=h/p=h/mv.
Matter waves
Waves associated with moving material particles such as electrons, protons and neutrons.

Macroscopic objects and electrons compared

Wavelength of a ball and of an electron
Find the de Broglie wavelength of (a) an object of mass 0.5\,\mathrm{kg} moving at 10\,\mathrm{m\,s^{-1}} and (b) an electron (m=9.1\times10^{-31}\,\mathrm{kg}) moving at 10^{6}\,\mathrm{m\,s^{-1}}.
  1. (a) Use \lambda=h/mv.
    \lambda=\dfrac{6.63\times10^{-34}}{(0.5)(10)}=1.3\times10^{-34}\,\mathrm{m}
  2. This is far smaller than even a nucleus (10^{-15}\,\mathrm{m}), so the wave nature of the object can never be observed.
  3. (b) Same formula for the electron.
    \lambda=\dfrac{6.63\times10^{-34}}{(9.1\times10^{-31})(10^{6})}=7.3\times10^{-10}\,\mathrm{m}
  4. This is comparable to the spacing between atoms in a crystal (about 10^{-10}\,\mathrm{m}), which is just right for diffraction and interference. So the wavelengths of very small particles are observable.
\lambda_{\text{object}}=1.3\times10^{-34}\,\mathrm{m},\qquad \lambda_{\text{electron}}=7.3\times10^{-10}\,\mathrm{m}
Example 21.4
A boy throws a stone of mass 5.0\,\mathrm{mg} with the help of a catapult. If the stone moves with a speed of 8.0\,\mathrm{m\,s^{-1}}, calculate the de Broglie wavelength associated with the stone.
Given
  • m = 5.0\,\mathrm{mg}=5.0\times10^{-6}\,\mathrm{kg}
  • v = 8.0\,\mathrm{m\,s^{-1}}
To find
  • \lambda
  1. de Broglie relation.
    \lambda=\dfrac{h}{mv}
  2. Substitute (1\,\mathrm{mg}=10^{-6}\,\mathrm{kg}).
    \lambda=\dfrac{6.63\times10^{-34}}{(5.0\times10^{-6})(8.0)}=1.66\times10^{-29}\,\mathrm{m}
\lambda=1.66\times10^{-29}\,\mathrm{m}
Assignment 21.4
The speed of a rifle bullet is 1500\,\mathrm{m\,s^{-1}}. If its mass is 20\,\mathrm{g}, find the de Broglie wavelength associated with it. Is this wavelength observable? Give reason.
  1. Convert the mass: 20\,\mathrm{g}=0.020\,\mathrm{kg}. Use \lambda=h/mv.
    \lambda=\dfrac{6.63\times10^{-34}}{(0.020)(1500)}=2.2\times10^{-35}\,\mathrm{m}
  2. No. To observe a wave we need diffraction or interference by an obstacle or slit about as small as the wavelength. Nothing in nature is this small (a nucleus is about 10^{-15}\,\mathrm{m}), so the bullet behaves purely as a particle.
\lambda=2.2\times10^{-35}\,\mathrm{m}\ \ (\text{not observable})

Wavelength of an electron accelerated through a potential difference

Derivation de Broglie wavelength of an accelerated electron

Aim. Find the wavelength of an electron accelerated from rest through a potential difference V.

Assumptions.
  • The electron starts from rest.
  • Its speed stays well below c, so p=mv can be used (true up to a few tens of kilovolts).
  1. The electron gains kinetic energy equal to the work done by the field.
    \tfrac12mv^{2}=eV
  2. Write kinetic energy in terms of momentum: \tfrac12mv^{2}=\dfrac{p^{2}}{2m}.
    \dfrac{p^{2}}{2m}=eV\ \Rightarrow\ p=\sqrt{2meV}
  3. Put this momentum into the de Broglie relation.
    \lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}
  4. Insert h, m=9.11\times10^{-31}\,\mathrm{kg} and e=1.6\times10^{-19}\,\mathrm{C} to get a handy form (V in volts).
    \lambda\approx\dfrac{1.23\times10^{-9}}{\sqrt{V}}\,\mathrm{m}
Result\lambda=\dfrac{h}{\sqrt{2meV}}

A larger accelerating voltage gives a shorter wavelength. For V=100\,\mathrm{V}, \lambda=1.23\times10^{-10}\,\mathrm{m} (about the size of an atom); for V=10\,\mathrm{kV}, \lambda=1.2\times10^{-11}\,\mathrm{m}=0.12\,\text{\AA}. This is the key to the electron microscope.

21.6Electron diffraction: evidence for matter waves

If electrons are waves they must show diffraction, just like light passing through a grating. A diffraction grating works best when its spacing is comparable to the wavelength. The atoms in a crystal are about 10^{-10}\,\mathrm{m} apart, and electrons accelerated through a few hundred to a few thousand volts have wavelengths of about 10^{-10} to 10^{-11}\,\mathrm{m}, comparable to this spacing. So a thin crystal acts as a natural grating for electrons.

Electron diffraction. A beam of electrons from an electron gun passes through a very thin graphite film. The electrons that reach the fluorescent screen form a pattern of bright concentric rings.

What the experiment shows

  • A pattern of bright and dark concentric rings appears on the screen, similar to the pattern made when X-rays pass through a powdered crystal. Rings of this kind are produced by constructive and destructive interference, which only waves can do.
  • When the accelerating voltage is increased, the electrons move faster, their wavelength \lambda=h/\sqrt{2meV} becomes shorter, and the rings become smaller (less diffraction).
  • The wavelength calculated from the size of the rings agrees with de Broglie's formula \lambda=h/mv.
  • A magnet brought near the tube shifts the whole pattern. So the pattern is made by charged electrons, not by X-rays or light produced in the tube.

21.7Electron microscopeOften examined

The electron microscope is a device based on the wave nature of electrons. Instead of light, it uses a beam of accelerated electrons as its source of illumination. It lets us observe objects in detail down to the nanometre scale, with remarkably high resolution. It is commonly used to study microscopic organisms such as viruses and bacteria, and the structure of metals and crystals.

Why the wavelength decides the resolution

Resolution is the ability of a microscope to show two very close points as separate. Because of diffraction, any microscope can only show details that are about as large as, or larger than, the wavelength of the radiation it uses. Shorter wavelength means finer detail and higher resolution.

  • Visible light has wavelengths from 4000\,\text{\AA} to 7000\,\text{\AA} (400 to 700\,\mathrm{nm}). The best optical microscope cannot resolve details smaller than about 200\,\mathrm{nm}.
  • Electrons accelerated through a p.d. V have wavelength \lambda=h/\sqrt{2meV}. Through 10\,000\,\mathrm{V} (10\,\mathrm{keV}) the wavelength is about 0.12\,\text{\AA} (0.012\,\mathrm{nm}), tens of thousands of times shorter than visible light. At 100\,\mathrm{kV} it is about 0.004\,\mathrm{nm}.
  • In practice the magnetic lenses are not perfect, so the resolution is limited to about 0.1 to 0.2\,\mathrm{nm}. This is still about a thousand times finer than an optical microscope, enough to see individual large molecules and rows of atoms.

Construction and working

A transmission electron microscope. Electrons from a hot filament are accelerated by the anode, focused by magnetic lenses (coils) through a thin specimen and magnified onto a fluorescent screen. The whole column is kept in a vacuum.
  1. Electron gun: a tungsten filament is heated and emits electrons. A high potential on the anode (tens of kilovolts) accelerates them into a narrow, fast beam, like a beam of light.
  2. Condenser lens: a magnetic coil that focuses the electron beam onto the specimen.
  3. Specimen (sample) stage: the sample must be very thin, so that electrons can pass through it, and carefully prepared. It is placed in the vacuum chamber.
  4. Objective lens: forms the first magnified image of the specimen.
  5. Intermediate and projector lenses: magnify the image further and project it downwards.
  6. Fluorescent screen or detector: at the very bottom. Electrons striking it make it glow, forming a visible image. A photograph of it is called an electron micrograph.
  7. Vacuum system: the whole column is evacuated so that the electrons are not scattered or stopped by air molecules.

The electron beam passes through a series of coil-shaped electromagnets that replace the glass lenses of an optical microscope. The magnetic field of a coil bends the paths of the electrons and brings them to a focus. Increasing the current in a coil makes its field stronger, so the magnetic lens becomes stronger (shorter focal length). The electron gun and condenser lens are in the upper part; the sample stage and objective lens are in the middle; the intermediate and projector lenses are in the lower part; the screen is at the bottom.

Optical and electron microscopes compared
FeatureOptical microscopeElectron microscope
IlluminationVisible lightBeam of fast electrons
Wavelength400 to 700\,\mathrm{nm}about 0.01\,\mathrm{nm} or less (depends on V)
LensesGlass lensesMagnetic lenses (current-carrying coils)
Best resolutionabout 200\,\mathrm{nm}about 0.1 to 0.2\,\mathrm{nm}
Useful magnificationup to about 1500 timesup to about a million times
MediumWorks in airNeeds a vacuum
SpecimenCan be livingThin, dry, non-living, in vacuum
ImageSeen directly by the eye, in colourFormed on a fluorescent screen or detector (black and white)

21.8Heisenberg's uncertainty principleOften examined

On the microscopic level we cannot measure any property of a particle without interacting with it in some way, for example by bouncing a photon off it. This interaction introduces an unavoidable uncertainty into the result. From Planck's point of view, E=hc/\lambda: a photon of short wavelength has a large energy and a large momentum, so it disturbs whatever it hits.

The German physicist and Nobel laureate Werner Heisenberg formulated the uncertainty principle in 1927. It is a consequence of the dual (wave and particle) nature of matter.

Heisenberg's uncertainty principle
It is impossible to find both the position and the momentum of a particle accurately at the same time. The product of the uncertainties in position and momentum, measured at the same instant, is approximately equal to \hbar (or greater).
\Delta x\,\Delta p\ \ge\ \hbar,\qquad \hbar=\dfrac{h}{2\pi}=1.054\times10^{-34}\,\mathrm{J\,s}

\Delta x: uncertainty in position, \Delta p: uncertainty in momentum, \hbar (read as h-cut or h-bar): the reduced Planck constant.

If the position were known exactly (\Delta x=0), the momentum would be completely unknown (\Delta p=\infty), and the other way round. The more precisely we fix one, the less precisely we can know the other.

The wave picture

In quantum mechanics a particle is described by a wave. Position means where the wave is concentrated; the uncertainty in position is how far the wave is spread out. Momentum is fixed by the wavelength (p=h/\lambda); the uncertainty in momentum is how unclear the wavelength is. A wave cannot be both short (definite position) and have one exact wavelength (definite momentum).

(a) A wave with one definite wavelength
The wavelength, and so the momentum p=h/\lambda, is exactly known (\Delta p\approx0). But the wave is spread everywhere, so the position is completely unknown.
(b) A short wave packet
The wave is squeezed into a small region \Delta x, so the position is well known. To build such a short pulse many wavelengths must be mixed, so the wavelength and the momentum become uncertain.

Why measuring position disturbs momentum

If we want accuracy in position we must use photons of shorter wavelength, because the best resolution we can get is about the wavelength of the radiation used. But short-wavelength radiation means high-frequency, high-energy photons with large momentum h/\lambda. When these collide with the electron they give it a large and unpredictable kick. If we use longer wavelength (less energetic) photons, the kick is gentler but the position becomes blurred.

Derivation Rough estimate of the uncertainty product

Aim. Show that locating an electron with light of wavelength \lambda gives \Delta x\,\Delta p\approx h.

  1. A photon of wavelength \lambda can locate the electron only to within about one wavelength.
    \Delta x\approx\lambda
  2. The photon has momentum h/\lambda. After it bounces off, the electron's momentum is uncertain by up to about this much, because the direction of the scattered photon is not known exactly.
    \Delta p\approx\dfrac{h}{\lambda}
  3. Multiply. The wavelength cancels, so the product cannot be made smaller by choosing a different \lambda.
    \Delta x\,\Delta p\approx\lambda\times\dfrac{h}{\lambda}=h
Result\Delta x\,\Delta p\approx h

This rough argument gives the order of magnitude. A careful treatment gives \Delta x\,\Delta p\ge\hbar (the book's form); the exact modern form is \Delta x\,\Delta p\ge\hbar/2. All versions say the same thing: the product has a lower limit of the order of Planck's constant.

