Medical imaging

Class 12 Physics, unit 25, book pages 201–210

Medical imaging lets doctors see inside the body without surgery. This unit explains how a piezoelectric transducer makes and detects ultrasound and how echoes are turned into images, how an X-ray tube produces characteristic and continuous X-rays with a minimum wavelength $λ_{m i n} = h c / e V$ , how X-ray photographs get their contrast, and how a CT scanner combines many X-ray views into 2-D slices and a 3-D image.

Learning outcomes (5)

Explain the piezoelectric effect and its application in medical science (ultrasound waves are generated and detected by a piezoelectric transducer).
Explain how ultrasound can be used to obtain diagnostic information about internal body structures.
Explain that X-rays are produced by electron bombardment of a metal target, and calculate the minimum wavelength of X-rays produced from the accelerating p.d.
Explain the use of X-rays in imaging internal body structures, including describing the term contrast in X-ray imaging.
Explain how computed tomography (CT) scanning works: it produces a 3-D image of an internal structure by first combining multiple X-ray images taken in the same section from different angles to obtain a 2-D image of the section, then repeating this process along an axis and combining 2-D images of multiple sections.

What is medical imaging

Doctors often need to see inside the body without cutting it open. Medical imaging is a range of tests used to create images of internal body parts. It reveals structures hidden by the skin and bones, helps to diagnose and treat disease, and builds a record of normal anatomy and physiology so that abnormal features can be recognised.

Every imaging method sends some form of energy into the body (or detects energy coming out of it) and records how the body changes it. Each method uses different physics:

Imaging methods named in this unit. The book explains ultrasound, X-rays and CT in detail; MRI and PET are only named.
Method	What it uses	Uses ionising radiation	Best for
X-ray imaging (radiography)	X-rays passing through the body	Yes	Bones, fractures, teeth, chest, foreign objects
CT scan	Many X-ray views combined by a computer	Yes (larger dose)	Detailed 3-D images of soft tissue, brain, lungs, tumors
Ultrasound (sonography)	Echoes of high-frequency sound	No	Unborn babies, heart, abdomen, blood flow
MRI	A strong magnet and radio waves	No	Brain, spinal cord, joints and other soft tissue
PET scan	Gamma rays from a radioactive tracer	Yes	Activity of organs, spread of cancer

25.1Piezoelectric effectOften examined

Some crystals, such as quartz, behave in a surprising way. Squeeze them and a voltage appears across their faces. Apply a voltage to them and they change shape. This two-way link between mechanical stress and electricity is used to make and to detect ultrasound.

Piezoelectric effect: The ability of certain materials to generate an electric charge in response to applied mechanical stress.
Inverse piezoelectric effect: The generation of mechanical stress (a change of shape) in a piezoelectric material when an electric field is applied to it.
Piezoelectric materials: Crystals and ceramics that show these effects, for example quartz, lead zirconate titanate (PZT) and lithium niobate ( $LiNb O_{3}$ ), which is used in optical and acoustic devices.
Transducer: A device that changes energy from one form to another. A piezoelectric transducer changes electrical energy into sound energy, and sound energy back into electrical energy.

How the effect happens

A piezoelectric crystal is built from positive and negative ions. In an unstressed crystal the centre of positive charge and the centre of negative charge of each small unit are at the same point, so the crystal is neutral everywhere. When the crystal is compressed or stretched, the ions shift slightly and the two centres separate. Each unit becomes a tiny electric dipole, positive charge collects on one face and negative charge on the opposite face, and an electric field (a p.d.) appears across the crystal, Fig. 25.1 (a).

The effect is reversible. If a p.d. is applied across the crystal, the electric field pushes the positive and negative ions in opposite directions, so the crystal becomes slightly thicker or thinner, Fig. 25.1 (b). This is the inverse piezoelectric effect. An alternating p.d. makes the crystal expand and contract again and again, so it vibrates at the frequency of the p.d. and sends out sound waves.

After Fig. 25.1. (a) Squeezing a piezoelectric crystal (PZT) between metal plates produces a p.d. (b) An alternating p.d. makes the crystal vibrate and send out ultrasound.

Direct and inverse piezoelectric effects compared.
Piezoelectric effect	Inverse piezoelectric effect
Mechanical stress on the crystal produces electric charge (a p.d.) across it.	An electric field (p.d.) applied to the crystal produces mechanical stress, so the crystal changes shape.
Mechanical energy is changed into electrical energy.	Electrical energy is changed into mechanical energy.
Used to detect sound: returning echoes squeeze the crystal and give an electrical signal.	Used to generate sound: an alternating p.d. makes the crystal vibrate.
Examples: ultrasound receivers, microphones, pressure sensors, spark igniters of gas lighters.	Examples: ultrasound transmitters, buzzers, the vibrating quartz crystal of a watch.

Applications of the piezoelectric effect

Production and detection of sound, especially ultrasound in medical scanners, and in sonar and ultrasonic cleaners.
Generation of high voltages: a sharp blow on a piezoelectric crystal gives several kilovolts, enough for the spark of a gas lighter or a stove igniter.
Electronic frequency generation: a quartz crystal vibrates at a very steady natural frequency, so it keeps time in watches, clocks and computers.
Microbalances: the vibration frequency of a crystal changes when a tiny mass settles on it, so extremely small masses can be measured.
Scanning probe microscopes: piezoelectric supports move the probe by fractions of a nanometre, which gives images with atomic resolution.

25.1Generation and detection of ultrasoundOften examined

A normal human ear can hear sound of frequency from about $20 Hz$ to $20000 Hz$ . Sound waves with frequency above $20000 Hz$ ( $20 kHz$ ) are inaudible and are called ultrasound (ultrasonic waves). Medical scanners use frequencies of a few megahertz, typically $1$ to $15 MHz$ .

Ultrasound is a longitudinal mechanical wave, like ordinary sound. It needs a medium and travels at about $1540 m s^{- 1}$ in soft tissue. Its high frequency gives it a very short wavelength, and a short wavelength can show fine detail.

λ = \frac{v}{f} = \frac{1540 m s ^{- 1}}{5 \times 1 0 ^{6} Hz} \approx 3 \times 1 0^{- 4} m = 0.3 mm

$v$ : speed of sound in tissue, $f$ : frequency, $λ$ : wavelength. Structures of about this size can be seen.

Generation: inverse piezoelectric effect

The probe of the scanner contains a piezoelectric transducer: a thin PZT crystal with electrodes on its faces. An alternating p.d. of ultrasonic frequency is applied to it. By the inverse piezoelectric effect the crystal expands and contracts at the same frequency, so it pushes on the skin and sends ultrasound into the body. These vibrations are far too rapid for our ears to hear. The p.d. is applied in short bursts, so the probe sends short pulses. The crystal is made so that its natural frequency equals the driving frequency; at resonance it vibrates with the largest amplitude.

Detection: direct piezoelectric effect

When a pulse meets a boundary between two different tissues, part of it is reflected or scattered back as an echo. Between pulses the same crystal acts as a receiver. The returning echo makes the crystal vibrate, and by the direct piezoelectric effect this produces a small alternating p.d. This electric signal is amplified and processed by a computer.

Pulse-echo method. Part of the pulse is reflected at the front boundary of the organ (echo 1) and part at its back boundary (echo 2). The time taken by each echo gives the depth of that boundary. Gel removes the air between the probe and the skin.

How echoes give diagnostic information

Echo time gives depth. The pulse goes to the boundary and back, so it covers twice the depth: $2 d = v t$ , which gives $d = \frac{v t}{2}$ .
Echo strength shows the type of boundary. A large change of material (for example soft tissue to bone) gives a strong echo, shown as a bright dot; a small change gives a weak, darker dot.
Sweeping the beam builds a picture. The probe has a row of many small crystals. Echoes from many directions are combined by a computer into a 2-D image of a slice of the body. The image is renewed many times a second, so movement (a beating heart, a moving baby) is seen in real time.
The detected signals show the shape, size and consistency (solid, liquid-filled or gas-filled) of an organ, as in the ultrasound image of the pancreas in Fig. 25.1 (c).

Depth of a boundary

An echo from the back of an organ returns to the probe

130 μ s

after the pulse was sent. The speed of ultrasound in soft tissue is

1540 m s^{- 1}

. How deep is this boundary below the skin?

