Cosmology

Class 12 Physics, unit 23, book pages 170–182

Cosmology is the study of the origin, structure, history and future of the whole universe. We cannot visit the stars, so almost everything we know about them comes from their radiation. This unit shows how the black body laws (Wien's law and the Stefan-Boltzmann law) give a star's temperature, luminosity and radius, how the inverse square law and standard candles give distances, and how the red shift of galaxies and Hubble's law show that the universe is expanding from a Big Bang about 13.8 billion years ago.

Learning outcomes (9)

Explain the term luminosity as the total power of radiation emitted by a star.
Apply the inverse square law for radiant flux intensity $F$ in terms of the luminosity $L$ of the source, $F = \frac{L}{4 π d ^{2}}$ .
Define and apply standard candles, and explain their use to determine distances to galaxies.
Explain black body radiation and apply Wien's displacement law, $λ_{m a x} T = constant$ , to estimate the surface temperature of a star.
Apply the Stefan-Boltzmann law, $L = 4 π r^{2} σ T^{4}$ , to solve problems.
Estimate the radius of a star by applying Wien's displacement law and the Stefan-Boltzmann law.
Explain that the lines in the emission and absorption spectra from distant objects show an increase in wavelength from their known values.
Explain why red shift leads to the idea that the universe is expanding, using $\frac{Δ λ}{λ} \approx \frac{Δ f}{f} \approx \frac{v}{c}$ for a source moving relative to an observer, and solve problems relating to the expanding universe.
State and explain Hubble's law and how it leads to the Big Bang theory.

23Cosmology and the role of radiation

Almost everything we know about stars and galaxies comes from the light and other electromagnetic radiation they send us. We cannot visit a star or take a sample of it, but its radiation carries information about its temperature, size, composition, distance and motion. This unit shows how to read that information.

Cosmology: The study of the origin, development, structure, history and future of the entire universe.

Electromagnetic (EM) energy travels as waves. The EM spectrum runs from very long radio waves to very short gamma rays. The human eye detects only a small part of it, called visible light, with wavelengths from about $400 nm$ (violet) to $700 nm$ (red). Space agencies such as NASA use instruments that detect the full range of the spectrum to study the Earth, the solar system and the universe beyond.

The electromagnetic spectrum. Visible light is only a narrow band. Moving to the right, the wavelength decreases while the frequency and photon energy increase.

Radiation plays a special role in cosmology. The universe is still filled with faint residual radiation left over from the Big Bang, and the light of distant galaxies tells us that the universe is expanding. These ideas come at the end of the unit.

The unit builds one idea on another:

Black body radiation and Wien's law give a star's temperature from its colour (its peak wavelength).
The Stefan-Boltzmann law links a star's luminosity (power output) to its temperature and size.
The inverse square law links luminosity to the radiant flux intensity measured on Earth. With a standard candle this gives the distance.
Combining these laws gives a star's radius.
Spectral lines show a star's composition, and their red shift gives its speed away from us.
Speed and distance together give Hubble's law, the expansion of the universe and its age (the Big Bang theory).

23.1Black body radiationOften examined

Hold an iron rod in a flame. As it gets hotter it glows dull red, then orange, then yellow-white. Every hot object gives out electromagnetic radiation, and the mix of wavelengths depends on its temperature. Physicists study this in its simplest, ideal form: the black body.

Black body: An ideal object which is perfectly opaque and non-reflecting. It absorbs all the radiation that falls on it, at every wavelength, and it is also the best possible emitter of radiation.
Black body radiation: The electromagnetic radiation given out by a black body because of its temperature. Its spectrum depends only on the temperature of the body, not on its material or shape.

Properties of a black body

It absorbs all radiation that falls on it; it reflects and transmits none. That is why a cold black body looks perfectly black.
It is also a perfect emitter: at a given temperature no object can emit more radiation than a black body.
It emits a continuous spectrum that contains all wavelengths.
The shape of this spectrum is fixed by its temperature alone.

A model of a black body

A perfect black body does not exist; it is a theoretical object. A good model is a hollow sphere with a dull black inner wall and a very small hole (Figure 23.1). Radiation that enters the hole is reflected again and again inside the cavity. At each reflection some of it is absorbed, so almost none finds its way back out through the tiny hole. The hole therefore behaves as a perfect absorber. When the cavity is heated, the radiation coming out of the hole is black body radiation.