Uncertainty in energy and time

Another kind of uncertainty concerns the energy of a quantum state and the time for which it lasts. If \Delta E is the uncertainty in energy and \Delta t is the uncertainty in the time to which that energy refers, then:

\Delta E\,\Delta t\ \ge\ \hbar

Derivation Energy-time uncertainty

Aim. Show how a limited time of observation makes the energy uncertain.

  1. To measure the frequency of a wave we count the number of waves in a time interval \Delta t. The count is uncertain by about one wave, so the uncertainty in frequency is about one wave divided by the time.
    \Delta f\ \ge\ \dfrac{1}{\Delta t}
  2. The energy of a photon is E=hf, so an uncertainty in frequency means an uncertainty in energy.
    \Delta E=h\,\Delta f
  3. Combine the two lines.
    \Delta E\ \ge\ \dfrac{h}{\Delta t}\ \Rightarrow\ \Delta E\,\Delta t\ \ge\ h\ \text{(order of magnitude)}
Result\Delta E\,\Delta t\ \ge\ \hbar\quad(\text{book's form})

A system that stays in a state for a very long time (a metastable state) can have a well-defined energy. A state that lasts only a short time has a correspondingly large uncertainty in energy. As with \Delta x\,\Delta p, the rough argument gives h and the book writes \hbar; they differ only by the factor 2\pi.

Why every measurement has an uncertainty

  • These uncertainties are built into the physical world. They have nothing to do with poor instruments or the skill of the observer, and better apparatus cannot remove them.
  • Every measurement needs an interaction (for example a photon hitting the object), and every interaction disturbs the object being measured.
  • Because Planck's constant is so small, the uncertainties are far too small to notice for everyday objects. For electrons inside atoms they are very important.
Example 21.5
Calculate the uncertainty in the momentum of an electron if the uncertainty in its position is 1\,\text{\AA} (10^{-10}\,\mathrm{m}).
Given
  • \Delta x = 10^{-10}\,\mathrm{m}
To find
  • \Delta p
  1. Uncertainty principle.
    \Delta x\,\Delta p\ \ge\ \dfrac{h}{2\pi}\ \Rightarrow\ \Delta p\ \ge\ \dfrac{h}{2\pi\,\Delta x}
  2. Substitute.
    \Delta p=\dfrac{6.63\times10^{-34}}{2(3.14)(10^{-10})}=1.06\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}
\Delta p\approx1.06\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}
Assignment 21.5
Consider a ball of mass 0.1\,\mathrm{kg} moving at a speed of 30\,\mathrm{m\,s^{-1}}. The uncertainty in its momentum is 0.000001 of its actual momentum. Determine the uncertainty in its position.
  1. Momentum of the ball.
    p=mv=0.1\times30=3\,\mathrm{kg\,m\,s^{-1}}
  2. Uncertainty in momentum is 10^{-6} of this.
    \Delta p=10^{-6}\times3=3\times10^{-6}\,\mathrm{kg\,m\,s^{-1}}
  3. Uncertainty principle, \Delta x\,\Delta p\ge\hbar.
    \Delta x\ \ge\ \dfrac{\hbar}{\Delta p}=\dfrac{1.054\times10^{-34}}{3\times10^{-6}}=3.5\times10^{-29}\,\mathrm{m}
  4. This is far smaller than a nucleus, so for a ball the uncertainty is completely negligible. The uncertainty principle matters only for tiny particles.
\Delta x\approx3.5\times10^{-29}\,\mathrm{m}

21.9Discrete energy levels in an atom and its spectraOften examined

The electrons in an atom can occupy only certain allowed orbits (also called shells). Each orbit has a definite energy, called an energy level: E_1, E_2, E_3 and so on. The energy of the atom is therefore quantized, as Planck's theory requires. An electron can move from one allowed level to another only by gaining or losing a definite amount of energy.

Energy level
One of the definite, allowed energies that an electron in an atom can have.
Ground state
The lowest energy level of an atom, E_1. Higher levels are called excited states.
Excitation
The transition of an electron from a lower to a higher energy level. It needs energy, for example the absorption of a photon.
De-excitation
The transition of an electron from a higher to a lower energy level. Energy is released as a photon.
Excitation and de-excitation in an isolated atom such as hydrogen. An electron absorbs a photon of exactly hf=E_3-E_1 to jump up, and emits a photon of the same energy when it falls back.

An electron moves to a higher or lower level by absorbing or emitting electromagnetic radiation of frequency f. The frequency depends only on the energy difference between the two levels:

hf=E_n-E_p\qquad\text{or}\qquad \lambda=\dfrac{hc}{E_n-E_p}

E_n: higher energy level, E_p: lower energy level, f: frequency and \lambda: wavelength of the photon absorbed or emitted.

Obtaining a spectrum

A spectrum is produced by using a prism or a diffraction grating to separate the different wavelengths in a beam of light. Photons emitted in such transitions are passed through a narrow slit and a prism (or grating); each wavelength is bent through a different angle and forms a separate line on the detector.

Set-up for obtaining a spectrum. Light from the gas passes through a slit and a lens; the prism (or diffraction grating) separates the wavelengths into separate lines on the detector.

A line spectrum contains a discrete set of wavelengths. It is a characteristic feature of an element, unique to it.

Emission line spectrum

An emission spectrum of an element consists of a series of separate bright lines of definite frequencies (or wavelengths) on a dark background. It is produced by the excited atoms of the element in a vapour or gas at low pressure, for example in a discharge tube. The photons are emitted as electrons fall from higher to lower excited states or to the ground state. Each line corresponds to one particular transition.

Since each element has a unique set of orbital electrons and energy levels, its emission line spectrum is also unique. The spectrum can therefore be used to identify the element, like a fingerprint. Different coloured lines show which elements a substance is made of, because each element radiates its own particular energies.

Absorption line spectrum

An absorption spectrum of an element consists of a series of separate dark lines of definite frequencies (or wavelengths) on a continuous coloured background. White light from a source such as a tungsten filament lamp (containing all wavelengths) is passed through the cool vapour of the element and then through the slit and grating. The atoms absorb only those photons whose energy exactly matches a jump between their levels. These wavelengths are removed from the beam, leaving dark lines. The excited atoms soon return to the ground state and re-emit the same photons, but in all directions, so very few of them continue in the original direction.

(a) Emission and (b) absorption line spectra of hydrogen in the visible region. The dark absorption lines lie at exactly the same wavelengths (410, 434, 486 and 656\,\mathrm{nm}) as the bright emission lines.
FeatureEmission line spectrumAbsorption line spectrum
AppearanceBright coloured lines on a dark backgroundDark lines on a continuous coloured background
Produced byHot, excited gas or vapour at low pressureWhite light passed through cool gas or vapour
Electron transitionsDownwards (de-excitation), photons emittedUpwards (excitation), photons absorbed
Line positionsCharacteristic of the elementSame wavelengths as that element's emission lines
Using hf=\Delta E for hydrogen
Three energy levels of the hydrogen atom are E_1=-13.6\,\mathrm{eV}, E_2=-3.40\,\mathrm{eV} and E_3=-1.51\,\mathrm{eV}. (a) Find the frequency and wavelength of the photon emitted when an electron falls from E_3 to E_2. (b) What photon energy is needed to raise an electron from the ground state to E_2?
  1. (a) Energy difference between the two levels.
    \Delta E=E_3-E_2=-1.51-(-3.40)=1.89\,\mathrm{eV}=1.89\times1.6\times10^{-19}=3.02\times10^{-19}\,\mathrm{J}
  2. Frequency from hf=\Delta E.
    f=\dfrac{\Delta E}{h}=\dfrac{3.02\times10^{-19}}{6.63\times10^{-34}}=4.56\times10^{14}\,\mathrm{Hz}
  3. Wavelength from \lambda=c/f. This is the red line of hydrogen (measured value 656\,\mathrm{nm}; the small difference comes from rounding the constants).
    \lambda=\dfrac{3\times10^{8}}{4.56\times10^{14}}=6.58\times10^{-7}\,\mathrm{m}\approx658\,\mathrm{nm}
  4. (b) Excitation needs exactly the energy difference; the photon must be absorbed whole.
    hf=E_2-E_1=-3.40-(-13.6)=10.2\,\mathrm{eV}\ \ (\text{ultraviolet, }\lambda\approx122\,\mathrm{nm})
\text{(a)}\ f=4.56\times10^{14}\,\mathrm{Hz},\ \lambda\approx658\,\mathrm{nm};\qquad \text{(b)}\ 10.2\,\mathrm{eV}

Formula sheet

Wave equation for EM waves
c=f\lambda
c: speed of light, f: frequency, \lambda: wavelength
Units: c in \mathrm{m\,s^{-1}}, f in Hz, \lambda in m
Planck's quantum theory
E=nhf
n=1,2,3,\ldots: quantum number, h: Planck's constant, f: frequency
Units: E in J; h=6.63\times10^{-34}\,\mathrm{J\,s}
Energy of a photon
E=hf=\dfrac{hc}{\lambda}
h: Planck's constant, f: frequency, \lambda: wavelength, c: speed of light
Units: J or eV
Electronvolt
1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}
energy gained by an electron moving through 1\,\mathrm{V}; 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}
Units: eV, J
Momentum of a photon
p=\dfrac{E}{c}=\dfrac{hf}{c}=\dfrac{h}{\lambda}
E: photon energy, c: speed of light, \lambda: wavelength
Unit: \mathrm{kg\,m\,s^{-1}}
Force exerted by light
F=\dfrac{P}{c}\ (\text{absorbed}),\qquad F=\dfrac{2P}{c}\ (\text{reflected})
P: power of the beam, c: speed of light
Unit: N
Maximum kinetic energy and stopping potential
K.E_{\max}=\tfrac12mv_{\max}^{2}=eV_0
e: electron charge, V_0: stopping potential, v_{\max}: speed of the fastest photoelectrons
Units: J (or eV when V_0 in volts)
Work function
\Phi=hf_0=\dfrac{hc}{\lambda_0}
f_0: threshold frequency, \lambda_0: threshold wavelength
Units: J or eV
Einstein's photoelectric equation
hf=\Phi+K.E_{\max}
hf: photon energy, \Phi: work function, K.E_{\max}: maximum kinetic energy of photoelectrons
Units: J or eV
Einstein's equation (other forms)
K.E_{\max}=h(f-f_0),\qquad V_0=\dfrac{h}{e}f-\dfrac{\Phi}{e}
f_0: threshold frequency; the V_0 against f graph has slope h/e
Units: J; V
Compton shift
\Delta\lambda=\lambda'-\lambda=\dfrac{h}{m_0c}\left(1-\cos\theta\right)
\lambda, \lambda': incident and scattered wavelengths, m_0: electron rest mass, \theta: scattering angle
Unit: m
Compton wavelength
\lambda_c=\dfrac{h}{m_0c}=2.43\times10^{-12}\,\mathrm{m}
m_0=9.11\times10^{-31}\,\mathrm{kg}
Unit: m
Energy conservation in Compton scattering
hf=hf'+K.E
f, f': incident and scattered frequencies, K.E: kinetic energy of the recoil electron
Unit: J
Pair production
hf=2m_0c^{2}+K.E_{e^-}+K.E_{e^+},\qquad hf_{\min}=2m_0c^{2}=1.02\,\mathrm{MeV}
m_0c^{2}=0.511\,\mathrm{MeV}: rest mass energy of an electron or positron
Units: J or MeV
Pair annihilation
\begin{aligned}2m_0c^{2}+K.E_{e^-}+K.E_{e^+}&=\text{energy of two photons}\\ hf&=m_0c^{2}=0.511\,\mathrm{MeV}\ \text{each (at rest)}\end{aligned}
two gamma photons move in opposite directions
Units: J or MeV
de Broglie wavelength
\lambda=\dfrac{h}{p}=\dfrac{h}{mv}
p: momentum, m: mass, v: speed of the particle
Unit: m
Wavelength of an accelerated electron
\lambda=\dfrac{h}{\sqrt{2meV}}\approx\dfrac{1.23\times10^{-9}}{\sqrt{V}}\,\mathrm{m}
V: accelerating potential difference in volts, m: electron mass, e: electron charge
Unit: m
Uncertainty in position and momentum
\Delta x\,\Delta p\ge\hbar
\Delta x: uncertainty in position, \Delta p: uncertainty in momentum
Unit: \hbar in \mathrm{J\,s}
Uncertainty in energy and time
\Delta E\,\Delta t\ge\hbar
\Delta E: uncertainty in energy, \Delta t: uncertainty in time (lifetime of the state)
Unit: \hbar in \mathrm{J\,s}
Reduced Planck constant
\hbar=\dfrac{h}{2\pi}=1.054\times10^{-34}\,\mathrm{J\,s}
h: Planck's constant
Unit: \mathrm{J\,s}
Photon emitted or absorbed by an atom
hf=E_n-E_p
E_n: higher level, E_p: lower level, f: frequency of the photon
Units: J or eV