The pulse travels to the boundary and back, a total distance $2 d$ , in time $t$ .
$2 d = v t \Rightarrow d = \frac{v t}{2}$
Substitute, with $t = 130 μ s = 130 \times 1 0^{- 6} s$ .
$d = \frac{( 1540 ) ( 130 \times 1 0 ^{- 6} )}{2} = 0.100 m$

d \approx 0.10 m = 10 cm

25.1Applications of ultrasound in medicineOften examined

Medical technology uses ultrasound for many purposes. Its great advantage over other imaging methods is that sound waves are comparatively harmless: ultrasound is not ionising radiation, so it can be used even for unborn babies.

Diagnosis

Sonography (echography). Images of tissues and organs such as the liver, kidneys, gall bladder, pancreas and uterus are made by the pulse-echo method. It is the standard way to check the growth and position of a baby before birth.
Echocardiography. Real-time images of the heart show its valves and walls moving and how well it pumps.
Blood flow (Doppler ultrasound). Ultrasound reflected from moving blood cells comes back with a slightly changed frequency (the Doppler shift). The shift gives the speed of the blood, so doctors can find blood clots and blocked arteries and check cardiac function in adults and in developing fetuses.

Treatment

Neurosurgery. Very intense ultrasonic waves can be focused on a small region to destroy unwanted tissue without disturbing the surrounding tissues, and without cutting the skin.
Physiotherapy. Special ultrasonic equipment heats deep tissues gently. It is used to treat arthritis, muscular pain, rheumatism and sciatica.

Industrial application: ultrasonic flow meter

Transducers fixed on a pipe send high-frequency sound into the flowing liquid. The waves reflect from particles or bubbles moving with the liquid, so their frequency is shifted (Doppler shift) by an amount that depends on the flow velocity. The meter measures this shift, finds the flow velocity $v$ , and uses the cross-sectional area $A$ of the pipe to give the flow rate:

Q = A v

$Q$ : volume flow rate ( $m^{3} s^{- 1}$ ), $A$ : cross-sectional area of the pipe ( $m^{2}$ ), $v$ : flow velocity ( $m s^{- 1}$ ). The same technique monitors blood flow through arteries.

Advantages of ultrasound imaging	Limitations
No ionising radiation; safe for pregnant women and babies	Cannot pass through bone or air (lungs, gas in the bowel)
Real-time moving images	Less detail than CT or MRI
Cheap, portable and quick	Image quality depends on the skill of the operator

25.2X-rays and characteristic X-raysOften examined

X-rays are electromagnetic waves of very short wavelength, roughly $0.01 nm$ to $10 nm$ , so their photons carry much more energy than photons of light. X-rays come from heavy atoms. They are produced when fast electrons strike a metal target, by two different processes:

Transitions of inner-shell electrons between the shells of heavy atoms give characteristic X-rays (sharp lines).
Coulomb interaction of a fast electron with the orbital electrons and the positive nucleus gives continuous X-rays (a smooth spread of wavelengths).

Inner-shell transitions

The inner-shell electrons of heavy atoms are tightly bound, so a large amount of energy is needed to move them from their normal energy levels. When a heavy target is bombarded with a beam of electrons accelerated through several kilovolts, some of these electrons collide with inner-shell electrons of the target atoms and knock them out of their atoms, Fig. 25.2 (a).

Let a K-shell electron be knocked out, leaving a vacancy in the K shell. An electron from the L, M or N shell quickly jumps down to fill it. The electron moves to a lower energy level, and the excess energy is emitted as one X-ray photon:

h f = E_{upper} - E_{K} \Rightarrow λ = \frac{h c}{E _{upper} - E _{K}}

$h$ : Planck's constant, $f$ : frequency of the photon, $E_{upper}$ : energy of the shell the electron falls from, $E_{K}$ : energy of the K shell, $c$ : speed of light.

The energy levels have fixed values that are different for every element, so these X-rays consist of a series of specific wavelengths (or frequencies), like a fingerprint of the target element. That is why they are called characteristic X-rays.

$K_{α}$ X-rays: X-ray photons emitted when an electron falls from the L shell to a vacancy in the K shell.
$K_{β}$ X-rays: X-ray photons emitted when an electron falls from the M shell to a vacancy in the K shell.
$K_{γ}$ X-rays: X-ray photons emitted when an electron falls from the N shell to a vacancy in the K shell.

Energy-level picture of characteristic X-rays (compare Fig. 25.2 (a)). A vacancy in the K shell is filled from the L, M or N shell, giving the

K_{α}

K_{β}

and

K_{γ}

lines.

Only heavy atoms give X-rays. In a heavy atom the nucleus has a large charge, so the K and L shells are very tightly bound and the gaps between them are thousands of electronvolts. Photons with energies of keV are X-rays. For the outer electrons of any atom the gaps are only a few electronvolts, which gives visible or ultraviolet light.

Energy of a characteristic X-ray photon

In Fig. 25.3 the

K_{α}

line of molybdenum is at

λ = 0.071 nm

. Find the energy of this photon in joules and in keV.

Use the photon energy equation.
$E = \frac{h c}{λ} = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{0.071 \times 1 0 ^{- 9}} = 2.80 \times 1 0^{- 15} J$
Convert to electronvolts by dividing by $1.6 \times 1 0^{- 19} J$ per eV.
$E = \frac{2.80 \times 1 0 ^{- 15}}{1.6 \times 1 0 ^{- 19}} eV = 1.75 \times 1 0^{4} eV = 17.5 keV$
This is the L to K energy gap of molybdenum, thousands of times larger than the gap for a visible-light transition (about $2 eV$ ).

E \approx 2.8 \times 1 0^{- 15} J \approx 17.5 keV

25.2Continuous X-rays (braking radiation) and minimum wavelengthOften examined

When a high-energy electron travels towards a target nucleus in the X-ray tube, it has a Coulomb interaction with the orbital electrons and with the positive nucleus. The interaction with the nucleus is very strong because of its concentrated positive charge. The attraction pulls the electron towards the nucleus and bends its path; as the electron swings past and moves away, it is decelerated, Fig. 25.2 (b).

A decelerated charge radiates electromagnetic waves. The radiation given out by the decelerated electron is called Bremsstrahlung, a German word meaning braking radiation. This radiation forms the continuous X-rays. According to quantum theory it comes out as photons. Each photon carries energy, so the electron must lose kinetic energy in its encounter with the nucleus:

h f = K.E._{before} - K.E._{after}

An electron can lose any amount of energy in one encounter, from almost nothing up to all of its kinetic energy, and most electrons lose their energy in several encounters. So the photons have every possible energy up to a maximum, and the X-rays have a continuous spectrum of wavelengths.

Bremsstrahlung (braking radiation): Electromagnetic radiation (X-rays) emitted when a fast electron is decelerated by the Coulomb attraction of a nucleus in the target.
Continuous X-rays: X-rays with a continuous range of wavelengths, produced by the Coulomb interaction of fast electrons with the orbital electrons and nuclei of the target.
Minimum (cut-off) wavelength: The shortest wavelength in the continuous spectrum, $λ_{m i n} = h c / e V$ , produced when an electron gives all its kinetic energy to one photon.

Derivation Minimum wavelength of X-rays

Aim. Show that the shortest wavelength of the X-rays from a tube with accelerating p.d. $V$ is $λ_{m i n} = h c / e V$ .

Assumptions.

Each electron leaves the hot cathode with negligible speed, and the whole p.d. $V$ accelerates it.
In the extreme case the electron is stopped in a single collision and all its kinetic energy becomes one photon.

After Fig. 25.2 (b). The attraction of the nucleus deflects and decelerates the fast electron; the kinetic energy it loses is emitted as an X-ray photon of energy

h f

The electric field does work on the electron as it moves through the p.d. $V$ . Work done equals charge times p.d., and it becomes the electron's kinetic energy when it reaches the target.
$K.E. = e V$
In one encounter with a nucleus the electron loses some kinetic energy, which appears as one X-ray photon of energy $h f$ . A photon cannot carry more energy than the electron has.
$h f \leq e V$
The photon of largest energy, and so of highest frequency $f_{m a x}$ , is produced when the electron is stopped in a single collision and gives all its kinetic energy to one photon.
$h f_{m a x} = e V$
For electromagnetic waves $c = f λ$ , so $f = c / λ$ . The highest frequency corresponds to the shortest wavelength.
$\frac{h c}{λ _{m i n}} = e V$
Rearrange for $λ_{m i n}$ . This is equation (25.1) of the book.
$λ_{m i n} = \frac{h c}{e V}$
Put in $h = 6.63 \times 1 0^{- 34} J s$ , $c = 3 \times 1 0^{8} m s^{- 1}$ and $e = 1.6 \times 1 0^{- 19} C$ to get a quick form for numericals.
$λ_{m i n} = \frac{1.24 \times 1 0 ^{- 6} V m}{V}$

Result

λ_{m i n} = \frac{h c}{e V}

The shortest wavelength depends only on the accelerating p.d., not on the target metal. A higher p.d. gives a shorter $λ_{m i n}$ (more energetic, more penetrating X-rays); doubling $V$ halves $λ_{m i n}$ . No photon from the tube can have a shorter wavelength, so the spectrum ends sharply at $λ_{m i n}$ .