Figure 23.1 (redrawn): an ideal black body. Radiation entering the small hole is reflected many times and absorbed, so it almost never comes back out.

Black body radiation curves

If we measure the intensity of the radiation from a black body at each wavelength and plot it against wavelength, we get a smooth hump-shaped curve (Figure 23.2). There is a different curve for each temperature.

Black body radiation curves

0.5

1.5

2.5

λ

I

UV

visible

infrared

λ_{m a x}

λ_{m a x}

T = 6000 K

T = 5000 K

T = 4000 K

T = 3000 K

Figure 23.2 (redrawn): intensity against wavelength

λ

(in

μ m

) for black bodies at four temperatures. Each curve rises to a peak at

λ_{m a x}

and then falls. Hotter bodies peak at shorter wavelengths and are more intense at every wavelength.

Features of the curves:

Each curve is continuous: the body emits all wavelengths, but not equally.
Each curve has a peak. The wavelength at the peak is called the peak wavelength $λ_{m a x}$ .
As the temperature rises, the peak moves to shorter wavelengths (from the infrared towards the visible and ultraviolet). This is Wien's displacement law.
As the temperature rises, the intensity rises at every wavelength, so a hotter curve lies completely above a cooler one.
The total area under a curve is the total power emitted per unit area. It grows very quickly with temperature, as $T^{4}$ . This is the Stefan-Boltzmann law.

23.1Wien's displacement lawOften examined

Wien's displacement law puts a number on the observation that hotter objects emit shorter wavelengths (such as blue light) while cooler objects emit longer wavelengths (such as red light). It connects the wavelength at which a star is brightest to the temperature of its surface.

Wien's displacement law: The black body radiation curve for different temperatures peaks at a wavelength which is inversely proportional to the absolute temperature.

λ_{m a x} \propto \frac{1}{T} ⟹ λ_{m a x} = constant \times \frac{1}{T}

λ_{m a x} T = b = 2.9 \times 1 0^{- 3} m K

(23.1)

$λ_{m a x}$ : the peak wavelength, the wavelength at which the intensity of the radiation is greatest (unit: m).
$T$ : the absolute (kelvin) temperature of the surface of the body (unit: K).
$b$ : Wien's constant, $2.9 \times 1 0^{- 3} m K$ (read as 'metre kelvin', not 'millikelvin').

What the law tells us

The higher the temperature of a body, the shorter the wavelength at the peak intensity. Doubling $T$ halves $λ_{m a x}$ .
Therefore hotter stars look white or blue, and cooler stars look red or yellow. The colour of a star is a thermometer.
The higher the temperature of a body, the greater the intensity of the radiation at each wavelength.

Hotter stars peak at shorter wavelengths.
Star	Surface temperature	Peak wavelength	Colour seen
Betelgeuse (red supergiant)	$\approx 3500 K$	$\approx 830 nm$ (infrared)	red
Sun	$\approx 5800 K$	$\approx 500 nm$ (visible)	yellow-white
Rigel (blue supergiant)	$\approx 11000 K$	$\approx 260 nm$ (ultraviolet)	blue-white

Peak wavelength against temperature

200

400

600

800

1000

1200

T

λ_{m a x}

visible

Betelgeuse

Sun

Rigel

λ_{m a x} = \frac{2.9 \times 1 0 ^{- 3}}{T}

Peak wavelength

λ_{m a x}

(in nm) against temperature

T

(in thousands of kelvin). The graph of

λ_{m a x} = b / T

is a hyperbola. The shaded band is visible light.

Using Wien's law to find the temperature of a star

Method

Record the star's spectrum and read off the peak wavelength $λ_{m a x}$ , where the intensity is greatest.
Convert $λ_{m a x}$ to metres: $1 nm = 1 0^{- 9} m$ .
Rearrange Wien's law for $T$ and substitute.
$T = \frac{b}{λ _{m a x}} = \frac{2.9 \times 1 0 ^{- 3} m K}{λ _{m a x}}$

Derivation Ratio form of Wien's law

Aim. Compare the temperatures of two bodies from their peak wavelengths without using the value of $b$ .

Assumptions.

Both bodies radiate as black bodies.