Key terms

Quantum physics
The branch of physics that studies the behaviour of matter and energy at the atomic and subatomic level.
Electromagnetic wave
A transverse wave of oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel, moving through vacuum at c=3\times10^{8}\,\mathrm{m\,s^{-1}}.
Quantum
The smallest amount of energy that can be emitted or absorbed as electromagnetic radiation, E=hf.
Photon
A quantum (packet) of electromagnetic radiation with energy E=hf, momentum p=h/\lambda and zero rest mass.
Planck's constant
h=6.63\times10^{-34}\,\mathrm{J\,s}, the constant linking photon energy and frequency.
Quantum number
The whole number n in E=nhf that labels the allowed (quantized) energy states.
Electronvolt
The energy gained by an electron moving through a potential difference of one volt; 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}.
Radiation pressure
The force per unit area exerted by light on a surface because the photons' momentum changes.
Photoelectric effect
The emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it.
Photoelectrons
Electrons emitted from a metal surface by the action of light.
Saturation current
The maximum photoelectric current, reached when all emitted photoelectrons are collected.
Stopping potential
The minimum retarding potential at which the photoelectric current becomes zero; eV_0=K.E_{\max}.
Threshold frequency
The lowest frequency of radiation that can eject electrons from a given metal, f_0=\Phi/h.
Threshold wavelength
The longest wavelength of radiation that can eject electrons from a given metal, \lambda_0=c/f_0.
Work function (threshold energy)
The minimum energy needed by an electron to break free from a metal surface, \Phi=hf_0.
Einstein's photoelectric equation
hf=\Phi+K.E_{\max}: the photon energy equals the work function plus the maximum kinetic energy of the photoelectron.
Photocell
A device based on the photoelectric effect that converts light energy into an electric current.
Compton effect
The increase in wavelength of X-rays or gamma rays scattered by electrons, because the photons give some energy and momentum to the electrons.
Compton shift
The change in wavelength \Delta\lambda=\lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta) of a scattered photon.
Compton wavelength
h/m_0c=2.43\times10^{-12}\,\mathrm{m}, the Compton shift at a scattering angle of 90^\circ.
Antiparticle
A particle with the same mass as another particle but opposite charge; the positron is the antiparticle of the electron.
Positron
The antiparticle of the electron: mass 9.11\times10^{-31}\,\mathrm{kg}, charge +e.
Rest mass energy
The energy m_0c^{2} equivalent to the rest mass of a particle; 0.511\,\mathrm{MeV} for an electron.
Pair production
The creation of an electron and a positron from a photon of energy at least 1.02\,\mathrm{MeV} near a heavy nucleus (energy converted into mass).
Pair annihilation
The combination of an electron and a positron into two gamma-ray photons moving in opposite directions (mass converted into energy).
Wave-particle duality
The property of light and matter of behaving as waves in some experiments and as particles in others.
de Broglie wavelength
The wavelength of the wave associated with a moving particle, \lambda=h/p=h/mv.
Matter waves
Waves associated with moving material particles such as electrons.
Electron diffraction
The spreading of an electron beam into a ring or spot pattern when it passes through a thin crystal; evidence that electrons behave as waves.
Electron microscope
A microscope that uses a beam of fast electrons, focused by magnetic lenses, instead of light; its very short electron wavelength gives very high resolution.
Resolution
The ability of an instrument to show two close points as separate; it is limited by the wavelength of the radiation used.
Electron micrograph
A photograph of the image formed in an electron microscope.
Heisenberg's uncertainty principle
The position and momentum of a particle cannot both be measured exactly at the same time; \Delta x\,\Delta p\ge\hbar.
Reduced Planck constant
\hbar=h/2\pi=1.054\times10^{-34}\,\mathrm{J\,s}, read as h-cut or h-bar.
Metastable state
An excited state in which an atom stays for a comparatively long time, so its energy is well defined.
Energy level
One of the definite allowed energies of an electron in an atom.
Ground state
The lowest energy level of an atom.
Excitation
The jump of an electron to a higher energy level by absorbing energy.
De-excitation
The fall of an electron to a lower energy level, emitting a photon of energy hf=E_n-E_p.
Line spectrum
A spectrum containing only a discrete set of wavelengths, characteristic of an element.
Emission line spectrum
Bright lines of definite wavelengths on a dark background, produced by excited atoms of a gas at low pressure.
Absorption line spectrum
Dark lines of definite wavelengths on a continuous background, produced when white light passes through a cool gas.

Multiple choice questionsp. 138–139

1
Which of the following spectral regions has the highest energy?
  1. AInfrared
  2. BVioletAnswer
  3. CRed
  4. DBlue
Why

Photon energy is E=hf, so the region with the highest frequency has the highest energy. Of these, violet light has the highest frequency (shortest wavelength, about 400\,\mathrm{nm}). Infrared has an even lower frequency than red, and blue lies below violet.

2
Which of the following statements is true about a photon?
  1. AA photon has zero mass and zero momentum.
  2. BA photon has finite mass and a finite value of momentum.
  3. CA photon has zero mass but finite value of momentum.Answer
  4. DA photon has finite mass but zero momentum.
Why

A photon has zero rest mass, but it carries momentum p=E/c=h/\lambda. The Compton effect proves that photons have momentum, so option A is wrong; B and D are wrong because a photon has no rest mass.

3
What happens to the kinetic energy of the emitted electrons when the light is incident on a metal surface?
  1. AIt varies with the frequency of light.Answer
  2. BIt varies with the light intensity.
  3. CIt varies with the speed of light.
  4. DIt varies irregularly.
Why

By Einstein's equation K.E_{\max}=hf-\Phi, the kinetic energy depends on the frequency f. Intensity changes only the number of electrons emitted per second, not their energy.

4
A photoelectric cell is a device which
  1. AConverts light energy into electricity.Answer
  2. BConverts electricity into light energy.
  3. CStores light energy.
  4. DStores electricity.
Why

In a photocell, light falling on the cathode ejects photoelectrons, which produce an electric current. So it converts light energy into electrical energy; it does not store energy.

5
Which property does the Compton Effect describe about photons?
  1. AMass
  2. BMomentumAnswer
  3. CWave properties
  4. DSpeed rates
Why

The Compton effect is explained only if the photon behaves as a particle carrying momentum p=h/\lambda that it shares with the electron in a collision. It shows the particle nature, not the wave properties.

6
What happens to a high energy photon after it strikes an electron?
  1. Adecreases frequencyAnswer
  2. Bdecreases wavelength
  3. Cincreases energy
  4. Dincreases momentum
Why

The photon gives part of its energy to the electron, so its energy hf' is smaller and its frequency decreases. Its wavelength increases and its momentum h/\lambda' decreases, so B, C and D are wrong.

7
Energy of gamma photon having a wavelength of 1\,\text{\AA} is:
  1. A12.4\times10^{1}\,\mathrm{eV}
  2. B12.4\times10^{2}\,\mathrm{eV}
  3. C12.4\times10^{3}\,\mathrm{eV}Answer
  4. D12.4\times10^{4}\,\mathrm{eV}
Why
E=\dfrac{hc}{\lambda}=\dfrac{(6.63\times10^{-34})(3\times10^{8})}{1\times10^{-10}}=1.99\times10^{-15}\,\mathrm{J}=\dfrac{1.99\times10^{-15}}{1.6\times10^{-19}}\,\mathrm{eV}=12.4\times10^{3}\,\mathrm{eV}

So the energy is 12.4\,\mathrm{keV} (1.24\times10^{4}\,\mathrm{eV}). A photon of this wavelength is usually called an X-ray photon, but the calculation is the same.

Note. Book misprint: all four options are printed as 12.4\times10^{4}\,\mathrm{eV}. The options here are restored with different powers of ten so that the question can be answered; the correct energy is 12.4\times10^{3}\,\mathrm{eV}.

8
If uncertainty in the position of an electron is zero, the uncertainty in its momentum will be
  1. Aless than h/4\pi
  2. Bgreater than h/4\pi
  3. Czero
  4. DinfiniteAnswer
Why

From \Delta x\,\Delta p\ge\hbar, \Delta p\ge\hbar/\Delta x. As \Delta x\to0, \Delta p\to\infty: knowing the position exactly makes the momentum completely unknown. Option B is true in general, but the exact consequence of \Delta x=0 is an infinite uncertainty.

9
What is the process by which a particle and its antiparticle annihilate each other?
  1. APair production
  2. BPair annihilationAnswer
  3. CNuclear fission
  4. DNuclear fusion
Why

When a particle meets its antiparticle (for example an electron and a positron) both disappear and their mass becomes gamma-ray photons. This is pair annihilation. Pair production is the reverse process.

10
Which of the following particles is produced during pair annihilation?
  1. APhotonAnswer
  2. BElectron
  3. CProton
  4. DNeutron
Why

An electron and a positron annihilate into two gamma-ray photons moving in opposite directions. Electrons are consumed, not produced; protons and neutrons are not involved.

Short questionsp. 139

13 marks
Prove that energy and momentum of photon are directly proportional to each other.
Solution

A photon moves with the speed of light c. By Einstein's mass-energy relation its energy E is equivalent to a mass m=E/c^{2} (not a rest mass). Its momentum is mass times velocity:

p=mc=\dfrac{E}{c^{2}}\times c=\dfrac{E}{c}\quad\Rightarrow\quad E=pc

Since c=3\times10^{8}\,\mathrm{m\,s^{-1}} is a constant, E\propto p: doubling the momentum of a photon doubles its energy. The same result follows from E=hf and p=h/\lambda:

\dfrac{E}{p}=\dfrac{hf}{h/\lambda}=f\lambda=c=\text{constant}

Hence the energy of a photon is directly proportional to its momentum, and the constant of proportionality is the speed of light.

E=pc\ \Rightarrow\ E\propto p
23 marks
Convert 800\,\mathrm{MeV} into joules.
Solution

One electronvolt is the energy gained by an electron moving through a p.d. of one volt: 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}. The prefix mega means 10^{6}, so 1\,\mathrm{MeV}=10^{6}\,\mathrm{eV}=1.6\times10^{-13}\,\mathrm{J}.

  1. Change MeV into eV.
    800\,\mathrm{MeV}=800\times10^{6}\,\mathrm{eV}=8\times10^{8}\,\mathrm{eV}
  2. Multiply by 1.6\times10^{-19}\,\mathrm{J} per eV.
    E=8\times10^{8}\times1.6\times10^{-19}=1.28\times10^{-10}\,\mathrm{J}

So 800\,\mathrm{MeV} is 1.28\times10^{-10}\,\mathrm{J}. Energies of this size occur in nuclear and particle physics; even so, in joules they are very small numbers.

800\,\mathrm{MeV}=1.28\times10^{-10}\,\mathrm{J}
33 marks
Write the two phenomena to describe photon-electron interaction.
Solution

The two phenomena are the photoelectric effect and the Compton effect. In both, a photon behaves as a particle and interacts with a single electron.