In practice all the radiation produced does not have the wavelength given by equation (25.1), because many of the electrons are not stopped in a single collision. They lose their energy bit by bit and give many photons of lower energy (longer wavelength). This results in the continuous spectrum. Fig. 25.3 shows the X-rays from a molybdenum target at $35 kV$ : a smooth continuous curve that starts at $λ_{m i n}$ , with two sharp characteristic peaks $K_{β}$ and $K_{α}$ on top of it.

Molybdenum X-ray spectrum at 35 kV (λ in nm)

0.02

0.04

0.06

0.08

0.1

0.12

λ

I

K_{β}

K_{α}

λ_{m i n}

continuous X-rays

After Fig. 25.3;

I

is the relative intensity. The continuous curve starts at

λ_{m i n} = h c / e V = 0.0355 nm

. The

K_{β}

(

0.063 nm

) and

K_{α}

(

0.071 nm

) lines are characteristic of molybdenum.

Effect of tube p.d. on the spectrum (λ in nm)

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.2

0.4

0.6

λ

I

0.036

0.050

V = 35 kV

V = 25 kV

I

is the relative intensity. Raising the p.d. from

25 kV

35 kV

moves

λ_{m i n}

from

0.050 nm

0.036 nm

and raises the intensity at every wavelength. Characteristic lines (not drawn) would stay at the same wavelengths, because they depend only on the target metal.

Characteristic and continuous X-rays compared.
Characteristic X-rays	Continuous X-rays
Produced when an electron from an outer shell (L, M or N) falls into a vacancy in an inner shell (K) of a target atom.	Produced when a fast electron is decelerated (braked) by the Coulomb attraction of a target nucleus (Bremsstrahlung).
Only certain fixed wavelengths: sharp lines such as $K_{α}$ , $K_{β}$ , $K_{γ}$ .	All wavelengths from $λ_{m i n}$ upwards: a smooth curve.
The wavelengths depend on the target material (its energy levels), not on the tube voltage.	The minimum wavelength depends on the accelerating p.d., $λ_{m i n} = h c / e V$ , not on the target.
Used to identify elements and to study atomic structure.	Forms most of the X-ray beam used in imaging.

Example 25.1

When electrons are accelerated through a potential difference of

1 0^{5} V

in an X-ray tube, calculate the minimum wavelength produced.

Given

$V = 1 0^{5} V$ accelerating p.d.
$h = 6.63 \times 1 0^{- 34} J s$
$c = 3 \times 1 0^{8} m s^{- 1}$
$e = 1.60 \times 1 0^{- 19} C$

To find

$λ_{m i n}$

Use equation (25.1).
$λ_{m i n} = \frac{h c}{e V}$
Substitute the values.
$λ_{m i n} = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{( 1.60 \times 1 0 ^{- 19} ) ( 1 0 ^{5} )} = 1.24 \times 1 0^{- 11} m$

λ_{m i n} = 1.24 \times 1 0^{- 11} m = 0.0124 nm

Assignment 25.1

What will be the minimum wavelength of X-rays produced, if the electrons were slowed down by the target? The potential difference across the X-ray tube is

4000 V

Given

$V = 4000 V$

To find

$λ_{m i n}$

The shortest wavelength comes from an electron that gives all its energy $e V$ to one photon.
$λ_{m i n} = \frac{h c}{e V}$
Substitute the values.
$λ_{m i n} = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{( 1.6 \times 1 0 ^{- 19} ) ( 4000 )} = 3.11 \times 1 0^{- 10} m$
Check with the quick form: $1.24 \times 1 0^{- 6} /4000 = 3.1 \times 1 0^{- 10} m$ . It agrees.

λ_{m i n} = 3.11 \times 1 0^{- 10} m = 0.311 nm

25.2Setup for production of X-rays (X-ray tube)Often examined

X-rays are produced by bombarding a metal target with fast electrons. The arrangement used, the X-ray tube, is shown in Fig. 25.4.

After Fig. 25.4. Electrons from the heated filament are accelerated by a high p.d. and strike the tungsten target; X-rays leave through a thin window. Heat from the target is conducted away along the copper rod, which is often cooled by water flowing behind the anode.

Construction

Cathode: a filament heated by a current from a battery (a low-voltage supply). The hot filament emits electrons (thermionic emission). A metal cup around it focuses the electrons towards the target.
Anode: a solid copper bar. A piece of a metal with a high melting point, such as tungsten or platinum, is embedded at the end of the copper rod and serves as the target. Its face is inclined so that the X-rays come out sideways towards the window.
Evacuated glass chamber: the cathode and anode are enclosed in a glass tube from which the air has been removed, so the electrons do not collide with air molecules.
High p.d.: a d.c. voltage of the order of 50\,000\,\mathrm{V} (50\,\mathrm{kV}) is maintained between the cathode (negative) and the anode (positive).
Cooling: the heat produced in the target is conducted away through the copper rod; sometimes water flows behind the anode to cool it.

Working

The electrons emitted from the cathode are accelerated by the high p.d. and gain kinetic energy eV. When these highly energetic electrons are suddenly stopped by the target, an intense beam of X-rays (continuous and characteristic) is produced and leaves through the window.

Only a small part of the kinetic energy of the electrons (about 1\%) is converted into X-rays; the rest (about 99\%) is converted into heat. The target becomes very hot, so it must have a high melting point and must be cooled.

Choice of target

The target should have (i) a high atomic number, because heavy atoms convert the energy of the electrons into X-rays more efficiently and their tightly bound inner shells give high-energy characteristic X-rays, and (ii) a high melting point, because it becomes very hot. Tungsten (Z=74, melting point about 3400^\circ\mathrm{C}) is the usual choice.

Controlling the X-ray beam

Quantity	Controlled by	Reason
Intensity (number of X-ray photons per second)	Filament current	A hotter filament emits more electrons per second, so more X-ray photons are produced.
Penetrating power (quality)	Accelerating p.d.	A higher p.d. gives the electrons more energy, so the photons have higher energy and shorter wavelength. These hard X-rays penetrate more; soft X-rays from a lower p.d. are absorbed more easily.

25.2Nature and properties of X-rays

X-rays are a form of high-energy electromagnetic radiation that passes through most objects. Their properties follow from their short wavelength and the large energy of each photon:

They are electromagnetic waves, so they travel through vacuum at the speed of light, c=3\times10^{8}\,\mathrm{m\,s^{-1}}, whatever their energy.
They carry no charge, so electric and magnetic fields do not deflect them.
They penetrate matter. Materials of low density and low atomic number (flesh) let them pass easily; dense materials of high atomic number (bone, lead) absorb them strongly.
They darken photographic film and make some substances glow (fluorescence), so they can be recorded and detected.
They ionise gases and can damage or kill living cells, so exposure must be kept as low as possible.
They show wave behaviour: crystals diffract them, because the spacing of atoms in a crystal (about 0.1\,\mathrm{nm}) is similar to their wavelength.
They show particle behaviour: X-ray photons eject electrons from metal surfaces (photoelectric effect).

Uses of X-rays: medical diagnosis (X-ray photographs and CT scans), radiotherapy for cancer, checking metal castings and welds for hidden cracks, security scanning of luggage at airports, and X-ray crystallography, which finds the arrangement of atoms in crystals.

25.2Use of X-rays in imaging internal body structuresOften examined

X-ray imaging (X-ray photography, or radiography) is the most common and oldest diagnostic imaging technique. During an X-ray test, X-rays are allowed to pass through the part of the body being examined, and a detector on the other side (a photographic plate or a digital detector) receives the X-rays that get through.

Why an image forms

Different organs absorb X-rays in different quantities. The calcium in bones (high atomic number, high density) absorbs X-rays much better than soft tissue such as fat and muscle, and the air in the lungs absorbs almost nothing. So the X-rays reaching the detector form a pattern of strong and weak regions: a shadow picture of the inside of the body, in shades of grey.

Principle of an X-ray photograph. Rays that pass through soft tissue reach the film and darken it; rays that meet bone are mostly absorbed, so bone casts a light shadow.