Write Wien's law for each body. The constant $b$ is the same for all black bodies.
$λ_{1} T_{1} = b and λ_{2} T_{2} = b$
The right-hand sides are equal, so the left-hand sides are equal.
$λ_{1} T_{1} = λ_{2} T_{2}$
Divide both sides by $λ_{1} T_{2}$ to get the ratio of the temperatures.
$\frac{T _{1}}{T _{2}} = \frac{λ _{2}}{λ _{1}}$

Result

\frac{T _{1}}{T _{2}} = \frac{λ _{2}}{λ _{1}} or T_{2} = T_{1} \frac{λ _{1}}{λ _{2}}

Temperature and peak wavelength are inversely proportional: a star whose peak wavelength is 3 times shorter is 3 times hotter. This form is used in numerical problem 2 of the exercise.

Example 23.2: temperatures of Rigel and Betelgeuse

The spectrum of the star Rigel in the constellation of Orion peaks at a wavelength of

263 nm

, while the spectrum of the star Betelgeuse peaks at a wavelength of

828 nm

. Find the surface temperature of these two stars. Which of these two stars is cooler?

Given

$λ_{R} = 263 nm = 263 \times 1 0^{- 9} m$ peak wavelength of Rigel
$λ_{B} = 828 nm = 828 \times 1 0^{- 9} m$ peak wavelength of Betelgeuse
$b = 2.9 \times 1 0^{- 3} m K$ Wien's constant

To find

$T_{R}$ and $T_{B}$
which star is cooler

Rearrange Wien's law for the temperature.
$λ_{m a x} T = 2.9 \times 1 0^{- 3} m K ⟹ T = \frac{2.9 \times 1 0 ^{- 3}}{λ _{m a x}}$
Rigel:
$T_{R} = \frac{2.9 \times 1 0 ^{- 3}}{263 \times 1 0 ^{- 9}} = 1.10 \times 1 0^{4} K (\approx 11026 K)$
Betelgeuse:
$T_{B} = \frac{2.9 \times 1 0 ^{- 3}}{828 \times 1 0 ^{- 9}} = 3.50 \times 1 0^{3} K (\approx 3502 K)$
Compare: the longer peak wavelength belongs to the cooler star.

Betelgeuse (about $3500 K$ ) is cooler than Rigel (about $11000 K$ ). This matches their colours: Betelgeuse looks red and Rigel looks blue-white.

T_{Rigel} \approx 1.1 \times 1 0^{4} K, T_{Betelgeuse} \approx 3.5 \times 1 0^{3} K (cooler)

23.2Luminosity and the Stefan-Boltzmann lawOften examined

Luminosity and radiant flux intensity are two different quantities that describe the radiation of a star. Luminosity belongs to the star itself; radiant flux intensity is what we measure on Earth. This section deals with luminosity.

Luminosity: The total amount of electromagnetic energy emitted per unit time by a star or other celestial object. It is a measure of the total power output of radiation emitted by the star. SI unit: joule per second ( $J s^{- 1}$ ), which is the watt ( $W$ ).

Luminosity is a property of the star: it does not depend on where the observer is. For example, the Sun's luminosity is $3.84 \times 1 0^{26} W$ : every second the Sun sends out $3.84 \times 1 0^{26} J$ of energy in all directions.

Luminosity depends on two things: the surface temperature of the star and its surface area (its size). The law that connects them is the Stefan-Boltzmann law.

Stefan-Boltzmann law: The total energy emitted by a black body per unit area per second is directly proportional to the fourth power of its absolute temperature.

\frac{P}{A} \propto T^{4} ⟹ \frac{P}{A} = σ T^{4}

$P / A$ : power emitted per unit area of the surface ( $W m^{- 2}$ ).
$T$ : absolute temperature of the surface (K).
$σ$ : the Stefan-Boltzmann constant, $σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ .

Derivation Luminosity of a star, $L = 4 π r^{2} σ T^{4}$

Aim. Find the total power emitted by a star of radius $r$ and surface temperature $T$ .

Assumptions.

The star behaves as a black body, so the Stefan-Boltzmann law applies.
The star is a sphere of radius $r$ with the same temperature $T$ over its whole surface.