Photoelectric effectCompton effect
What happens to the photonCompletely absorbed by one electronScattered with less energy and longer wavelength
ElectronBound electron is ejected from a metal surfaceFree (loosely bound) electron recoils
Energy equationhf=\Phi+K.E_{\max}hf=hf'+K.E
Typical radiationVisible or ultraviolet (a few eV)X-rays or gamma rays (keV and above)
ShowsEnergy of light comes in quantaPhotons carry momentum h/\lambda

At still higher photon energies (above 1.02\,\mathrm{MeV}) a photon can also create an electron-positron pair near a nucleus (pair production).

43 marks
In the interpretation of the photoelectric effect, how is it known that an electron does not absorb more than one photon?
Solution

Einstein assumed that one photon gives all its energy to one electron, so K.E_{\max}=hf-\Phi. The experimental results confirm that an electron does not absorb two or more photons:

  • No emission below the threshold frequency. If an electron could collect the energy of several photons, very intense light with hf<\Phi would eventually eject electrons. It never does, however intense the light is.
  • Maximum kinetic energy does not depend on intensity. With multi-photon absorption, brighter light (more photons) would give electrons energies such as 2hf-\Phi. In fact K.E_{\max} and the stopping potential stay the same for all intensities.
  • Slope of the graph: the K.E_{\max} against f graph has a slope of exactly h, the energy per unit frequency of one photon; two-photon absorption would give a slope of 2h.

Hence each photoelectron is released by exactly one photon.

53 marks
How you can determine the work function from a plot of the stopping potential versus the frequency of the incident radiation in a photoelectric effect experiment? Can you determine the value of Planck's constant from this plot?
Solution

Einstein's equation eV_0=hf-\Phi can be written as a straight line of the form y=mx+c:

V_0=\left(\dfrac{h}{e}\right)f-\dfrac{\Phi}{e}
Stopping potential against frequency
Two metals give parallel lines: the slope h/e is the same for every metal. The frequency intercept is the threshold frequency f_0 and the extended line cuts the V_0 axis at -\Phi/e.
  • Work function: the line cuts the frequency axis at the threshold frequency f_0 (where V_0=0), so \Phi=hf_0. Or extend the line back to f=0: it cuts the V_0 axis at -\Phi/e, so \Phi=e\times|\text{intercept}|.
  • Planck's constant: yes. The slope of the line is h/e, so h=e\times\text{slope}=e\,\dfrac{\Delta V_0}{\Delta f}. The slope is the same for every metal.
63 marks
Speculate how increasing the temperature of a photo electrode affects the outcomes of the photoelectric effect experiment.
Solution

The free electrons in a metal already have some thermal energy. Heating the photo-electrode (emitter) gives them more, which can change the results in these ways:

  • The work function decreases slightly as the metal is heated, so the threshold frequency becomes a little lower and the stopping potential (and K.E_{\max}) for a given frequency becomes a little larger.
  • At high temperature some electrons escape by thermionic emission even in the dark. This adds a dark current to the photoelectric current, and the cut-off at the stopping potential becomes less sharp.
  • At ordinary temperatures the thermal energy of an electron (about 0.03\,\mathrm{eV}) is tiny compared with \Phi (2 to 5\,\mathrm{eV}), so the effect is very small.

So the basic laws (threshold frequency, K.E_{\max}=hf-\Phi, current proportional to intensity) still hold; only small shifts and an extra thermal current appear when the electrode is made very hot.

73 marks
Which aspects of the photoelectric effect cannot be explained by classical physics?
Solution

Classical wave theory treats light as a continuous wave whose energy depends on its intensity (amplitude). It cannot explain these observations:

  1. Threshold frequency: no electrons are emitted below f_0 however intense the light, whereas wave theory predicts emission at any frequency if the light is bright enough.
  2. Kinetic energy independent of intensity: K.E_{\max} does not increase with intensity, although a stronger wave should give electrons more energy.
  3. Kinetic energy depends on frequency: K.E_{\max} increases linearly with f, which wave theory cannot explain.
  4. No time lag: emission is instant (about 10^{-9}\,\mathrm{s}) even in dim light, whereas wave theory predicts a delay while the electron collects energy.

All four are explained by the photon theory, hf=\Phi+K.E_{\max}.

83 marks
What is the physical significance of Compton's effect?
Solution

In the Compton effect X-ray photons scattered by electrons come out with a longer wavelength, \Delta\lambda=\dfrac{h}{m_0c}(1-\cos\theta). Its significance is:

  • It gives strong evidence for the particle nature of electromagnetic radiation: the photon collides with an electron like a billiard ball.
  • It confirms that a photon carries momentum p=h/\lambda as well as energy E=hf.
  • It shows that energy and momentum are conserved in each individual photon-electron collision.
  • Classical wave theory predicts no change in wavelength, so the effect supports quantum theory.
93 marks
Differentiate between Compton's shift and Compton's wavelength.
Solution
Compton shiftCompton wavelength
The change in wavelength of a scattered photon: \Delta\lambda=\lambda'-\lambda.The constant \lambda_c=\dfrac{h}{m_0c} that appears in the Compton equation.
Depends on the scattering angle: \Delta\lambda=\lambda_c(1-\cos\theta).Fixed value for the electron: 2.43\times10^{-12}\,\mathrm{m}=0.00243\,\mathrm{nm}.
Varies from 0 (at \theta=0^\circ) to 2\lambda_c=4.86\,\mathrm{pm} (at 180^\circ).Equals the Compton shift at \theta=90^\circ.
A measured change, different for each angle.A constant with the dimension of length, depending only on h, m_0 and c.

In short, the Compton wavelength is the scale factor, and the Compton shift is that factor multiplied by (1-\cos\theta).

103 marks
How can you say that scattering angle of photon plays effective role in measurement of Compton's shift?
Solution

The Compton shift is given by

\Delta\lambda=\dfrac{h}{m_0c}\left(1-\cos\theta\right)

Here h, m_0 and c are constants, so the only variable is the scattering angle \theta. The shift does not depend on the incident wavelength or on the scattering material. Changing \theta changes the shift:

\theta0^\circ60^\circ90^\circ180^\circ
\Delta\lambda01.21\,\mathrm{pm}2.43\,\mathrm{pm}4.86\,\mathrm{pm} (maximum)

So the scattering angle alone decides the size of the shift: it is zero for no deflection and largest when the photon bounces straight back.

113 marks
What are the conditions required for pair production to occur?
Solution

In pair production a gamma-ray photon changes into an electron and a positron. The conditions are:

  1. Minimum energy: the photon energy must be at least the rest mass energy of the two particles, hf\ge2m_0c^{2}=2(0.511)=1.02\,\mathrm{MeV} (a gamma-ray photon, \lambda\le1.2\times10^{-12}\,\mathrm{m}). Any extra energy becomes kinetic energy of the pair.
  2. A heavy nucleus nearby: the process must happen close to a nucleus, which takes up part of the momentum, so that momentum can be conserved.
  3. Conservation laws: charge, energy and momentum must all be conserved. Since the photon has no charge, a particle and its antiparticle (electron and positron, total charge zero) must be created together.
123 marks
What is meant by de Broglie wavelength?
Solution

In 1924 de Broglie suggested that every moving particle has a wave associated with it (a matter wave). The de Broglie wavelength is the wavelength of this wave. It equals Planck's constant divided by the momentum of the particle:

\lambda=\dfrac{h}{p}=\dfrac{h}{mv}

Greater momentum means a shorter wavelength. For a 0.5\,\mathrm{kg} ball at 10\,\mathrm{m\,s^{-1}}, \lambda=1.3\times10^{-34}\,\mathrm{m} (unobservable), but for an electron at 10^{6}\,\mathrm{m\,s^{-1}}, \lambda=7.3\times10^{-10}\,\mathrm{m}, similar to atomic spacing, so electron diffraction can be observed.

133 marks
Which phenomenon provides evidence for particulate nature of electromagnetic radiation?
Solution

The photoelectric effect provides the clearest evidence. It shows a threshold frequency, a maximum kinetic energy that depends on frequency but not on intensity, and no time lag. These results can be explained only if radiation consists of particles (photons) of energy hf, each absorbed by one electron: hf=\Phi+K.E_{\max}.

The Compton effect (photons colliding with electrons and transferring momentum h/\lambda) and pair production (a single photon turning into an electron and a positron) give further evidence of the particle nature of electromagnetic radiation.

Comprehensive questionsp. 139–140

12026 Annual I, Set B, Q.5 OR2026 Annual I, Set D, Q.5 OR2026 Annual I, Old syllabus (Curriculum 2006), Q.4, old syllabus8 marks
What is photoelectric effect? How quantum physics explain its results.
Solution
Definition

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons and the current they produce is the photoelectric current.

Experimental arrangement
Photoelectric set-up: emitter E, collector C, ammeter A and a reversible battery in an evacuated tube.

Light falls on the emitter E inside an evacuated tube. Photoelectrons move to the collector C and the ammeter records the current. The p.d. V between C and E can be changed and reversed.

Experimental results
  1. No electrons are emitted if the frequency is below a certain threshold frequency f_0, however intense the light.
  2. Above f_0, emission starts instantly, even in very dim light.
  3. The photoelectric current (number of electrons per second) is proportional to the intensity of light.
  4. The current becomes zero at a negative collector potential -V_0, the stopping potential, and K.E_{\max}=eV_0. V_0 is the same for all intensities, so K.E_{\max} does not depend on intensity.
  5. K.E_{\max} increases linearly with frequency.
Photoelectric current against collector potential
Light of one frequency at three intensities. Brighter light gives a larger saturation current, but every curve reaches zero at the same stopping potential -V_0.
Failure of classical wave theory

Wave theory says a more intense wave carries more energy, so brighter light of any frequency should eventually eject electrons, and should give them more kinetic energy after some delay. It cannot explain the threshold frequency, the independence of K.E_{\max} from intensity, its dependence on frequency, or the absence of a time lag.

Quantum (photon) explanation

In 1905 Einstein used Planck's quantum idea: light is a stream of photons, each of energy hf. A photon interacts with a single electron and is absorbed completely (one photon, one electron).

Derivation Einstein's photoelectric equation

Aim. Relate photon energy, work function and maximum kinetic energy.

  1. Light consists of photons. One photon of frequency f carries energy hf and gives all of it to one electron.
    E=hf
  2. Part of this energy, the work function \Phi, is used to pull the electron out of the metal. By conservation of energy the rest is the electron's kinetic energy, which is maximum for an electron from the surface.
    hf=\Phi+K.E_{\max}
  3. At the threshold frequency f_0 the electron just escapes with zero kinetic energy.
    hf_0=\Phi
  4. Substitute for \Phi and use K.E_{\max}=\tfrac12mv_{\max}^{2}=eV_0.
    K.E_{\max}=\tfrac12mv_{\max}^{2}=h(f-f_0)=eV_0
Resulthf=\Phi+\tfrac12mv_{\max}^{2}

Photoemission happens only if hf\ge\Phi, and the maximum kinetic energy rises linearly with frequency with slope h.

How the equation explains the results
  • Threshold frequency: if hf<\Phi one photon cannot free an electron, and an electron cannot combine the energy of two photons, so there is no emission however many photons arrive.
  • No time lag: the energy arrives in one packet, so the electron leaves as soon as it absorbs a photon.
  • Current proportional to intensity: a brighter beam contains more photons per second, which free more electrons per second.
  • Kinetic energy independent of intensity: each photon still has the same energy hf, so K.E_{\max}=hf-\Phi is unchanged.
  • Kinetic energy increases with frequency: K.E_{\max}=hf-\Phi is a straight line in f with slope h and intercept f_0.
Maximum kinetic energy against frequency
No electrons are emitted below the threshold frequency f_0. Above f_0 the graph is a straight line of slope h; extended back (dashed) it cuts the energy axis at -\Phi. The line is drawn to scale for sodium (\Phi=2.3\,\mathrm{eV}, f_0=5.6\times10^{14}\,\mathrm{Hz}).

So the photoelectric effect is explained completely by the photon (quantum) theory, and it is strong evidence that electromagnetic radiation has a particle nature.

28 marks
What is Compton's effect? Discuss in detail.
Solution
Definition

When a beam of X-rays of wavelength \lambda is scattered by the electrons of a material (Compton used graphite in 1923), the scattered X-rays have a slightly longer wavelength \lambda'. The increase depends on the scattering angle. This is the Compton effect; the increase \Delta\lambda=\lambda'-\lambda is the Compton shift.