Where many X-rays arrive, the film is darkened: soft tissue and air appear dark grey or black.
Where few X-rays arrive (behind bone), the film stays light: bones appear white.
The result is an image in shades of grey, like the X-ray image of a hand in Fig. 25.5. Bone casts its shadow on the photographic plate because bone is more opaque to X-rays than flesh.

What X-ray photographs reveal

Fractures and diseases of bones.
Foreign bodies, such as a swallowed coin or pin, or a bullet.
Teeth and jaws (dental X-rays).
The chest: lung infections and the size of the heart.

25.2Contrast in X-ray imagingOften examined

A conventional X-ray image is sufficient for bones and other dense structures, but soft tissues are difficult to see. Neighbouring soft tissues (for example the wall of the stomach and the liver) have almost the same density and atomic number, so they absorb X-rays almost equally and appear in nearly the same shade of grey.

Contrast: The range of brightness, from lightest to darkest, in an image. In an X-ray image it is the difference in brightness (degree of blackening) between neighbouring structures.

An image with good contrast shows a large difference in brightness between neighbouring structures, so their edges are easy to see. Bone against soft tissue gives high contrast; soft tissue against soft tissue gives low contrast. Contrast in X-ray imaging allows the radiologist to evaluate structures that are not clearly evident on conventional X-ray exams.

(a) Low contrast: two soft tissues give almost the same grey. (b) High contrast: a strongly absorbing structure stands out clearly from its surroundings.

How contrast is improved

Contrast media (contrast agents). A substance that absorbs X-rays strongly (high atomic number) is put into the organ. The patient swallows a barium meal (barium sulfate) to show the stomach and intestines, or an iodine-based dye is injected to show blood vessels (angiography) and kidneys. The filled organ then appears white against the surrounding tissue.
Choice of tube voltage. Soft tissues differ more in how they absorb lower-energy (soft) X-rays, so a lower p.d. is used for soft-tissue images such as breast X-rays (mammography), and a higher p.d. for bones.
CT scanning. The computer can detect very small differences in absorption, so a CT scan shows soft tissues far better than an ordinary X-ray film (section 25.3).

25.3Computed tomography (CT) scanOften examined

A CT scan (computed tomography scan) is a medical imaging method that is useful to get a detailed 3-D image of certain parts of the body, such as soft tissues, blood vessels, the lungs, the brain, the abdomen and bones. A CT machine uses X-rays and a computer to get the image. 'Tomography' means imaging a body in slices.

Parts of a CT scanner

X-ray source: an X-ray tube that sends a thin, fan-shaped beam through the patient.
Detectors: a large number of X-ray detectors on the opposite side. Each detector records how much of the beam gets through along its line, that is, an X-ray image from that direction.
Gantry: the circular part of the machine. The source and detectors are mounted on a large ring inside it.
Couch: the table on which the patient lies. It slowly slides the patient through the gantry.
Computer: combines the huge number of readings into images (Fig. 25.6 shows a CT scanner).

CT scanning. Left: the X-ray tube and the detectors rotate together around the patient and record one slice from many directions. Right: the couch moves the patient along the axis, and the computer stacks the 2-D slices into a 3-D image.

How a CT scan works

The patient lies on the couch, which slides slowly through the gantry.
Beams of X-rays are aimed at the body. The source and the detectors rotate around the patient, so X-ray images of one section (slice) are taken from many different directions.
The computer combines these multiple X-ray images of the same section, taken from different angles, to obtain a 2-D cross-sectional image of that section, called a tomographic image.
The couch and patient are then moved a small step along the axis of the machine, and another set of images is taken for the next section.
Repeating this process along the axis and combining the 2-D images of many sections, the computer gives a merged, detailed 3-D image of the organ under investigation. It can be turned and viewed from any direction.

Fig. 25.7 shows a CT scan of the brain: (a) the dotted line shows the position of one cross-section, whose image is taken from different angles; (b) the 2-D images of different sections, which the computer combines into a 3-D image of the brain.

The idea behind combining views (simplified)

One slice is divided into four small squares with unknown absorptions a, b (top row) and c, d (bottom row). Beams along the rows, the columns and one diagonal measure these totals: a+b=5, c+d=9, a+c=6, b+d=8 and a+d=7. Find the absorption of each square.

Add the two totals that contain a and one other unknown each, then subtract c+d so that c and d cancel.
(a+c)+(a+d)-(c+d)=6+7-9\;\Rightarrow\;2a=4\;\Rightarrow\;a=2
Use the row, column and diagonal totals that contain a.
b=5-a=3,\qquad c=6-a=4,\qquad d=7-a=5
Check with the total not yet used: b+d=3+5=8. It agrees.

A single view gives only totals along lines, like the shadow of an ordinary X-ray. Views from several directions are needed to find the value of each square. A real CT slice has hundreds of thousands of squares (pixels) and needs hundreds of views, which is why a computer is essential.

a=2,\quad b=3,\quad c=4,\quad d=5

Importance of CT scan over normal X-ray film

Normal X-ray film	CT scan
A single shadow picture taken from one direction.	Many views from all directions combined by a computer.
Organs lie on top of each other, so fine detail can be invisible when one organ lies in front of the region being studied.	Each thin slice is shown separately, so overlapping organs are separated.
Poor for soft tissues (low contrast).	Shows small differences between soft tissues clearly.
Gives a 2-D image only.	Gives 2-D slices and a 3-D image.
Small radiation dose; cheap.	Larger radiation dose; expensive.

Because it gives high-quality images, a CT scanner helps in the study of tumors in cancer patients, where images of high quality are essential. It is also used for injuries and bleeding in the brain, lung disease and internal injuries after accidents.

Formula sheet

Ultrasound frequency

f>20\,000\,\mathrm{Hz}

f: frequency of the sound wave; the human ear hears 20\,\mathrm{Hz} to 20\,000\,\mathrm{Hz}

Unit: \mathrm{Hz}

Wavelength of ultrasound

\lambda=\dfrac{v}{f}

v: speed of sound in the medium (about 1540\,\mathrm{m\,s^{-1}} in soft tissue), f: frequency

Unit: \mathrm{m}

Depth from echo time (pulse-echo)

d=\dfrac{vt}{2}

d: depth of the boundary, v: speed of ultrasound, t: time between sending the pulse and receiving its echo

Unit: \mathrm{m}

Flow rate (flow meter)

Q=Av

A: cross-sectional area of the pipe, v: flow velocity found from the Doppler shift

Unit: \mathrm{m^3\,s^{-1}}

Characteristic X-ray photon

hf=E_{\text{upper}}-E_K,\qquad hf_{K_\alpha}=E_L-E_K,\qquad hf_{K_\beta}=E_M-E_K

E_K, E_L, E_M: energies of the K, L, M shells of the target atom, h: Planck's constant

Units: \mathrm{J} (or \mathrm{eV})

Photon energy

E=hf=\dfrac{hc}{\lambda}

h=6.63\times10^{-34}\,\mathrm{J\,s}, c=3\times10^{8}\,\mathrm{m\,s^{-1}}, \lambda: wavelength

Units: \mathrm{J}; 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}

Kinetic energy of an accelerated electron

\mathrm{K.E.}=eV

e=1.6\times10^{-19}\,\mathrm{C}: electron charge, V: accelerating p.d.

Unit: \mathrm{J}

Maximum frequency of X-rays

f_{\max}=\dfrac{eV}{h}

V: accelerating p.d., h: Planck's constant

Unit: \mathrm{Hz}

Minimum wavelength of X-rays

\lambda_{\min}=\dfrac{hc}{eV}

h: Planck's constant, c: speed of light, e: electron charge, V: accelerating p.d.