By the Stefan-Boltzmann law, each square metre of the surface emits a power $σ T^{4}$ .
$\frac{P}{A} = σ T^{4}$
The luminosity is the power from the whole surface, so multiply by the total surface area $A$ .
$L = A σ T^{4} (23.2 a)$
The surface area of a sphere of radius $r$ is $4 π r^{2}$ . Substitute it.
$A = 4 π r^{2} ⟹ L = 4 π r^{2} σ T^{4} (23.2 b)$

Result

L = 4 π r^{2} σ T^{4}

Luminosity grows as the square of the radius and as the fourth power of the temperature, so a big, hot star is extremely luminous. Temperature has the stronger effect: doubling $T$ multiplies $L$ by $2^{4} = 16$ , while doubling $r$ multiplies $L$ by $2^{2} = 4$ .

Comparing two stars

Write $L = 4 π r^{2} σ T^{4}$ for two stars A and B and divide. The constants $4 π$ and $σ$ cancel.

\frac{L _{A}}{L _{B}} = (\frac{r _{A}}{r _{B}})^{2} (\frac{T _{A}}{T _{B}})^{4}

Change	Effect
Radius doubled, same $T$	$L$ becomes $2^{2} = 4$ times larger
Temperature doubled, same $r$	$L$ becomes $2^{4} = 16$ times larger
Radius doubled and temperature halved	$L$ becomes $4 \times \frac{1}{16} = \frac{1}{4}$ of its value
Same $L$ , radius 10 times larger	$T$ becomes $1/ 10 \approx 0.32$ times (a bigger star of the same luminosity is cooler)

Assignment 23.1: relative luminosity

What is the relative luminosity of the objects A and B if

T_{A} = 100 K

T_{B} = 200 K

R_{A} = 10 m

and

R_{B} = 5 m

Treat both objects as black bodies, write $L = 4 π R^{2} σ T^{4}$ for each and divide, so that $4 π$ and $σ$ cancel.
$\frac{L _{A}}{L _{B}} = \frac{4 π R _{A}^{2} σ T _{A}^{4}}{4 π R _{B}^{2} σ T _{B}^{4}} = (\frac{R _{A}}{R _{B}})^{2} (\frac{T _{A}}{T _{B}})^{4}$
Substitute the values.
$\frac{L _{A}}{L _{B}} = (\frac{10}{5})^{2} (\frac{100}{200})^{4} = 4 \times \frac{1}{16} = \frac{1}{4}$
Check with the actual luminosities.
$L_{A} L_{B} = 4 π (10)^{2} (5.67 \times 1 0^{- 8}) (100)^{4} = 7.1 \times 1 0^{3} W = 4 π (5)^{2} (5.67 \times 1 0^{- 8}) (200)^{4} = 2.9 \times 1 0^{4} W$

Object A is twice as big but half as hot. The factor of 16 from temperature beats the factor of 4 from size, so B is 4 times more luminous than A.

L_{A} : L_{B} = 1 : 4 (L_{B} = 4 L_{A})

23.2Radiant flux intensity and the inverse square lawOften examined

A torch looks dazzling close up but faint from across a field, although it sends out the same power. Stars behave in the same way. A star's light leaves its surface and spreads out in all directions through a growing spherical shell. The farther the light travels, the larger the area it is spread over, so each square metre receives less power. This is why distant stars look fainter.

Radiant flux intensity: The luminosity per unit area measured on the surface of the Earth, that is, the radiation power received per square metre of a surface facing the star. Symbol F; SI unit \mathrm{W\,m^{-2}}.

F=\dfrac{L}{A}

(23.3 a)

Here L is the luminosity of the star and A is the area over which its power is spread (the illuminated surface).

Derivation Inverse square law, F=\dfrac{L}{4\pi d^{2}}

Aim. Find the radiant flux intensity at a distance d from a star of luminosity L.

Assumptions.

The star is a point source that radiates equally in all directions.
No radiation is absorbed or scattered between the star and the observer.

Figure 23.3 (redrawn): the same power passes through areas A, 4A and 9A at distances r, 2r and 3r, so the radiant flux intensity falls to F, F/4 and F/9.