Diagram
An X-ray photon collides with an electron at rest; the scattered photon leaves at angle \theta with wavelength \lambda'>\lambda and the electron recoils at angle \phi.
Explanation

Classical theory predicts no change of wavelength. Compton treated the X-ray as photons, each of energy hf and momentum h/\lambda, colliding with free electrons like billiard balls. In each collision energy and momentum are conserved. The electron recoils and takes some energy, so the scattered photon has less energy: lower frequency f' and longer wavelength \lambda'.

Conservation of energy (K.E is the kinetic energy of the recoil electron):

hf=hf'+K.E

Conservation of momentum along and perpendicular to the incident direction (p_e is the momentum of the electron):

\dfrac{h}{\lambda}=\dfrac{h}{\lambda'}\cos\theta+p_e\cos\phi,\qquad 0=\dfrac{h}{\lambda'}\sin\theta-p_e\sin\phi
Compton shift

Derivation Compton equation

Aim. Obtain \Delta\lambda from conservation of energy and momentum.

  1. With p=h/\lambda and p'=h/\lambda', energy conservation gives the electron's total energy after the collision.
    E_e=(p-p')c+m_0c^{2}
  2. Momentum conservation, squared (angle \theta between \vec p and \vec p\,'):
    p_e^{2}=p^{2}+p'^{2}-2pp'\cos\theta
  3. For the electron E_e^{2}=p_e^{2}c^{2}+m_0^{2}c^{4}. Substituting steps 1 and 2 and cancelling equal terms leaves:
    (p-p')\,m_0c=pp'(1-\cos\theta)
  4. Divide by pp'm_0c and use 1/p=\lambda/h, 1/p'=\lambda'/h.
    \lambda'-\lambda=\dfrac{h}{m_0c}(1-\cos\theta)
Result\Delta\lambda=\dfrac{h}{m_0c}(1-\cos\theta)

The shift depends only on the scattering angle.

The constant h/m_0c=2.43\times10^{-12}\,\mathrm{m} is the Compton wavelength.

Main features
  • \Delta\lambda depends only on \theta: zero at 0^\circ, 2.43\,\mathrm{pm} at 90^\circ, maximum 4.86\,\mathrm{pm} at 180^\circ.
  • It is independent of the incident wavelength and of the scattering material.
  • The effect is noticeable only for X-rays and gamma rays; for visible light a shift of a few picometres is negligible.
Compton shift against scattering angle
The shift is zero for \theta=0, equals the Compton wavelength \lambda_c at 90^\circ and is largest, 2\lambda_c=4.86\,\mathrm{pm}, for a photon bounced straight back (180^\circ).
Significance
  • It proves the particle nature of electromagnetic radiation.
  • It shows that photons carry momentum p=h/\lambda.
  • It confirms conservation of energy and momentum in single photon-electron collisions.
38 marks
State and discuss the Heisenberg Uncertainty Principle.
Solution
Statement

Heisenberg's uncertainty principle (1927): it is impossible to measure accurately both the position and the momentum of a particle at the same time. The product of the uncertainties in position and momentum at the same instant is approximately equal to \hbar or greater:

\Delta x\,\Delta p\ge\hbar,\qquad \hbar=\dfrac{h}{2\pi}=1.054\times10^{-34}\,\mathrm{J\,s}

Here \Delta x is the uncertainty in position and \Delta p the uncertainty in momentum. If \Delta x\to0 then \Delta p\to\infty, and the other way round.

Explanation: measurement disturbs the particle

To observe an electron we must bounce at least one photon off it. The position can be found only to within about one wavelength, \Delta x\approx\lambda. To locate it precisely we need a short wavelength, but a short-wavelength photon has a large momentum h/\lambda and gives the electron a large, unknown kick, \Delta p\approx h/\lambda. Using a longer wavelength gives a smaller kick but a blurred position. In every case

\Delta x\,\Delta p\approx\lambda\times\dfrac{h}{\lambda}=h

which agrees in order of magnitude with \Delta x\,\Delta p\ge\hbar.

Explanation: wave nature

A particle is described by a wave. Position means where the wave is concentrated; momentum is fixed by the wavelength, p=h/\lambda. A long wave has a definite wavelength (definite momentum) but no definite position. A short pulse has a definite position, but it is made by mixing many wavelengths, so its momentum is uncertain.

(b) A short wave packet
The wave is squeezed into a small region \Delta x, so the position is well known. To build such a short pulse many wavelengths must be mixed, so the wavelength and the momentum become uncertain.
Energy and time

A similar relation holds for energy and time. If a state lasts for a time \Delta t, its energy is uncertain by \Delta E, where

\Delta E\,\Delta t\ge\hbar

A metastable state that lasts a long time has a sharply defined energy; a short-lived state has a large uncertainty in energy.

Examples
  • Electron located within \Delta x=10^{-10}\,\mathrm{m} (size of an atom): \Delta p\ge\hbar/\Delta x=1.05\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}, which corresponds to a speed uncertainty of about 10^{6}\,\mathrm{m\,s^{-1}}. This is huge for an electron.
  • Ball of mass 0.1\,\mathrm{kg} with \Delta p=3\times10^{-6}\,\mathrm{kg\,m\,s^{-1}}: \Delta x\ge3.5\times10^{-29}\,\mathrm{m}, far too small to notice.
Conclusion

These uncertainties are a property of nature, not of poor instruments or of the observer's skill. Every measurement needs an interaction that disturbs the object, so every empirical measurement must carry some uncertainty. Because h is so small, the effect matters only for atomic particles.

42026 Annual I, Set B, Q.2 (vii)8 marks
Discuss Pair production of electron as practical implementation of conversion of energy into mass correspondences to of Eisenstein energy-mass equation.
Solution
Definition

Pair production is the creation of a particle and its antiparticle from energy. When a high-energy gamma-ray photon passes close to a heavy nucleus, it disappears and an electron (e^-) and a positron (e^+) are created. The photon is completely absorbed.

A gamma-ray photon changes into an electron-positron pair near a heavy nucleus.
Energy converted into mass

Before the event there is only electromagnetic energy (the photon has no rest mass). Afterwards there are two particles, each of rest mass m_0. Energy has been converted into mass, exactly as Einstein's mass-energy relation E=mc^{2} predicts: creating a rest mass m_0 needs energy m_0c^{2}.

Conservation laws
  • Charge: the photon has zero charge; the electron and positron have charges -e and +e, total zero. So the photon can never produce a single electron, a single positron, or two particles of the same charge.
  • Momentum: the photon's momentum hf/c must equal the total momentum of the products, \dfrac{hf}{c}=mv_{e^+}+mv_{e^-} (plus the recoil of the nucleus). A pair alone cannot carry all of the photon's momentum, so the heavy nucleus takes up the rest; this is why the process needs a nucleus.
  • Energy: the photon energy equals the total energy (rest energy plus kinetic energy) of the two particles.
hf=(K.E_{e^-}+m_0c^{2})+(K.E_{e^+}+m_0c^{2})=2m_0c^{2}+K.E_{e^-}+K.E_{e^+}
Minimum (threshold) energy

Derivation Threshold energy for pair production

Aim. Find the least photon energy that can create a pair.

  1. The photon energy is least when the pair has no kinetic energy.
    hf_{\min}=2m_0c^{2}
  2. Rest energy of an electron.
    m_0c^{2}=(9.11\times10^{-31})(3\times10^{8})^{2}=8.2\times10^{-14}\,\mathrm{J}=0.511\,\mathrm{MeV}
  3. Double it.
    hf_{\min}=1.02\,\mathrm{MeV}=1.64\times10^{-13}\,\mathrm{J}
  4. Corresponding largest wavelength.
    \lambda_{\max}=\dfrac{hc}{hf_{\min}}=1.21\times10^{-12}\,\mathrm{m}
Resulthf_{\min}=2m_0c^{2}=1.02\,\mathrm{MeV}

Only gamma-ray photons can produce pairs; any energy above 1.02\,\mathrm{MeV} becomes kinetic energy of the electron and positron.

Example

A 3.0\,\mathrm{MeV} photon: 1.02\,\mathrm{MeV} becomes the rest mass of the pair and the remaining 1.98\,\mathrm{MeV} becomes their kinetic energy.

Pair production is therefore a practical demonstration of Einstein's equation E=mc^{2}: electromagnetic energy is turned into matter.

Note. The book prints Eisenstein for Einstein; the question refers to Einstein's mass-energy equation E=mc^{2}.

52026 Annual I, Set D, Q.2 (vii)8 marks
Explain Annihilation process as practical implementation of conversion of mass into energy correspondences to of Eisenstein energy-mass equation.
Solution
Definition

Pair annihilation is the process in which an electron and its antiparticle, the positron, combine and disappear, and their mass is converted into two gamma-ray photons. It is the reverse of pair production.

An electron and a positron annihilate into two 0.511\,\mathrm{MeV} photons moving in opposite directions.
Mass converted into energy

Before annihilation there are two particles, each of rest mass m_0. Afterwards there is no rest mass at all, only electromagnetic energy. The whole mass 2m_0 has been converted into energy E=(2m_0)c^{2}, as Einstein's relation predicts.

Conservation laws
  • Charge: (+e)+(-e)=0 before; photons have no charge, so charge is conserved.
  • Momentum: the pair is nearly at rest, so its total momentum is about zero. One photon always has momentum hf/c\ne0, so a single photon cannot be produced. Two photons of equal energy moving in opposite directions have a total momentum of zero, so momentum is conserved.
  • Energy: the total energy of the pair (rest energy plus kinetic energy) equals the energy of the two photons.
2m_0c^{2}+K.E_{e^-}+K.E_{e^+}=\text{energy of two gamma photons}
Energy of each photon

Derivation Photon energy in annihilation at rest

Aim. Find the energy and wavelength of each photon when the pair is at rest.

  1. Total energy available is the rest energy of the two particles.
    2hf=2m_0c^{2}
  2. Each photon gets half.
    hf=m_0c^{2}=0.511\,\mathrm{MeV}=8.2\times10^{-14}\,\mathrm{J}
  3. Wavelength of each photon.
    \lambda=\dfrac{h}{m_0c}=2.43\times10^{-12}\,\mathrm{m}
Resulthf=0.511\,\mathrm{MeV}\ \text{each}

The photons are gamma rays. Their energy comes entirely from the mass of the particles.

Examples and uses

Proton-antiproton annihilation is another example. In medicine, PET scanners detect the pairs of 0.511\,\mathrm{MeV} photons produced when positrons from a tracer annihilate with electrons in the body.

Annihilation is therefore a practical demonstration of E=mc^{2} in the direction mass into energy, while pair production shows energy into mass. Together they confirm the photon model of radiation and Einstein's idea of rest mass energy.

Note. The book prints Eisenstein for Einstein.

68 marks
What do you understand by the term 'Dual nature of light'? Discuss.
Solution
Meaning

Dual nature of light means that light behaves as a wave in some experiments and as a stream of particles (photons) in others. Neither picture alone explains everything, so light has both wave and particle characteristics (wave-particle duality).

Evidence for the wave nature
  • Interference: in Young's double slit experiment light from two slits produces bright and dark fringes. Only waves can add and cancel in this way.
  • Diffraction: light spreads round the edges of narrow slits and a diffraction grating gives sharp maxima.
  • Polarization: light can be polarized, which shows that it is a transverse wave.
Evidence for the particle nature
  • Photoelectric effect: threshold frequency, instant emission and K.E_{\max}=hf-\Phi independent of intensity can be explained only by photons of energy hf, each absorbed by one electron.
  • Compton effect: X-ray photons collide with electrons like particles and transfer momentum h/\lambda, so the scattered wavelength increases.
  • Pair production and annihilation: a single photon turns into an electron-positron pair, and a pair turns into photons, conserving energy and momentum like particles.
Linking the two natures

The particle properties of a photon are fixed by its wave properties:

E=hf,\qquad p=\dfrac{h}{\lambda}

The two natures complement each other: light travels (spreads, interferes, diffracts) as a wave, but it is emitted and absorbed as particles, one photon at a time. Interference and diffraction experiments show the wave behaviour; photoelectric and Compton experiments show the particle behaviour.