Unit: \mathrm{m}

Minimum wavelength (quick form)

\lambda_{\min}=\dfrac{1.24\times10^{-6}}{V}\ \mathrm{m}

V in volts; obtained by putting in h, c and e

Unit: \mathrm{m}

Key terms

Medical imaging: A range of tests used to create images of internal body parts.
Piezoelectric effect: The ability of certain materials to generate an electric charge in response to applied mechanical stress.
Inverse piezoelectric effect: The generation of stress (a change of shape) in a piezoelectric material when an electric field is applied to it.
Piezoelectric transducer: A piezoelectric crystal with electrodes that changes electrical energy into ultrasound and ultrasound back into electrical signals.
Ultrasound: Sound waves of frequency above 20\,000\,\mathrm{Hz}, inaudible to the normal human ear.
Sonography (echography): Making images of tissues and organs with ultrasound echoes.
Pulse-echo method: Sending a short ultrasound pulse into the body and timing its echoes to find the depth of boundaries, d=vt/2.
Doppler shift: The change in frequency of a wave reflected from a moving object; used to measure blood flow and liquid flow.
X-rays: Electromagnetic waves of very short wavelength (about 0.01 to 10\,\mathrm{nm}) and high photon energy, produced when fast electrons strike a metal target.
Characteristic X-rays: X-rays of specific wavelengths produced by transitions of electrons between the inner shells of heavy atoms.
K_\alpha, K_\beta, K_\gamma X-rays: Characteristic X-rays emitted when an electron falls into a K-shell vacancy from the L, M and N shell respectively.
Continuous X-rays: X-rays with a continuous range of wavelengths produced by the Coulomb interaction of a fast-moving electron with orbital electrons and the positive nucleus.
Bremsstrahlung (braking radiation): Radiation emitted by a fast electron when it is decelerated by the attraction of a nucleus; it forms the continuous X-rays.
Minimum (cut-off) wavelength: The shortest X-ray wavelength from a tube, \lambda_{\min}=hc/eV, when an electron gives all its kinetic energy to one photon.
X-ray tube: An evacuated tube in which electrons from a heated filament are accelerated by a high p.d. and strike a metal target to produce X-rays.
Target: The metal of high atomic number and high melting point (tungsten or platinum) that the electrons strike in an X-ray tube.
Hard and soft X-rays: Hard X-rays (from a high p.d.) have short wavelength and high penetrating power; soft X-rays (from a low p.d.) are absorbed easily.
Contrast: The range of brightness, from lightest to darkest, in an image; in an X-ray image, the difference in brightness between neighbouring structures.
Contrast medium: A substance of high atomic number (barium sulfate, iodine compounds) put into an organ so that it absorbs X-rays strongly and shows up clearly.
CT scan (computed tomography): A medical imaging method that uses X-rays and a computer to get a detailed 3-D image of certain parts of the body.
Tomographic image: A 2-D cross-sectional image of one section (slice) of the body, obtained by combining X-ray images taken from many angles.
Gantry and couch: The gantry is the circular part of the CT machine that carries the X-ray source and detectors; the couch is the table that slides the patient through it.

Multiple choice questionsp. 209

Frequency of X-rays depends upon

ANumber of electrons striking target
BAccelerating potential
CNature of the target.
DBoth B and CAnswer

Why

The highest frequency of the continuous X-rays is f_{\max}=eV/h, so it depends on the accelerating potential V. The frequencies of the characteristic lines depend on the energy levels of the target atoms, that is, on the nature of the target. The number of electrons changes only the intensity, not the frequency.

Target material used in X-rays tube must have following properties:

AHigh atomic number and high melting point.Answer
BHigh atomic number and low melting point.
CLow atomic number and low melting point.
DHigh atomic number only.

Why

A high atomic number gives efficient production of high-energy X-rays, and a high melting point is needed because about 99\% of the electrons' energy becomes heat in the target. Option D is incomplete: a metal of high atomic number but low melting point (such as lead) would melt.

Which of the following is drawback while conducting a medical imaging test:

AEarly detection of problem.
BAccurate diagnosis.
CContribution to choose of effective treatment.
DExposure to radiation.Answer

Why

Options A, B and C are benefits of medical imaging. X-ray, CT and PET tests expose the patient to ionising radiation, which can damage living cells, so exposure to radiation is the drawback.

In an experiment, high speed electrons emitted from the cathode are accelerated through a potential difference of 20\,\mathrm{kV}. Find the minimum wavelength of the X-rays produced. (take Planck's constant, h=6.6\times10^{-34}\,\mathrm{J\,s})

A0.36782\,\mathring{\mathrm{A}}
B0.61875\,\mathring{\mathrm{A}}Answer
C0.45867\,\mathring{\mathrm{A}}
D0.72854\,\mathring{\mathrm{A}}

Why

\lambda_{\min}=\dfrac{hc}{eV}=\dfrac{(6.6\times10^{-34})(3\times10^{8})}{(1.6\times10^{-19})(2\times10^{4})}=6.1875\times10^{-11}\,\mathrm{m}=0.61875\,\mathring{\mathrm{A}}

Here 1\,\mathring{\mathrm{A}}=10^{-10}\,\mathrm{m}. The options were calculated with h=6.6\times10^{-34}\,\mathrm{J\,s}; with h=6.63\times10^{-34}\,\mathrm{J\,s} the answer is 0.622\,\mathring{\mathrm{A}}, still option B.

In an X-ray tube, electrons accelerated through a very high potential difference strike a metal target. If the potential difference is increased, the speed of the emitted X-rays:

Aincreases.
Bdecreases.
Cis always equal to 3\times10^{8}\,\mathrm{m\,s^{-1}}.Answer
Dincreases or decreases depending on the target material.

Why

X-rays are electromagnetic waves, and all electromagnetic waves travel in vacuum (and almost exactly in air) at c=3\times10^{8}\,\mathrm{m\,s^{-1}}. A higher p.d. increases the energy and frequency of the photons, not their speed. Option A is the tempting mistake: it is the electrons that move faster.

Short questionsp. 210

13 marks

Differentiate between piezoelectric effect and inverse piezoelectric effect.

Solution

Piezoelectric effect: the ability of certain materials (quartz, PZT) to generate an electric charge in response to applied mechanical stress. Inverse piezoelectric effect: the same material generates mechanical stress (changes shape) when an electric field is applied to it.

Direct and inverse piezoelectric effects compared.
Piezoelectric effect	Inverse piezoelectric effect
Mechanical stress on the crystal produces electric charge (a p.d.) across it.	An electric field (p.d.) applied to the crystal produces mechanical stress, so the crystal changes shape.
Mechanical energy is changed into electrical energy.	Electrical energy is changed into mechanical energy.
Used to detect sound: returning echoes squeeze the crystal and give an electrical signal.	Used to generate sound: an alternating p.d. makes the crystal vibrate.
Examples: ultrasound receivers, microphones, pressure sensors, spark igniters of gas lighters.	Examples: ultrasound transmitters, buzzers, the vibrating quartz crystal of a watch.

In an ultrasound probe one crystal uses both: the inverse effect to send out ultrasound pulses and the direct effect to change the returning echoes into electrical signals.

23 marks

What is the distinction between a continuous X-ray and a characteristic X-ray?

Solution

Both are produced when fast electrons strike the target of an X-ray tube, but by different processes:

Characteristic and continuous X-rays compared.
Characteristic X-rays	Continuous X-rays
Produced when an electron from an outer shell (L, M or N) falls into a vacancy in an inner shell (K) of a target atom.	Produced when a fast electron is decelerated (braked) by the Coulomb attraction of a target nucleus (Bremsstrahlung).
Only certain fixed wavelengths: sharp lines such as K_\alpha, K_\beta, K_\gamma.	All wavelengths from \lambda_{\min} upwards: a smooth curve.
The wavelengths depend on the target material (its energy levels), not on the tube voltage.	The minimum wavelength depends on the accelerating p.d., \lambda_{\min}=hc/eV, not on the target.
Used to identify elements and to study atomic structure.	Forms most of the X-ray beam used in imaging.

On a graph of intensity against wavelength, continuous X-rays form a smooth curve starting at \lambda_{\min}, and characteristic X-rays appear as sharp peaks (K_\alpha, K_\beta) on top of it.

33 marks

What is the main purpose of X-rays?

Solution

The main purpose of X-rays is to see inside objects that light cannot pass through, above all inside the human body without surgery. X-rays pass easily through soft tissue but are strongly absorbed by bone, so an X-ray photograph shows bones, fractures, teeth and foreign objects such as a swallowed coin, and a CT scan gives detailed 3-D images of internal organs.

Other uses: radiotherapy (high-energy X-rays destroy cancer cells), checking metal castings and welds for hidden cracks, security scanning of luggage at airports, and X-ray crystallography, which finds the arrangement of atoms in crystals by diffraction.

43 marks

X-rays can emit electrons from metal surface and X-rays can be diffracted. Comment.

Solution

The two facts together show the dual nature of X-rays: they behave as particles in some experiments and as waves in others.

Emission of electrons shows particle nature. X-rays consist of photons of energy E=hf. An X-ray photon carries thousands of electronvolts, far more than the work function of a metal (a few eV), so it can give its energy to one electron and eject it from the surface (photoelectric effect). This one-photon, one-electron process can only be explained by photons.

Diffraction shows wave nature. Diffraction is a property of waves. X-rays have wavelengths of about 0.1\,\mathrm{nm}, similar to the spacing of atoms in a crystal, so a crystal acts as a three-dimensional diffraction grating and gives a diffraction pattern (Bragg's law):

2d\sin\theta=n\lambda

d: spacing of crystal planes, \theta: glancing angle, n: order, \lambda: X-ray wavelength.