Imagine a hollow sphere centred on the star, with radius equal to the distance d between the star and the Earth. All the power L leaving the star passes through this sphere and is shared equally over its inner surface.
\text{area of the sphere: }A=4\pi d^{2}
Radiant flux intensity is power per unit area, so divide the luminosity by this area.
F=\dfrac{L}{A}=\dfrac{L}{4\pi d^{2}}\qquad(23.3\,\text{b})
For a given star, L and 4\pi are constant, so F depends only on the distance.
F\propto\dfrac{1}{d^{2}}
To compare two distances, divide the two expressions; L and 4\pi cancel.
\dfrac{F_2}{F_1}=\dfrac{d_1^{2}}{d_2^{2}}

ResultF=\dfrac{L}{4\pi d^{2}}\qquad\Longrightarrow\qquad F\propto\dfrac{1}{d^{2}}

The radiant flux intensity is inversely proportional to the square of the distance from the source: this is the inverse square law. At twice the distance the same power is spread over 4 times the area, so F falls to 1/4; at three times the distance it falls to 1/9.

Radiant flux intensity against distance

F against distance d (in units of r). Doubling the distance gives one-fourth of the flux, tripling it gives one-ninth.

Luminosity and radiant flux intensity compared.
	Luminosity L	Radiant flux intensity F
Meaning	total power emitted by the star	power received per unit area at the observer
Where it is defined	at the star (a property of the star)	at the observer (measured on Earth)
Depends on	radius and surface temperature, L=4\pi r^{2}\sigma T^{4}	luminosity and distance, F=L/4\pi d^{2}
Changes with distance	no	yes, F\propto1/d^{2}
SI unit	\mathrm{W}	\mathrm{W\,m^{-2}}

For stars of equal luminosity, a greater radiant flux intensity (larger F) means the star is closer to the Earth (smaller d).
Two stars can look equally bright although one is far more luminous, if it is also much farther away.
Measuring F alone does not give L or d; we must know one of them to find the other.

Example 23.1: distance of a star

A star has a known luminosity of 9.7\times10^{27}\,\mathrm{W}. Observations of the star show that the radiant flux intensity of light received on Earth from the star is 114\,\mathrm{nW\,m^{-2}}. Determine the distance of the star from Earth.

Given

L = 9.7\times10^{27}\,\mathrm{W}
F = 114\,\mathrm{nW\,m^{-2}}=114\times10^{-9}\,\mathrm{W\,m^{-2}}

To find

distance d

Start from the inverse square law.
F=\dfrac{L}{4\pi d^{2}}
Rearrange for d: multiply both sides by d^{2}, divide by F, then take the square root.
d^{2}=\dfrac{L}{4\pi F}\quad\Longrightarrow\quad d=\sqrt{\dfrac{L}{4\pi F}}
Substitute the values.
d=\sqrt{\dfrac{9.7\times10^{27}}{4\pi\,(114\times10^{-9})}}=\sqrt{\dfrac{9.7\times10^{27}}{1.43\times10^{-6}}}=\sqrt{6.77\times10^{33}}
Evaluate.
d=8.2\times10^{16}\,\mathrm{m}

This is about 8.7 light-years (1\,\text{light-year}=9.46\times10^{15}\,\mathrm{m}).

d\approx8.2\times10^{16}\,\mathrm{m}

23.3Standard candles: a distance indicatorOften examined

Measuring astronomical distances accurately is extremely difficult. A direct measurement is possible only when an object is close enough to the Earth. For more distant objects, such as other galaxies, indirect methods must be used. Standard candles are the most important of these.

Standard candle: An astronomical object which has a known luminosity due to a characteristic quality possessed by that class of object.

In other words, a standard candle is a source whose luminosity L we know without knowing its distance. The name comes from everyday life: if you know how bright a candle really is, how faint it looks tells you how far away it is.

How a standard candle gives a distance

Recognise the object as a member of a standard candle class (for example, from the way its brightness changes). This gives its luminosity L.
Measure the radiant flux intensity F received from it on Earth (how bright it appears).
Use the inverse square law, Eq. (23.3 b), rearranged for the distance.
F=\dfrac{L}{4\pi d^{2}}\quad\Longrightarrow\quad d=\sqrt{\dfrac{L}{4\pi F}}
If the standard candle is inside a galaxy, this is also the distance of that galaxy.