Wave natureParticle nature
InterferencePhotoelectric effect
DiffractionCompton effect
PolarizationPair production and annihilation
Described by \lambda and fDescribed by E=hf and p=h/\lambda
Extension to matter

In 1924 de Broglie suggested that the duality works both ways: particles such as electrons have a wavelength \lambda=h/mv. Electron diffraction confirmed this, so wave-particle duality is a general property of both radiation and matter.

78 marks
Explain that electromagnetic radiation has a particulate nature.
Solution
Photon idea

In 1900 Planck showed that atoms emit and absorb radiation only in discrete packets (quanta) of energy hf. In 1905 Einstein proposed that radiation itself travels as these packets, called photons. A photon is a particle of electromagnetic radiation with:

  • energy E=hf=hc/\lambda,
  • momentum p=E/c=h/\lambda,
  • zero rest mass and speed c.

A beam of light is a stream of photons; its intensity is set by the number of photons per second, and the energy of each photon by its frequency.

Evidence 1: photoelectric effect

Electrons are emitted only if f\ge f_0, emission is instant, and K.E_{\max} depends on frequency but not on intensity. These results follow from hf=\Phi+K.E_{\max}, where one photon is absorbed by one electron. A continuous wave cannot explain them.

Evidence 2: Compton effect

X-ray photons scattered by electrons have a longer wavelength, \Delta\lambda=\dfrac{h}{m_0c}(1-\cos\theta). This is explained only by treating each photon as a particle with momentum h/\lambda colliding with an electron, conserving energy and momentum.

Evidence 3: pair production and annihilation

A single gamma photon of energy at least 1.02\,\mathrm{MeV} turns into an electron and a positron, and an electron-positron pair turns into two photons of 0.511\,\mathrm{MeV} each. Radiation behaves as discrete particles of definite energy and momentum.

Evidence 4: radiation pressure and line spectra

Light exerts a force F=P/c on a surface because each photon carries momentum. Atoms emit and absorb light only at definite wavelengths because each transition creates or destroys one photon of energy hf=E_n-E_p.

Conclusion

All these phenomena show that electromagnetic radiation is emitted, absorbed and scattered in discrete units (photons). Interference and diffraction show its wave nature, so radiation has a dual nature, but its interaction with matter is particulate.

82026 Annual I, Set B, Q.2 (viii) OR2026 Annual I, Set D, Q.2 (viii) OR8 marks
Explain how electron microscopes achieve very high resolution.
Solution
Principle

An electron microscope uses a beam of fast electrons instead of light. It works because moving electrons behave as waves (de Broglie), and their wavelength is extremely short.

Resolution depends on wavelength

Resolution is the ability to show two close points as separate. Because of diffraction, no microscope can resolve details much smaller than the wavelength of the radiation it uses. Visible light has \lambda=400 to 700\,\mathrm{nm}, so an optical microscope cannot resolve details below about 200\,\mathrm{nm}.

Wavelength of the electrons

Derivation Wavelength of an accelerated electron

Aim. Find the de Broglie wavelength of electrons accelerated through a p.d. V.

  1. Kinetic energy gained from the field.
    \tfrac12mv^{2}=eV\ \Rightarrow\ p=mv=\sqrt{2meV}
  2. de Broglie relation.
    \lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}
  3. For V=10\,000\,\mathrm{V}:
    \lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2(9.11\times10^{-31})(1.6\times10^{-19})(10^{4})}}=1.2\times10^{-11}\,\mathrm{m}=0.12\,\text{\AA}
Result\lambda=\dfrac{h}{\sqrt{2meV}}

Higher accelerating voltage gives shorter wavelength. At 10\,\mathrm{kV} the electron wavelength is tens of thousands of times shorter than that of visible light.

In practice imperfect magnetic lenses limit the resolution to about 0.1 to 0.2\,\mathrm{nm}, still about a thousand times better than an optical microscope. Very small objects such as viruses, and even rows of atoms in crystals, can be seen.

Construction and working
Electron microscope: electron gun, magnetic condenser, objective and projector lenses, thin specimen and fluorescent screen, all in a vacuum.
  1. A heated tungsten filament emits electrons; a high positive potential on the anode accelerates them into a narrow beam.
  2. A magnetic condenser lens (current-carrying coil) focuses the beam onto a very thin specimen.
  3. The objective lens forms a magnified image; the intermediate and projector lenses magnify it further.
  4. The image is formed on a fluorescent screen or detector and can be photographed (electron micrograph).
  5. The whole column is kept in a vacuum so the electrons are not scattered by air.
Comparison
Optical microscopeElectron microscope
RadiationLight, \lambda\approx500\,\mathrm{nm}Electrons, \lambda\approx0.01\,\mathrm{nm}
LensesGlassMagnetic coils
Resolutionabout 200\,\mathrm{nm}about 0.1\,\mathrm{nm}
MediumAirVacuum

Thus the very short de Broglie wavelength of fast electrons is what gives the electron microscope its very high resolution.

98 marks
There are discrete electron energy levels in isolated atoms. Explain this statement by the appearance and formation of emission and absorption line spectra.
Solution
Discrete energy levels

The electrons of an isolated atom (for example in a gas at low pressure) can occupy only certain allowed orbits. Each orbit has a definite energy E_1, E_2, E_3, ... These are the energy levels; E_1 is the ground state. An electron can change level only by gaining or losing exactly the energy difference between two levels.

Excitation by absorbing a photon and de-excitation by emitting a photon; hf=E_3-E_1 in both cases.
hf=E_n-E_p

E_n: higher level, E_p: lower level.

Because the levels are discrete, only certain energy differences exist, so only photons of certain definite frequencies can be emitted or absorbed. The spectra prove this.

Formation of the emission line spectrum

Atoms of a gas in a discharge tube are excited (by electric current or heating) to higher levels. When the electrons fall back to lower levels, each transition emits a photon of energy hf=E_n-E_p. The light is passed through a slit and a prism or diffraction grating.

Slit, lens and prism (or grating) separate the light into lines on a detector.

Appearance: a set of separate bright lines of definite wavelengths on a dark background. There is no continuous band, only lines, because only certain energy jumps are possible.

Formation of the absorption line spectrum

White light (containing all wavelengths) from a tungsten lamp is passed through the cool vapour of the element. Atoms in the ground state absorb only those photons whose energy exactly equals E_n-E_1 for some level, and jump up. The rest of the light passes through. The excited atoms re-emit the absorbed energy in all directions, so little of it continues in the original direction.

Appearance: a continuous coloured spectrum crossed by dark lines at definite wavelengths.

Hydrogen: (a) emission lines and (b) absorption lines appear at the same wavelengths.
Conclusion
  • Both spectra contain only discrete wavelengths, never a continuous range, so the energies of the electrons in the atom must be discrete.
  • The dark absorption lines occur at exactly the same wavelengths as the bright emission lines of the same element, because both come from jumps between the same fixed energy levels.
  • Each element has its own set of levels, so its line spectrum is unique and can be used to identify it.

Numerical problemsp. 140

Constants used, as in the book: h=6.626\times10^{-34}\,\mathrm{J\,s} (or 6.63\times10^{-34}), c=3\times10^{8}\,\mathrm{m\,s^{-1}}, e=1.6\times10^{-19}\,\mathrm{C}, m_0=9.11\times10^{-31}\,\mathrm{kg}, m_0c^{2}=0.511\,\mathrm{MeV}.

1
In an experiment, the work function of potassium surface is found to be 2.1\,\mathrm{eV}. What should be the wavelength of incident radiation if the stopping potential for the electron is 0.43\,\mathrm{V}?
Solution
Given
  • \Phi = 2.1\,\mathrm{eV} work function
  • V_0 = 0.43\,\mathrm{V} stopping potential
  • h = 6.626\times10^{-34}\,\mathrm{J\,s}
  • c = 3\times10^{8}\,\mathrm{m\,s^{-1}}
To find
  • wavelength \lambda
  1. The stopping potential gives the maximum kinetic energy: K.E_{\max}=eV_0. With V_0 in volts the energy in eV has the same number.
    K.E_{\max}=eV_0=0.43\,\mathrm{eV}
  2. Einstein's equation: photon energy = work function + maximum kinetic energy.
    E=\Phi+K.E_{\max}=2.1+0.43=2.53\,\mathrm{eV}
  3. Change into joules.
    E=2.53\times1.6\times10^{-19}=4.05\times10^{-19}\,\mathrm{J}
  4. Use E=hc/\lambda and make \lambda the subject.
    \lambda=\dfrac{hc}{E}=\dfrac{6.626\times10^{-34}\times3\times10^{8}}{4.05\times10^{-19}}=4.91\times10^{-7}\,\mathrm{m}
\lambda=4.91\times10^{-7}\,\mathrm{m}\ (491\,\mathrm{nm})

Book answer. 5.91\times10^{-7}\,\mathrm{m}

Note. The book's answer 5.91\times10^{-7}\,\mathrm{m} is wrong: it is hc/\Phi, the threshold wavelength of potassium, which ignores the stopping potential. Including K.E_{\max}=0.43\,\mathrm{eV} gives 4.91\times10^{-7}\,\mathrm{m}.

2
Light of wavelength 5000\,\text{\AA} falls on a metal surface having work function of 1.9\,\mathrm{eV}. Find (a) Energy of photon in electron-volt. (b) Kinetic energy of photo-electrons emitted. (c) Stopping potential.
Solution
(a) Energy of photon in electron-volt.
Given
  • \lambda = 5000\,\text{\AA}=5000\times10^{-10}\,\mathrm{m}=5\times10^{-7}\,\mathrm{m}
  • \Phi = 1.9\,\mathrm{eV}
  • h = 6.626\times10^{-34}\,\mathrm{J\,s}
To find
  • E in eV
  1. Photon energy in joules.
    E=\dfrac{hc}{\lambda}=\dfrac{6.626\times10^{-34}\times3\times10^{8}}{5\times10^{-7}}=3.98\times10^{-19}\,\mathrm{J}
  2. Divide by 1.6\times10^{-19}\,\mathrm{J} per eV.
    E=\dfrac{3.976\times10^{-19}}{1.6\times10^{-19}}=2.48\,\mathrm{eV}
E=2.48\,\mathrm{eV}
(b) Kinetic energy of photo-electrons emitted.
  1. Einstein's equation, working in eV.
    K.E_{\max}=E-\Phi=2.48-1.9=0.58\,\mathrm{eV}
  2. In joules, if required.
    K.E_{\max}=0.58\times1.6\times10^{-19}=9.3\times10^{-20}\,\mathrm{J}
K.E_{\max}=0.58\,\mathrm{eV}
(c) Stopping potential.
  1. K.E_{\max}=eV_0, so V_0=K.E_{\max}/e.
    V_0=\dfrac{0.58\times1.6\times10^{-19}\,\mathrm{J}}{1.6\times10^{-19}\,\mathrm{C}}=0.58\,\mathrm{V}
V_0=0.58\,\mathrm{V}

Book answer. 2.48\,\mathrm{eV}, 0.58\,\mathrm{eV}, 0.58\,\mathrm{V}

Note. With h=6.63\times10^{-34}\,\mathrm{J\,s} the photon energy comes out as 2.49\,\mathrm{eV} and K.E_{\max}=0.59\,\mathrm{eV}; the book's values use h=6.626\times10^{-34}\,\mathrm{J\,s}.

3
Calculate the energies of the photons associated with the (i) violet light of 413\,\mathrm{nm}. (ii) X-rays of 0.1\,\mathrm{nm}. (iii) radio waves of 10\,\mathrm{m}.
Solution
(i) Violet light of 413\,\mathrm{nm}.

For each part use E=hc/\lambda with hc=6.626\times10^{-34}\times3\times10^{8}=1.988\times10^{-25}\,\mathrm{J\,m}, then divide by 1.6\times10^{-19} to get eV.