53 marks

Why X-rays have different properties from light even though both originate from orbital transition of electrons in excited atoms?

Solution

Both are photons emitted when an electron falls from a higher to a lower energy level, hf=E_2-E_1. The difference lies in the size of the energy gap.

Light comes from transitions of the outer (valence) electrons, whose levels are only a few electronvolts apart, so the photons have about 2 to 3\,\mathrm{eV} and wavelengths of 400 to 700\,\mathrm{nm}. X-rays come from transitions of the inner (K, L) electrons of heavy atoms, which are tightly bound by the large nuclear charge; their levels are thousands of electronvolts apart, so the photons have energies of keV and wavelengths of about 0.01 to 1\,\mathrm{nm}.

\lambda=\dfrac{hc}{E}:\qquad 2\,\mathrm{eV}\;\to\;620\,\mathrm{nm},\qquad 20\,\mathrm{keV}\;\to\;0.062\,\mathrm{nm}

Because each X-ray photon carries about ten thousand times more energy, X-rays penetrate matter, ionise atoms and damage cells, while light does not.

62026 Annual I, Set D, Q.2 (v)3 marks

What is meant by breaking radiation?

Solution

Braking radiation (Bremsstrahlung, a German word meaning braking radiation) is the electromagnetic radiation (X-rays) emitted when a fast-moving electron is slowed down, or braked, by the Coulomb attraction of a nucleus in the target of an X-ray tube.

As the electron passes close to a nucleus, the attraction deflects and decelerates it. A decelerated charge radiates, so the kinetic energy lost is emitted as an X-ray photon, hf=\mathrm{K.E.}_{\text{before}}-\mathrm{K.E.}_{\text{after}}. An electron can lose any amount of energy, so braking radiation has a continuous range of wavelengths (continuous X-rays). The shortest wavelength occurs when all the kinetic energy eV goes into one photon: \lambda_{\min}=hc/eV.

A fast electron slowed and deflected by a nucleus emits a braking-radiation photon.

Note. 'Breaking' is a misprint; the correct term is braking radiation (Bremsstrahlung).

73 marks

What are K_\beta characteristic X-rays? Show with the help of diagram.

Solution

When a fast electron knocks a K-shell electron out of a heavy target atom, a vacancy is left in the K shell. If an electron from the M shell jumps down to fill this vacancy, the energy difference is emitted as an X-ray photon. This photon is called a **K_\beta characteristic X-ray**.

hf_{K_\beta}=E_M-E_K\qquad\Rightarrow\qquad\lambda_{K_\beta}=\dfrac{hc}{E_M-E_K}

K_\beta emission. A fast electron has removed one K electron; an M-shell electron falls into the vacancy and a K_\beta photon is emitted.

The M to K gap is larger than the L to K gap, so K_\beta photons have more energy and a shorter wavelength than K_\alpha photons. Their wavelength is fixed for each element, so it is characteristic of the target material.

Comprehensive questionsp. 210

12026 Annual I, Set B, Q.2 (viii)7 marks

What is piezoelectric effect? How is it used to generate ultrasonic waves? Write its application in medical science.

Solution

Piezoelectric effect

The piezoelectric effect is the ability of certain materials, such as quartz and lead zirconate titanate (PZT), to generate an electric charge in response to applied mechanical stress.

In such a crystal the centres of positive and negative charge coincide when it is unstressed. Compressing or stretching the crystal shifts its ions, the two centres separate, and positive and negative charges appear on opposite faces, so a p.d. appears across the crystal. The effect is reversible: an electric field applied across the crystal makes it change shape. This is the inverse piezoelectric effect.

(a) Direct effect: stress produces a p.d. (b) Inverse effect: an alternating p.d. makes the crystal vibrate and emit ultrasound.

Generation of ultrasonic waves

A thin piezoelectric crystal (the transducer) is placed between two metal electrodes inside the probe.
An alternating p.d. of frequency above 20\,\mathrm{kHz} (a few megahertz for medical use) is applied across the electrodes.
By the inverse piezoelectric effect the crystal expands during one half-cycle and contracts during the next, so it vibrates at the frequency of the applied p.d.
The vibrating face pushes and pulls the neighbouring medium and produces compressions and rarefactions: ultrasonic waves of the same frequency.
The crystal is made so that its natural frequency equals the applied frequency. Resonance then gives vibrations of large amplitude and a strong ultrasonic beam.

The same crystal also detects ultrasound: returning echoes squeeze and stretch it, and by the direct piezoelectric effect it produces a small alternating p.d., which is amplified and processed.

Applications in medical science

Sonography: images of organs (liver, kidneys, pancreas, uterus) and of the unborn baby, made from echoes, with no ionising radiation.
Echocardiography: moving images of the heart and its valves.
Doppler ultrasound: the frequency shift of echoes from moving blood gives the speed of blood flow and reveals blood clots and blocked arteries.
Neurosurgery: intense ultrasound focused on a small region destroys unwanted tissue without disturbing the surrounding tissues.
Physiotherapy: ultrasonic heating of deep tissues treats arthritis, muscular pain, rheumatism and sciatica.

27 marks

How ultrasound can be used to obtain diagnostic information about internal body structures? Explain.

Solution

Principle: the pulse-echo method

Ultrasound (frequency above 20\,\mathrm{kHz}; a few MHz in medicine) is sent into the body in short pulses. Whenever a pulse meets a boundary between two different tissues, part of it is reflected back as an echo. The echoes are detected, timed and measured, and a computer turns them into an image.

Production and detection

The probe contains a piezoelectric transducer. An alternating p.d. makes the crystal vibrate (inverse piezoelectric effect) and send out a pulse. Between pulses the same crystal acts as a receiver: returning echoes make it vibrate and it produces an electrical signal (direct piezoelectric effect). A gel is spread on the skin to remove air, because an air gap would reflect almost all the ultrasound before it enters the body.

Echoes from the front and back boundaries of an organ return after different times; the times give the depths d_1 and d_2.

Information obtained from the echoes

Depth of each boundary from the echo time t: the pulse travels to the boundary and back, so d=vt/2, with v\approx1540\,\mathrm{m\,s^{-1}} in soft tissue. For example, an echo after 130\,\mu\mathrm{s} comes from a depth of 10\,\mathrm{cm}.
Nature of the boundary from the strength of the echo: a large difference between tissues (tissue and bone, or tissue and a gallstone) gives a strong echo, shown bright; fluid-filled regions (cysts, the bladder, the fluid around a baby) give almost no echoes and look dark.
Shape, size and consistency of organs: an array of crystals sweeps the beam across the body, and the computer places every echo at the right depth and direction to build a 2-D image of a slice.
Movement in real time: the image is renewed many times a second, so a beating heart or a moving fetus can be watched.
Blood flow: echoes from moving blood cells show a Doppler shift in frequency, which gives the speed of the blood and reveals clots or blocked arteries.

Advantages and limitations

Ultrasound uses no ionising radiation, so it is safe even for unborn babies; it is cheap, portable and gives real-time images. It cannot pass through bone or air (lungs, gas in the bowel), and its detail is lower than that of CT or MRI.

37 marks

What are X-rays? Draw and explain the setup that produces X-rays.

Solution

X-rays

X-rays are electromagnetic waves of very short wavelength (about 0.01 to 10\,\mathrm{nm}) and high photon energy. They pass through most objects, affect photographic film and ionise matter. They are produced when fast-moving electrons are suddenly stopped by a metal target of high atomic number.

Diagram

X-ray tube: heated filament (cathode), tungsten target on a copper anode, evacuated glass tube, high p.d. and window.

Construction

Cathode: a filament heated by a current from a low-voltage battery; it emits electrons by thermionic emission.
Anode and target: a solid copper bar with a piece of tungsten or platinum (high melting point, high atomic number) embedded at its end as the target.
Evacuated glass tube: encloses the cathode and anode, so the electrons do not collide with air molecules.
High p.d.: a d.c. voltage of about 50\,000\,\mathrm{V} between cathode (-) and anode (+).
Window and cooling: X-rays leave through a thin window; water flowing behind the anode, or cooling fins, remove heat.

Working

The electrons emitted by the hot filament are accelerated by the high p.d. and gain kinetic energy \mathrm{K.E.}=eV. When these highly energetic electrons strike the target and are suddenly stopped, X-rays are produced in two ways: continuous X-rays (braking radiation) when electrons are decelerated by the nuclei of the target, and characteristic X-rays when electrons knock out inner-shell electrons of the target atoms and outer electrons fall into the vacancies. The shortest wavelength produced is \lambda_{\min}=hc/eV.