Examples of standard candles

Cepheid variable stars: Very luminous pulsating stars whose brightness rises and falls regularly, with a period of a few days to about 100 days. The longer the period, the greater the luminosity (the period-luminosity relation). Measuring the period gives L, so Cepheids are standard candles for our own and nearby galaxies.
Type Ia supernovae: Explosions of white dwarf stars in binary systems. They explode on reaching nearly the same critical mass (about 1.4 times the Sun's mass), so they all reach nearly the same peak luminosity, several billion times that of the Sun. They are so bright that they can be seen in very distant galaxies. (The book writes 'Type 1A'.)

The cosmic distance ladder

Each standard candle method can measure distances only within a certain range. Nearer methods are used to calibrate the next, more distant method, like the rungs of a ladder. Organising the data and measurements from each method allows astronomers to build up a larger picture of the universe, from nearby stars to distant galaxies. This is known as the cosmic distance ladder.

The cosmic distance ladder. Each method reaches farther than the one before it and is calibrated by it.

23.3Estimating the radius of a starOften examined

Stars are so far away that even in large telescopes most look like points of light, so their radius cannot be measured directly. Instead the radius is calculated by combining the two black body laws: Wien's law gives the temperature, and the Stefan-Boltzmann law then links temperature and luminosity to the radius.

Procedure

Find the surface temperature T of the star by using Wien's displacement law.
T=\dfrac{b}{\lambda_{\max}}
Find the luminosity L of the star by using the inverse square law of flux, if the radiant flux intensity F and the distance d of the star are known.
L=4\pi d^{2}F
Use the Stefan-Boltzmann law to obtain the radius r of the star.
L=4\pi r^{2}\sigma T^{4}\;\Longrightarrow\;r=\sqrt{\dfrac{L}{4\pi\sigma T^{4}}}

The three steps: temperature from the colour, luminosity from the brightness and distance, then the radius from both.

Derivation Radius of a star from measured quantities

Aim. Express the radius r of a star in terms of its peak wavelength and its luminosity (or its flux and distance).

Assumptions.

The star radiates as a spherical black body.
Its distance d is known, for example from a standard candle or parallax.

Rearrange the Stefan-Boltzmann relation for r.
L=4\pi r^{2}\sigma T^{4}\;\Longrightarrow\;r^{2}=\dfrac{L}{4\pi\sigma T^{4}}\;\Longrightarrow\;r=\dfrac{1}{T^{2}}\sqrt{\dfrac{L}{4\pi\sigma}}
Wien's law gives the temperature from the peak wavelength.
T=\dfrac{b}{\lambda_{\max}}\;\Longrightarrow\;\dfrac{1}{T^{2}}=\dfrac{\lambda_{\max}^{2}}{b^{2}}
Substitute this into the result of step 1.
r=\dfrac{\lambda_{\max}^{2}}{b^{2}}\sqrt{\dfrac{L}{4\pi\sigma}}
If L is not known, use L=4\pi d^{2}F from the inverse square law. The 4\pi cancels.
r=\dfrac{\lambda_{\max}^{2}}{b^{2}}\sqrt{\dfrac{4\pi d^{2}F}{4\pi\sigma}}=\dfrac{d\,\lambda_{\max}^{2}}{b^{2}}\sqrt{\dfrac{F}{\sigma}}

Resultr=\sqrt{\dfrac{L}{4\pi\sigma T^{4}}}=\dfrac{\lambda_{\max}^{2}}{b^{2}}\sqrt{\dfrac{L}{4\pi\sigma}}

For the same luminosity, a cooler star (longer \lambda_{\max}) must be larger, because each square metre emits less and so it needs more surface area. This is why very luminous cool red stars are giants.

Radius of the Sun (all three steps)

The Sun's spectrum peaks at 500\,\mathrm{nm}. The radiant flux intensity of sunlight just outside Earth's atmosphere is 1.36\times10^{3}\,\mathrm{W\,m^{-2}} and the Sun is 1.5\times10^{11}\,\mathrm{m} away. Estimate the Sun's surface temperature, luminosity and radius.