  1. Energy in joules.
    E=\dfrac{1.988\times10^{-25}}{413\times10^{-9}}=4.81\times10^{-19}\,\mathrm{J}
  2. Energy in eV.
    E=\dfrac{4.81\times10^{-19}}{1.6\times10^{-19}}=3.0\,\mathrm{eV}
E=4.81\times10^{-19}\,\mathrm{J}=3.0\,\mathrm{eV}
(ii) X-rays of 0.1\,\mathrm{nm}.
  1. Energy in joules, with \lambda=0.1\times10^{-9}=10^{-10}\,\mathrm{m}.
    E=\dfrac{1.988\times10^{-25}}{1\times10^{-10}}=1.988\times10^{-15}\,\mathrm{J}
  2. Energy in eV.
    E=\dfrac{1.988\times10^{-15}}{1.6\times10^{-19}}=1.24\times10^{4}\,\mathrm{eV}\approx12\,424\,\mathrm{eV}
E=1.99\times10^{-15}\,\mathrm{J}=1.24\times10^{4}\,\mathrm{eV}
(iii) Radio waves of 10\,\mathrm{m}.
  1. Energy in joules.
    E=\dfrac{1.988\times10^{-25}}{10}=1.99\times10^{-26}\,\mathrm{J}
  2. Energy in eV.
    E=\dfrac{1.988\times10^{-26}}{1.6\times10^{-19}}=1.24\times10^{-7}\,\mathrm{eV}
E=1.99\times10^{-26}\,\mathrm{J}=1.24\times10^{-7}\,\mathrm{eV}

Book answer. 3\,\mathrm{eV}; 12424\,\mathrm{eV}; 1.24\times10^{-7}\,\mathrm{eV}

4
Calculate the minimum energy required by a photon to transfer half of its energy to an electron at rest.
Solution

The photon gives energy to a free electron by Compton scattering. We need the smallest photon energy E for which the scattered photon can be left with only half its energy, E'=E/2.

Given
  • E' = \tfrac12E energy of scattered photon
  • m_0c^{2} = 0.511\,\mathrm{MeV} rest energy of the electron
To find
  • minimum photon energy E
  1. Half the energy means the scattered photon has half the frequency, so its wavelength doubles: \lambda'=2\lambda.
    E'=\dfrac{E}{2}\ \Rightarrow\ \lambda'=2\lambda\ \Rightarrow\ \Delta\lambda=\lambda'-\lambda=\lambda
  2. Compton equation: the shift needed equals the incident wavelength.
    \lambda=\dfrac{h}{m_0c}(1-\cos\theta)
  3. The largest possible shift is at \theta=180^\circ, where (1-\cos\theta)=2. So the incident wavelength can be at most 2h/m_0c.
    \lambda\le\dfrac{2h}{m_0c}
  4. Energy is inversely proportional to wavelength, so the minimum energy corresponds to this maximum wavelength (photon scattered straight back).
    E_{\min}=\dfrac{hc}{\lambda_{\max}}=\dfrac{hc}{2h/m_0c}=\dfrac{m_0c^{2}}{2}
  5. Substitute the rest energy of the electron.
    E_{\min}=\dfrac{0.511}{2}=0.256\,\mathrm{MeV}=4.1\times10^{-14}\,\mathrm{J}
E_{\min}=\dfrac{m_0c^{2}}{2}\approx0.26\,\mathrm{MeV}

Book answer. 0.25\,\mathrm{MeV}

Note. The exact value is m_0c^{2}/2=0.2555\,\mathrm{MeV}; the book rounds it down to 0.25\,\mathrm{MeV}.

5
An electron is placed in a box about the size of an atom that is about 1\,\text{\AA}. What is the velocity of the electron?
Solution
Given
  • \Delta x = 1\,\text{\AA}=1\times10^{-10}\,\mathrm{m} size of the box
  • m = 9.1\times10^{-31}\,\mathrm{kg}
  • h = 6.63\times10^{-34}\,\mathrm{J\,s}
To find
  • speed v
  1. The electron is somewhere inside the box, so the uncertainty in its position is the size of the box.
    \Delta x\approx1\times10^{-10}\,\mathrm{m}
  2. By the uncertainty principle (rough form \Delta x\,\Delta p\approx h), its momentum cannot be known better than h/\Delta x. A particle whose momentum is uncertain by this much must have a momentum of at least this size.
    p\approx\Delta p\approx\dfrac{h}{\Delta x}=\dfrac{6.63\times10^{-34}}{1\times10^{-10}}=6.63\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}
  3. Speed from p=mv.
    v=\dfrac{p}{m}=\dfrac{6.63\times10^{-24}}{9.1\times10^{-31}}=7.29\times10^{6}\,\mathrm{m\,s^{-1}}
v\approx7.29\times10^{6}\,\mathrm{m\,s^{-1}}

Book answer. 7.29\times10^{6}\,\mathrm{m\,s^{-1}}

6
A 50\,\mathrm{keV} X-ray is scattered through an angle of 90^\circ. What is the energy of the X-ray after Compton scattering?
Solution
Given
  • E = 50\,\mathrm{keV}=50\times10^{3}\times1.6\times10^{-19}=8.0\times10^{-15}\,\mathrm{J}
  • \theta = 90^\circ
  • \dfrac{h}{m_0c} = 2.43\times10^{-12}\,\mathrm{m} Compton wavelength
To find
  • energy E' of the scattered X-ray
  1. Wavelength of the incident X-ray from E=hc/\lambda.
    \lambda=\dfrac{hc}{E}=\dfrac{6.626\times10^{-34}\times3\times10^{8}}{8.0\times10^{-15}}=2.485\times10^{-11}\,\mathrm{m}
  2. Compton shift at 90^\circ, where \cos90^\circ=0.
    \Delta\lambda=\dfrac{h}{m_0c}(1-\cos90^\circ)=2.43\times10^{-12}\,\mathrm{m}
  3. Wavelength of the scattered X-ray.
    \lambda'=\lambda+\Delta\lambda=2.485\times10^{-11}+0.243\times10^{-11}=2.728\times10^{-11}\,\mathrm{m}
  4. Energy is inversely proportional to wavelength, so scale the incident energy.
    E'=E\,\dfrac{\lambda}{\lambda'}=50\times\dfrac{2.485}{2.728}=45.5\,\mathrm{keV}
E'\approx45.5\,\mathrm{keV}

Book answer. 45.5\,\mathrm{keV}

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 111 mark
In photoelectric effect, which factor increases by increasing the intensity of incident light?
  1. AKinetic energy of electron
  2. BStopping potential
  3. CNumber of emitted electronsAnswer
  4. DWork function
Why

Brighter light means more photons per second, and each photon frees one electron, so more electrons are emitted per second (larger current). The kinetic energy and stopping potential depend on frequency, not intensity, and the work function is a property of the metal.

2026 Annual I, Set B, MCQ 121 mark
If a particle of mass m is moving with speed v then De-Broglie wavelength \lambda is given by:
  1. A\lambda=\dfrac{3h}{mv}
  2. B\lambda=\dfrac{h}{mv}Answer
  3. C\lambda=\dfrac{2h}{mv}
  4. D\lambda=\dfrac{h}{2mv}
Why

de Broglie's relation is \lambda=h/p with momentum p=mv, so \lambda=h/mv. The other options put wrong numerical factors in front.

2026 Annual I, Set B, MCQ 171 mark
Which color of visible light has the lowest frequency?
  1. ARedAnswer
  2. BGreen
  3. CBlue
  4. DViolet
Why

Red light has the longest wavelength in the visible band (about 750\,\mathrm{nm}). Since f=c/\lambda, it has the lowest frequency. Violet has the highest.

2026 Annual I, Set B, Q.2 (vii)1+2 = 3 marks
What is meant by pair production? Also calculate the minimum energy required in this process.
Solution
Pair production (1 mark)

Pair production is the creation of a particle and its antiparticle from energy. When a high-energy gamma-ray photon passes near a heavy nucleus, it is completely absorbed and an electron-positron pair is created. Charge, energy and momentum are conserved; energy is converted into mass.

Minimum energy (2 marks)
  1. Energy conservation: the photon supplies the rest mass energy of both particles and their kinetic energies.
    hf=2m_0c^{2}+K.E_{e^-}+K.E_{e^+}
  2. The energy is minimum when the particles are created with zero kinetic energy.
    hf_{\min}=2m_0c^{2}
  3. Substitute m_0=9.11\times10^{-31}\,\mathrm{kg} and c=3\times10^{8}\,\mathrm{m\,s^{-1}}.
    hf_{\min}=2(9.11\times10^{-31})(3\times10^{8})^{2}=1.64\times10^{-13}\,\mathrm{J}
  4. Change into MeV (1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}).
    hf_{\min}=\dfrac{1.64\times10^{-13}}{1.6\times10^{-13}}\,\mathrm{MeV}=1.02\,\mathrm{MeV}
hf_{\min}=2m_0c^{2}=1.64\times10^{-13}\,\mathrm{J}=1.02\,\mathrm{MeV}
2026 Annual I, Set B, Q.2 (viii) OR3 marks
What is the role of electron wavelength in achieving high resolution of electron microscope?
Solution

The resolution of a microscope (its ability to show two close points separately) is limited by diffraction: it cannot show details much smaller than the wavelength of the radiation used. So the shorter the wavelength, the higher the resolution.

An electron microscope uses fast electrons, which behave as waves with de Broglie wavelength

\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}

where V is the accelerating potential difference. For V=10\,\mathrm{kV}, \lambda\approx1.2\times10^{-11}\,\mathrm{m} (0.12\,\text{\AA}), whereas visible light has \lambda\approx4000 to 7000\,\text{\AA}. The electron wavelength is thousands of times shorter, so the electron microscope resolves details down to about 0.1 to 0.2\,\mathrm{nm}, compared with about 200\,\mathrm{nm} for an optical microscope. Increasing V makes \lambda even shorter and the resolution even higher.

2026 Annual I, Set B, Q.5 OR1+2+3 = 6 marks
What is meant by photo electric effect? Why no emission of electrons occurs below threshold frequency? Discuss the effects of intensity of light on photoelectric current with graph.
Solution
Photoelectric effect (1 mark)

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.

No emission below threshold frequency (2 marks)

According to Einstein, light consists of photons of energy hf, and one photon is absorbed by one electron. The electron needs at least the work function \Phi to escape:

hf=\Phi+K.E_{\max},\qquad \Phi=hf_0

If f<f_0, the energy of each photon is less than the work function (hf<\Phi), so no single photon can free an electron. An electron cannot store the energy of one photon and add the energy of another, so increasing the intensity (sending more photons) still gives no emission. Hence no electrons are emitted below the threshold frequency, however bright the light.

Effect of intensity on photoelectric current (3 marks)

Keeping the frequency (above f_0) and the potential difference fixed, the photoelectric current increases linearly with the intensity. Intensity is the number of photons arriving per second; each photon releases one electron, so the number of photoelectrons per second, and hence the current, is proportional to intensity.

Photoelectric current against intensity
S is the intensity of the light. Frequency and voltage fixed. Doubling the intensity doubles the current, so the number of photoelectrons per second is proportional to the intensity.

When the current is plotted against the collector potential for different intensities, brighter light gives a larger saturation current, but the current falls to zero at the same stopping potential V_0 for all intensities. So intensity increases the number of electrons but not their maximum kinetic energy, K.E_{\max}=eV_0.

Photoelectric current against collector potential
Light of one frequency at three intensities. Brighter light gives a larger saturation current, but every curve reaches zero at the same stopping potential -V_0.
2026 Annual I, Set D, MCQ 21 mark
Which of the following electro-magnetic waves has the highest frequency?
  1. AGamma raysAnswer
  2. BUltraviolet
  3. CMicrowaves
  4. DInfrared
Why

Gamma rays have the shortest wavelength (about 10^{-12}\,\mathrm{m}) in the electromagnetic spectrum, so from f=c/\lambda they have the highest frequency (and photon energy). The order of increasing frequency is microwaves, infrared, ultraviolet, gamma rays.