Heating of the target

Only about 1\% of the electrons' kinetic energy becomes X-rays; about 99\% becomes heat. So the target must have a high melting point, and the heat is conducted away through the copper rod, which is often cooled by water.

Control of the beam

The intensity (number of photons) is controlled by the filament current, which sets the number of electrons. The penetrating power is controlled by the accelerating p.d.: a higher p.d. gives shorter-wavelength, more penetrating hard X-rays.

47 marks

Discuss the phenomenon of producing continuous X-rays. Also derive an expression for calculating the minimum wavelength of X-rays produced during the process.

Solution

Production of continuous X-rays

In an X-ray tube, electrons accelerated through a p.d. V strike a metal target with kinetic energy eV. When a fast electron passes close to a nucleus of the target, the strong Coulomb attraction of the concentrated positive charge deflects it and decelerates it. According to electromagnetic theory a decelerated charge radiates, and according to quantum theory the radiation is emitted as photons. This radiation is called Bremsstrahlung or braking radiation. The energy of the photon comes from the kinetic energy lost by the electron:

hf=\mathrm{K.E.}_{\text{before}}-\mathrm{K.E.}_{\text{after}}

Why the spectrum is continuous

An electron may lose any fraction of its kinetic energy in one encounter, and most electrons lose their energy in several encounters. So photons of every energy from almost zero up to a maximum are produced, and the X-rays have a continuous range of wavelengths. They are called continuous X-rays.

Derivation of the minimum wavelength

Derivation Minimum wavelength of X-rays

Aim. Show that the shortest wavelength of the X-rays from a tube with accelerating p.d. V is \lambda_{\min}=hc/eV.

Assumptions.

Each electron leaves the hot cathode with negligible speed, and the whole p.d. V accelerates it.
In the extreme case the electron is stopped in a single collision and all its kinetic energy becomes one photon.

After Fig. 25.2 (b). The attraction of the nucleus deflects and decelerates the fast electron; the kinetic energy it loses is emitted as an X-ray photon of energy hf.

The electric field does work on the electron as it moves through the p.d. V. Work done equals charge times p.d., and it becomes the electron's kinetic energy when it reaches the target.
\mathrm{K.E.}=eV
In one encounter with a nucleus the electron loses some kinetic energy, which appears as one X-ray photon of energy hf. A photon cannot carry more energy than the electron has.
hf\le eV
The photon of largest energy, and so of highest frequency f_{\max}, is produced when the electron is stopped in a single collision and gives all its kinetic energy to one photon.
hf_{\max}=eV
For electromagnetic waves c=f\lambda, so f=c/\lambda. The highest frequency corresponds to the shortest wavelength.
\dfrac{hc}{\lambda_{\min}}=eV
Rearrange for \lambda_{\min}. This is equation (25.1) of the book.
\lambda_{\min}=\dfrac{hc}{eV}
Put in h=6.63\times10^{-34}\,\mathrm{J\,s}, c=3\times10^{8}\,\mathrm{m\,s^{-1}} and e=1.6\times10^{-19}\,\mathrm{C} to get a quick form for numericals.
\lambda_{\min}=\dfrac{1.24\times10^{-6}\,\mathrm{V\,m}}{V}

Result\lambda_{\min}=\dfrac{hc}{eV}

The shortest wavelength depends only on the accelerating p.d., not on the target metal. A higher p.d. gives a shorter \lambda_{\min} (more energetic, more penetrating X-rays); doubling V halves \lambda_{\min}. No photon from the tube can have a shorter wavelength, so the spectrum ends sharply at \lambda_{\min}.

Graph

Molybdenum X-ray spectrum at 35 kV (λ in nm)

After Fig. 25.3; I is the relative intensity. The continuous curve starts at \lambda_{\min}=hc/eV=0.0355\,\mathrm{nm}. The K_\beta (0.063\,\mathrm{nm}) and K_\alpha (0.071\,\mathrm{nm}) lines are characteristic of molybdenum.

Result

The continuous spectrum has a sharp lower limit \lambda_{\min}=hc/eV that depends only on the accelerating p.d. and not on the target material. Increasing V shifts \lambda_{\min} to shorter wavelengths and increases the intensity. For example, at V=10^{5}\,\mathrm{V}, \lambda_{\min}=1.24\times10^{-11}\,\mathrm{m}.

52026 Annual I, Set B, Q.2 (vi) OR7 marks

Explain the use of X-rays in imaging internal body structures.

Solution

Principle

X-ray imaging is the most common and oldest diagnostic imaging technique. It works because different tissues absorb (attenuate) X-rays by different amounts. Bone contains calcium, which has a high atomic number and density, so it absorbs X-rays strongly. Soft tissue such as fat and muscle absorbs much less, and air in the lungs hardly absorbs at all.

Rays through soft tissue reach the film and darken it; bone absorbs most rays and casts a light shadow.

Formation of the image

A beam of X-rays is passed through the part of the body being examined.
A photographic plate or digital detector on the other side receives the X-rays that pass through.
Where many X-rays arrive (behind soft tissue or air) the image is dark; where few arrive (behind bone) it stays light. Bones therefore cast a white shadow, as in the X-ray of a hand (Fig. 25.5).
The result is a shadow image of the internal structures in shades of grey.

Uses

X-ray photographs reveal fractures of bones, the presence of foreign bodies (a swallowed coin, a bullet), dental problems, and lung and heart problems in chest X-rays.

Contrast

Contrast is the range of brightness, from lightest to darkest, in an image. Bone on soft tissue gives high contrast, but neighbouring soft tissues absorb almost equally and give low contrast, so they are hard to see on a conventional X-ray. To improve contrast, a contrast medium of high atomic number is used: a barium meal for the stomach and intestines, or an iodine-based dye injected into blood vessels. Contrast in X-ray imaging allows the radiologist to evaluate structures that are not clearly evident on conventional X-ray exams. For very detailed images of soft tissue a CT scan is used.

Precaution

X-rays are ionising and can damage cells, so the dose is kept as small as possible, other parts of the body are shielded with lead, and X-rays are avoided in pregnancy unless essential.

67 marks

Explain how computed tomography (CT) scanning works?

Solution

CT scan

A CT scan (computed tomography scan) is a medical imaging method that uses X-rays and a computer to get a detailed 3-D image of parts of the body such as the brain, lungs, abdomen, blood vessels, soft tissues and bones.

Main parts

An X-ray source giving a thin, fan-shaped beam.
A large number of detectors opposite the source; each records an X-ray image (the X-rays getting through along its line).
The gantry, a large ring on which the source and detectors are mounted.
The couch, a table that slides the patient through the gantry.
A computer that processes the readings.

The X-ray tube and detectors rotate around the patient to image one slice from many angles; the couch then moves and the 2-D slices are stacked into a 3-D image.

Working

The patient lies on the couch, which slowly slides into the gantry.
The X-ray source and the detectors rotate around the patient, so X-ray images of the same section (slice) are taken from many different directions.
The computer combines these multiple images of the same section to obtain a 2-D cross-sectional (tomographic) image of that section.
The couch moves the patient a small step along the axis of the machine, and the process is repeated for the next section.
The computer combines the 2-D images of many sections to give a detailed 3-D image of the organ, which can be viewed from any direction.

One X-ray view gives only the total absorption along each line through the body. By combining the totals from many directions, the computer can work out the absorption of every small element of the slice, so structures that overlap on an ordinary film are separated.

Importance over normal X-ray film

A normal X-ray film is like a shadow: fine detail may be invisible when one organ lies in front of the region being studied. CT shows each slice separately.
CT shows small differences between soft tissues, so it gives high-quality images of the brain, lungs and abdomen.
It gives 3-D images and is essential in the study of tumors in cancer patients and of internal injuries.

Its drawbacks are a larger radiation dose than an ordinary X-ray and a high cost.

Numerical problemsp. 210

If potential difference of 3000\,\mathrm{V} is applied across an X-ray tube. What will be the minimum wavelength of X-rays produced, if the electrons were slowed down by the target.