Given

\lambda_{\max} = 500\,\mathrm{nm}=5.0\times10^{-7}\,\mathrm{m}
F = 1.36\times10^{3}\,\mathrm{W\,m^{-2}}
d = 1.5\times10^{11}\,\mathrm{m}
\sigma = 5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}

To find

T, L and r

Step 1, Wien's law:
T=\dfrac{2.9\times10^{-3}}{5.0\times10^{-7}}=5.8\times10^{3}\,\mathrm{K}
Step 2, inverse square law:
L=4\pi d^{2}F=4\pi(1.5\times10^{11})^{2}(1.36\times10^{3})=3.85\times10^{26}\,\mathrm{W}
Step 3, Stefan-Boltzmann law:
r=\sqrt{\dfrac{L}{4\pi\sigma T^{4}}}=\sqrt{\dfrac{3.85\times10^{26}}{4\pi(5.67\times10^{-8})(5800)^{4}}}=\sqrt{4.77\times10^{17}}=6.9\times10^{8}\,\mathrm{m}

The luminosity agrees with the book's value of 3.84\times10^{26}\,\mathrm{W}, and the radius is very close to the measured radius of the Sun, 6.96\times10^{8}\,\mathrm{m}. The method works.

T\approx5800\,\mathrm{K},\quad L\approx3.85\times10^{26}\,\mathrm{W},\quad r\approx6.9\times10^{8}\,\mathrm{m}

Assignment 23.2: radius of Betelgeuse

Betelgeuse is our nearest red giant star. It has a luminosity of 4.49\times10^{31}\,\mathrm{W} and emits radiation with a peak wavelength of 850\,\mathrm{nm}. Calculate the ratio of the radius of Betelgeuse r_B to the radius of the Sun r_S.

Given

L_B = 4.49\times10^{31}\,\mathrm{W}
\lambda_B = 850\,\mathrm{nm}=8.5\times10^{-7}\,\mathrm{m}
L_S = 3.84\times10^{26}\,\mathrm{W} Sun's luminosity (from the book)
T_S = 5800\,\mathrm{K} Sun's surface temperature (standard value; not given in the question)

To find

r_B/r_S

Temperature of Betelgeuse from Wien's law.
T_B=\dfrac{2.9\times10^{-3}}{8.5\times10^{-7}}=3.41\times10^{3}\,\mathrm{K}
Write L=4\pi r^{2}\sigma T^{4} for both stars and divide; 4\pi\sigma cancels. Then solve for the ratio of the radii.
\dfrac{L_B}{L_S}=\left(\dfrac{r_B}{r_S}\right)^{2}\left(\dfrac{T_B}{T_S}\right)^{4}\;\Longrightarrow\;\dfrac{r_B}{r_S}=\sqrt{\dfrac{L_B}{L_S}}\left(\dfrac{T_S}{T_B}\right)^{2}
Substitute.
\dfrac{r_B}{r_S}=\sqrt{\dfrac{4.49\times10^{31}}{3.84\times10^{26}}}\left(\dfrac{5800}{3412}\right)^{2}=342\times2.89\approx9.9\times10^{2}
Check: the radius of Betelgeuse on its own.
r_B=\sqrt{\dfrac{4.49\times10^{31}}{4\pi(5.67\times10^{-8})(3412)^{4}}}=6.8\times10^{11}\,\mathrm{m}

Betelgeuse is about 1000 times wider than the Sun. It is cooler than the Sun but about 10^{5} times more luminous, so it must be enormous.

\dfrac{r_B}{r_S}\approx1\times10^{3}

23.4Spectra of light

Astronomers are very limited in how they can investigate objects in space. All of their techniques involve analysing the light emitted by a star or galaxy. One of the most powerful is to spread the light into a spectrum (with a prism or a diffraction grating) and study its emission and absorption lines.

Three types of spectra

Continuous spectrum: An unbroken band containing all wavelengths (all colours, like a rainbow) with no gaps. It is given out by a hot, dense object such as a heated solid or the dense surface of a star. Black body radiation is a continuous spectrum.
Emission spectrum: A set of bright lines of particular colours (wavelengths) on a dark background. A hot gas at low pressure glows and gives off only these very specific colours, called emission lines.
Absorption spectrum: A rainbow (continuous spectrum) crossed by dark lines. When light passes through a cooler gas, the atoms and molecules of the gas absorb certain colours, or wavelengths, of that light.

The three types of spectra. For hydrogen, the dark absorption lines are at exactly the same wavelengths as the bright emission lines (dashed guides).