2026 Annual I, Set D, MCQ 131 mark
De-Broglie wavelength of a body having mass 0.5\,\mathrm{kg} moving with a speed of 36\,\mathrm{km/hr} is:
  1. A1.3\times10^{-34}\,\mathrm{mm}
  2. B1.3\times10^{-24}\,\mathrm{mm}
  3. C1.3\times10^{-24}\,\mathrm{m}
  4. D1.3\times10^{-34}\,\mathrm{m}Answer
Why
\begin{aligned}v&=36\,\mathrm{km\,h^{-1}}=\dfrac{36\,000\,\mathrm{m}}{3600\,\mathrm{s}}=10\,\mathrm{m\,s^{-1}}\\ \lambda&=\dfrac{h}{mv}=\dfrac{6.63\times10^{-34}}{0.5\times10}=1.3\times10^{-34}\,\mathrm{m}\end{aligned}

This is the book's own example in section 21.6. Watch the unit: the answer is in metres, not millimetres.

2026 Annual I, Set D, MCQ 141 mark
Number of electrons emitted in photoelectric effect depends upon:
  1. AEnergy of incident light
  2. BWavelength of incident light
  3. CIntensity of Incident lightAnswer
  4. DFrequency of incident light
Why

Each photon releases one electron, so the number of electrons emitted per second is proportional to the number of photons per second, that is, to the intensity. Frequency, wavelength and photon energy decide the kinetic energy of the electrons (and whether emission happens at all), not their number.

2026 Annual I, Set D, Q.2 (vii)1+2 = 3 marks
Briefly describe pair annihilation. Also justify the conservation of energy and momentum in this process.
Solution
Pair annihilation (1 mark)

Pair annihilation is the process in which an electron and a positron (its antiparticle) combine and disappear, producing two gamma-ray photons. Mass is converted into energy according to E=mc^{2}. It is the reverse of pair production.

Conservation of energy and momentum (2 marks)

Energy: the total energy of the pair (rest mass energy plus kinetic energy) appears as the energy of the photons:

2m_0c^{2}+K.E_{e^-}+K.E_{e^+}=\text{energy of two photons}

For a pair at rest each photon has hf=m_0c^{2}=0.511\,\mathrm{MeV}, so the total energy before and after is 1.02\,\mathrm{MeV}.

Momentum: the electron and positron are nearly at rest, so their total momentum is about zero. A single photon always carries momentum hf/c, so one photon cannot be produced. Two photons of equal energy moving in opposite directions have equal and opposite momenta (+hf/c and -hf/c), giving zero total momentum. Hence momentum is conserved. Charge is also conserved: (+e)+(-e)=0.

2026 Annual I, Set D, Q.2 (viii) OR1.5+1.5 = 3 marks
Why does electron microscope have higher resolution than optical microscope? (Compare both.)
Solution
Reason (1.5 marks)

The resolution of any microscope is limited by the wavelength of the radiation it uses: details smaller than about one wavelength cannot be seen. An electron microscope uses fast electrons whose de Broglie wavelength \lambda=h/\sqrt{2meV} is extremely short (about 0.01\,\mathrm{nm} at 10\,\mathrm{kV}), thousands of times shorter than visible light (400 to 700\,\mathrm{nm}). So it can resolve much finer detail.

Comparison (1.5 marks)
Optical microscopeElectron microscope
IlluminationVisible lightBeam of fast electrons
Wavelength400 to 700\,\mathrm{nm}about 0.01\,\mathrm{nm}
LensesGlass lensesMagnetic lenses (coils)
Resolutionabout 200\,\mathrm{nm}about 0.1 to 0.2\,\mathrm{nm}
Medium and specimenAir; living specimens possibleVacuum; thin, non-living specimen
2026 Annual I, Set D, Q.5 OR1+3+2 = 6 marks
What is photoelectric effect? Explain work function of metal and Einstein photo electric equation. What is the effect of frequency of radiation on K.E of photoelectrons?
Solution
Photoelectric effect (1 mark)

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.

Work function and Einstein's equation (3 marks)

The work function \Phi of a metal is the minimum energy needed to remove an electron from its surface. It is different for different metals (for example about 2.3\,\mathrm{eV} for sodium and 4.3\,\mathrm{eV} for zinc). It is related to the threshold frequency f_0 by \Phi=hf_0.

Derivation Einstein's photoelectric equation

Aim. Relate photon energy, work function and maximum kinetic energy.

  1. Light is a stream of photons; one photon of frequency f gives its whole energy hf to one electron.
    E=hf
  2. Part of the energy is used as the work function to free the electron; the rest is its kinetic energy (maximum for surface electrons).
    hf=\Phi+K.E_{\max}
  3. At threshold, K.E_{\max}=0 and f=f_0, so \Phi=hf_0. Substituting:
    K.E_{\max}=\tfrac12mv_{\max}^{2}=hf-hf_0
Resulthf=\Phi+\tfrac12mv_{\max}^{2}

Emission occurs only when hf\ge\Phi; the extra energy appears as kinetic energy.

Effect of frequency on K.E of photoelectrons (2 marks)

From K.E_{\max}=hf-\Phi: below the threshold frequency f_0 no electrons are emitted. Above f_0, the maximum kinetic energy increases linearly with the frequency; the graph is a straight line of slope h, cutting the frequency axis at f_0. Increasing the frequency (for example from red to violet to ultraviolet light) gives faster photoelectrons, whatever the intensity.

Maximum kinetic energy against frequency
No electrons are emitted below the threshold frequency f_0. Above f_0 the graph is a straight line of slope h; extended back (dashed) it cuts the energy axis at -\Phi. The line is drawn to scale for sodium (\Phi=2.3\,\mathrm{eV}, f_0=5.6\times10^{14}\,\mathrm{Hz}).
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (i), old syllabus1+2 = 3 marks
What is electron volt (eV)? Derive its relation with SI unit of energy.
Solution
Definition (1 mark)

One electronvolt is the energy gained (or lost) by an electron when it moves through a potential difference of one volt.

Relation with the joule (2 marks)
  1. Work done (energy gained) when a charge q moves through a p.d. V.
    W=qV
  2. For an electron, q=e=1.6\times10^{-19}\,\mathrm{C}, and V=1\,\mathrm{V}.
    1\,\mathrm{eV}=(1.6\times10^{-19}\,\mathrm{C})(1\,\mathrm{V})
  3. Since 1\,\mathrm{C}\times1\,\mathrm{V}=1\,\mathrm{J}:
    1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}

Also 1\,\mathrm{keV}=1.6\times10^{-16}\,\mathrm{J} and 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}. The electronvolt is a convenient small unit for atomic and nuclear energies.

1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (i) OR, old syllabus3 marks
How can we calculate kinetic energy of photoelectrons?
Solution

The maximum kinetic energy of photoelectrons is found by measuring the stopping potential V_0. In the photoelectric set-up the collector is made negative with respect to the emitter (battery reversed), and the retarding potential is increased until the photoelectric current just becomes zero. At this potential even the fastest electrons are stopped, so their kinetic energy has all been used in doing work eV_0 against the field:

K.E_{\max}=\tfrac12mv_{\max}^{2}=eV_0
Photoelectric current against collector potential
Light of one frequency at three intensities. Brighter light gives a larger saturation current, but every curve reaches zero at the same stopping potential -V_0.

For example, if V_0=2\,\mathrm{V} then K.E_{\max}=2\,\mathrm{eV}=3.2\times10^{-19}\,\mathrm{J}. If the frequency and the work function are known instead, the same energy follows from Einstein's equation K.E_{\max}=hf-\Phi.

2026 Annual I, Old syllabus (Curriculum 2006), Q.4, old syllabus2+2+3 = 7 marks
What is photoelectric effect? Why classical Physics fails to explain photoelectric effect? Derive Einstein photoelectric equation.
Solution
Photoelectric effect (2 marks)

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons and the current they form is the photoelectric current.

Light falling on the emitter E ejects photoelectrons, which are collected by C and recorded by the ammeter.
Failure of classical physics (2 marks)

Classical physics treats light as a continuous wave whose energy depends on intensity. It fails to explain:

  • Threshold frequency: no emission below f_0 however intense the light (wave theory predicts emission at any frequency if the light is bright enough).
  • Kinetic energy independent of intensity: K.E_{\max} does not increase with intensity.
  • Kinetic energy depends on frequency: K.E_{\max} increases linearly with f.
  • No time lag: emission is instantaneous even in very dim light.
Einstein's photoelectric equation (3 marks)

Derivation Einstein's photoelectric equation

Aim. Derive the relation between photon energy, work function and maximum kinetic energy.

  1. Light consists of photons, each of energy hf. One photon is absorbed completely by one electron.
    E=hf
  2. The electron uses energy \Phi (work function) to escape from the surface; by conservation of energy the remainder is its maximum kinetic energy.
    hf=\Phi+K.E_{\max}
  3. At the threshold frequency the kinetic energy is zero, so the work function is hf_0.
    \Phi=hf_0
  4. Substitute and write the kinetic energy in terms of speed and stopping potential.
    \tfrac12mv_{\max}^{2}=hf-hf_0=eV_0
Resulthf=\Phi+\tfrac12mv_{\max}^{2}

This equation explains every observation: no emission if hf<\Phi; K.E_{\max} rises linearly with f (slope h) and is independent of intensity; the number of electrons depends on the number of photons.

Exam focus

  1. Photoelectric effect: definition, observations with graphs, failure of classical physics and derivation of Einstein's equation hf=\Phi+K.E_{\max}. (long)
    The most examined topic of the unit: a Section C question in all three 2026 papers (Set B Q.5 OR, Set D Q.5 OR, old syllabus Q.4).
    Go to answer
  2. Draw and explain the graphs: current against collector potential for different intensities, current against intensity, and K.E_{\max} against frequency. (long)
    Graphs earn marks in Section C (Set B Q.5 OR asked for the effect of intensity with a graph; Set D asked for the effect of frequency).
    Go to answer
  3. Stopping potential and K.E_{\max}=eV_0; how to find \Phi and h from a V_0 against f graph. (short)
    Old-syllabus 2026 Q.2 (i) OR asked how kinetic energy of photoelectrons is calculated; book short question 5 asks about the graph.
    Go to answer
  4. Photoelectric numericals: photon energy in eV, K.E_{\max}, stopping potential and wavelength. (numerical)
    Direct use of hf=\Phi+eV_0; book numericals 1 and 2 (note the misprinted answer of numerical 1).
    Go to answer
  5. Pair production: definition, conservation laws and minimum energy 2m_0c^{2}=1.02\,\mathrm{MeV}. (short)
    Asked in 2026 Set B Q.2 (vii) (1 + 2 marks); also a comprehensive question in the book.
    Go to answer
  6. Pair annihilation: why two photons, conservation of energy and momentum, 0.511\,\mathrm{MeV} per photon. (short)
    Asked in 2026 Set D Q.2 (vii) (1 + 2 marks).
    Go to answer
  7. Electron microscope: why electron wavelength gives very high resolution; comparison with an optical microscope. (short)
    Asked in both 2026 new-syllabus sets (Set B and Set D Q.2 (viii) OR).
    Go to answer
  8. de Broglie wavelength \lambda=h/mv and its numericals (convert km/h and mg to SI units). (mcq)
    Two 2026 MCQs (Set B MCQ 12 and Set D MCQ 13) and book examples 21.4 and assignment 21.4.
    Go to answer
  9. Compton effect: explanation with photon collisions, Compton shift \Delta\lambda=\dfrac{h}{m_0c}(1-\cos\theta) and its significance. (long)
    Book comprehensive question 2 and several short questions (8, 9, 10) and numericals 4 and 6 come from this section.
    Go to answer
  10. Heisenberg's uncertainty principle: statement, \Delta x\,\Delta p\ge\hbar, \Delta E\,\Delta t\ge\hbar and simple numericals. (long)
    Book comprehensive question 3, MCQ 8 and numerical 5; state which form of the principle you use.
    Go to answer
  11. Electronvolt: definition and derivation of 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}; converting MeV to joules. (short)
    Old-syllabus 2026 Q.2 (i) (1 + 2 marks) and book short question 2.
    Go to answer
  12. Discrete energy levels and the formation of emission and absorption line spectra, with hf=E_n-E_p. (long)
    Book comprehensive question 9 and two learning outcomes; energy level diagrams earn marks.
    Go to answer

Mark this unit as done to fill the Class 12 Physics ring.

Next, unit 22
Nuclear physics

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