Solution

Given

V = 3000\,\mathrm{V} accelerating p.d.
h = 6.63\times10^{-34}\,\mathrm{J\,s}
c = 3\times10^{8}\,\mathrm{m\,s^{-1}}
e = 1.6\times10^{-19}\,\mathrm{C}

To find

\lambda_{\min}

The shortest wavelength is produced when an electron gives all its kinetic energy eV to one photon.
\lambda_{\min}=\dfrac{hc}{eV}
Substitute the values.
\lambda_{\min}=\dfrac{(6.63\times10^{-34})(3\times10^{8})}{(1.6\times10^{-19})(3000)}=\dfrac{1.989\times10^{-25}}{4.8\times10^{-16}}
Evaluate and convert to nanometres (1\,\mathrm{nm}=10^{-9}\,\mathrm{m}).
\lambda_{\min}=4.14\times10^{-10}\,\mathrm{m}=0.414\,\mathrm{nm}

\lambda_{\min}=4.14\times10^{-10}\,\mathrm{m}=0.414\,\mathrm{nm}

Book answer. 0.414\,\mathrm{nm}

Find the voltage across the X-ray tube through which an electron must be accelerates in order to generate the continuous X-ray of exactly 0.1\,\mathrm{nm}.

Solution

Given

\lambda_{\min} = 0.1\,\mathrm{nm}=1\times10^{-10}\,\mathrm{m}
h = 6.63\times10^{-34}\,\mathrm{J\,s}
c = 3\times10^{8}\,\mathrm{m\,s^{-1}}
e = 1.6\times10^{-19}\,\mathrm{C}

To find

An X-ray of 0.1\,\mathrm{nm} can just be produced when it is the shortest wavelength of the spectrum, so use the minimum-wavelength relation and solve for V.
\lambda_{\min}=\dfrac{hc}{eV}\quad\Rightarrow\quad V=\dfrac{hc}{e\lambda_{\min}}
Substitute the values.
V=\dfrac{(6.63\times10^{-34})(3\times10^{8})}{(1.6\times10^{-19})(1\times10^{-10})}
Evaluate.
V=1.24\times10^{4}\,\mathrm{V}\approx12\,400\,\mathrm{V}

V=1.24\times10^{4}\,\mathrm{V}\approx12.4\,\mathrm{kV}

Book answer. 12\,400\,\mathrm{V}

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, Q.2 (vi) OR3 marks

How are X-rays used in imaging internal body structure? Explain briefly.

Solution

Principle. Different tissues absorb X-rays by different amounts. Bone (calcium, high atomic number and density) absorbs X-rays strongly; soft tissue such as fat and muscle absorbs much less; air in the lungs hardly absorbs at all.

Image. A beam of X-rays is passed through the body onto a photographic plate or detector. Where many X-rays arrive (behind soft tissue) the image is dark; behind bone few arrive, so bones cast a white shadow. The result is an image in shades of grey.

Bone absorbs most X-rays and casts a light shadow on the film.

Uses. It shows fractures, foreign bodies (a swallowed coin), teeth and chest problems. For soft tissues a contrast medium (barium meal, iodine dye) is used to increase contrast, or a CT scan gives a detailed image.

2026 Annual I, Set B, Q.2 (viii)1+2 = 3 marks

What is meant by Piezoelectric effect? Enlist its two applications.

Solution

Piezoelectric effect (1 mark)

The piezoelectric effect is the ability of certain materials (quartz, lead zirconate titanate) to generate an electric charge in response to applied mechanical stress. Squeezing the crystal separates its positive and negative charge centres, so a p.d. appears across its faces. The reverse also happens: an applied electric field changes the shape of the crystal (inverse piezoelectric effect).

Two applications (2 marks)

Production and detection of ultrasound: in medical ultrasound scanners an alternating p.d. makes the crystal vibrate and emit ultrasound, and the returning echoes make it produce electrical signals.
Generation of high voltage: a sharp blow on a piezoelectric crystal gives a spark in gas lighters and stove igniters.

Other applications: quartz crystals keep time in watches and computers (frequency generation), microbalances, and scanning probe microscopes.

2026 Annual I, Set D, Q.2 (v)2+1 = 3 marks

What is meant by breaking radiation? Also write the relation for minimum wavelength of x-rays.

Solution

Braking radiation (2 marks)

Braking radiation (Bremsstrahlung) is the X-radiation emitted when a fast-moving electron is decelerated (braked) by the Coulomb attraction of a nucleus in the target of an X-ray tube. The attraction deflects and slows the electron, and the kinetic energy it loses is emitted as an X-ray photon, hf=\mathrm{K.E.}_{\text{before}}-\mathrm{K.E.}_{\text{after}}. Since an electron can lose any amount of energy, braking radiation forms the continuous X-ray spectrum.

Minimum wavelength (1 mark)

The shortest wavelength is produced when an electron, accelerated through p.d. V, gives all its kinetic energy eV to one photon:

eV=hf_{\max}=\dfrac{hc}{\lambda_{\min}}\qquad\Rightarrow\qquad\lambda_{\min}=\dfrac{hc}{eV}

Here h is Planck's constant, c the speed of light, e the charge of the electron and V the accelerating potential difference.

Note. The paper writes 'breaking'; the correct term is braking radiation (Bremsstrahlung).

2026 Annual I, Set D, Q.2 (vi) OR3 marks

What is meant by contrast in X-ray imaging?

Solution

Contrast is the range of brightness, from lightest to darkest, in an image. In an X-ray image it is the difference in brightness between neighbouring structures, which comes from the difference in how much X-radiation they absorb.

Bone absorbs X-rays much more than soft tissue, so bone on soft tissue gives high contrast and shows clearly. Neighbouring soft tissues absorb almost equally, so they appear in nearly the same shade of grey (low contrast) and are hard to see on a conventional X-ray.

Contrast is improved with a contrast medium of high atomic number: a barium meal (barium sulfate) for the stomach and intestines, or an iodine-based dye injected into blood vessels. The filled organ then appears white. Contrast in X-ray imaging allows the radiologist to evaluate structures that are not clearly evident on conventional X-ray exams.

2026 Annual I, Set D, Q.2 (viii)3 marks

Briefly discuss two applications of ultrasound in medical technology.

Solution

1. Diagnostic imaging (sonography). Short pulses of ultrasound from a piezoelectric probe enter the body and are partly reflected at boundaries between tissues. The time of each echo gives the depth of the boundary (d=vt/2) and its strength shows the type of tissue, so a computer builds a real-time image of organs such as the liver, kidneys, heart and the unborn baby. It uses no ionising radiation, so it is safe in pregnancy.

2. Measuring blood flow (Doppler ultrasound). Ultrasound reflected from moving blood cells returns with a changed frequency (Doppler shift). The shift gives the speed of the blood, so doctors can find blood clots and blocked arteries and check the working of the heart in adults and fetuses.

Other applications that may be given instead: focused intense ultrasound in neurosurgery to destroy a small region of tissue without cutting, and ultrasonic heating in physiotherapy for arthritis, rheumatism and sciatica.

Exam focus

Derive \lambda_{\min}=hc/eV for the continuous X-ray spectrum. (long)
The only derivation in the unit; it is Comprehensive Q4, and its result was asked in 2026 (Set D Q.2 (v)).
Go to answer
Draw and explain the setup (X-ray tube) that produces X-rays. (long)
Standard labelled-diagram question (Comprehensive Q3); learn the labels and the 1% to 99% energy split.
Go to answer
What is meant by the piezoelectric effect? Enlist two applications. (short)
Asked in 2026 (Set B Q.2 (viii)); the book definition earns the first mark.
Go to answer
What is meant by braking radiation? Write the relation for minimum wavelength. (short)
Asked in 2026 (Set D Q.2 (v)); also Short Q6.
Go to answer
What is meant by contrast in X-ray imaging? (short)
Asked in 2026 (Set D Q.2 (vi) OR); the SLO names this term.
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How are X-rays used in imaging internal body structures? (short)
Asked in 2026 (Set B Q.2 (vi) OR) and set as Comprehensive Q5.
Go to answer
Two applications of ultrasound in medical technology. (short)
Asked in 2026 (Set D Q.2 (viii)).
Go to answer
How can ultrasound give diagnostic information about internal body structures? (long)
An SLO and Comprehensive Q2; the pulse-echo method and d=vt/2 earn the marks.
Go to answer
Explain how CT scanning works. (long)
An SLO and Comprehensive Q6; learn the sequence: many angles, 2-D slice, move along the axis, stack into 3-D.
Go to answer
Characteristic and continuous X-rays; K_\alpha and K_\beta lines with a diagram. (short)
Short Q2 and Q7; a comparison table and a shell diagram secure full marks.
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Numericals on \lambda_{\min}=hc/eV and on the tube voltage. (numerical)
The only numerical type in the unit (Example 25.1, Assignment 25.1, MCQ 4, Numericals 1 and 2).
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