Spectral signatures

Every element has a unique set of emission and absorption lines, fixed by the energy levels of its atoms. This pattern of lines is known as the spectral signature of the element, like a fingerprint.

The absorption and emission spectra of an element are inverses of each other: the wavelengths of an element's absorption lines are the same as the wavelengths of its emission lines.
The elements in a star's outer layers (mostly hydrogen and helium) absorb some of the wavelengths emitted by the hot surface below, so the star's spectrum shows characteristic dark lines.
By comparing the spectrum of a celestial object with the spectra of known elements and molecules measured in the laboratory, astronomers find what the object is made of.

What each type of spectrum tells astronomers.
Spectrum	Source	Information it gives
Continuous	hot dense body (a star's surface)	temperature, from \lambda_{\max} by Wien's law; with the total power, the luminosity and radius
Emission (bright lines)	hot thin gas (a glowing nebula)	which elements are present in the gas
Absorption (dark lines)	cooler gas in front of a hot source (a star's atmosphere)	composition of the star's atmosphere; a shift of the lines gives the motion (red shift)

23.4Red shift of light from galaxiesOften examined

When astronomers look at light from distant galaxies, they see the same familiar patterns of spectral lines (for example the hydrogen pattern), so the elements can still be identified. But the lines in the emission and absorption spectra show an increase in wavelength from their known laboratory values: every line is shifted towards the red end of the spectrum. This is called red shift (Figure 23.4).

Figure 23.4 (redrawn): (a) hydrogen lines from a laboratory source; (b) the same lines in the light from a distant galaxy. The pattern is the same, but every line has moved towards the red (longer wavelength) end.

The Doppler effect for light

Doppler effect: The apparent change in the wavelength or frequency of the radiation from a source due to its relative motion towards or away from the observer.

You know the Doppler effect for sound: the horn of a passing car sounds higher as it approaches and lower as it moves away. Light behaves in the same way. When the spectra of distant stars and galaxies are observed, the Doppler effect appears as:

Red shift, if the object is moving away from the Earth: the wavelength increases and the frequency decreases.
Blue shift, if the object is moving towards the Earth: the wavelength decreases and the frequency increases.

Red shift: The fractional increase in wavelength (or decrease in frequency) due to the source and the observer receding from each other.

\dfrac{\Delta\lambda}{\lambda}=\dfrac{\Delta f}{f}=\dfrac{v}{c}

(23.4)

\Delta\lambda: shift in wavelength (observed wavelength minus emitted wavelength).
\lambda: wavelength emitted by the source (the laboratory value).
\Delta f: shift in frequency (emitted frequency minus observed frequency).
f: frequency emitted by the source.
v: speed of recession, the speed of the source away from the observer.
c: speed of light in a vacuum, 3\times10^{8}\,\mathrm{m\,s^{-1}}.

Derivation Doppler red shift, \dfrac{\Delta\lambda}{\lambda}=\dfrac{v}{c}

Aim. Show that a source moving away at speed v has its wavelength increased by the fraction v/c, and that the fractional change in frequency is the same.

Assumptions.

The source moves directly away from the observer at a steady speed v that is much smaller than c.

Wavefronts from a source S moving to the left with speed v. Behind the source (towards the observer on the right) the waves are stretched to \lambda'=\lambda+vT: red shift. Ahead of it they are squeezed: blue shift.

The source emits waves of frequency f and period T=1/f. A new crest leaves the source once every period, and in that time a crest travels a distance cT, which is one wavelength.
\lambda=cT=\dfrac{c}{f}
During one period the source itself moves a distance vT away from the observer. So when the next crest is emitted, it starts vT farther back, and the gap between successive crests reaching the observer is larger.
\lambda'=\lambda+vT
The increase in wavelength is vT. Divide it by \lambda=cT; the period cancels.
\Delta\lambda=\lambda'-\lambda=vT\quad\Longrightarrow\quad\dfrac{\Delta\lambda}{\lambda}=\dfrac{vT}{cT}=\dfrac{v}{c}
Now the frequency. The observed frequency is f'=c/\lambda'. Write \lambda'=\lambda(1+\Delta\lambda/\lambda) and use the binomial approximation (1+x)^{-1}\approx1-x for small x.
f'=\dfrac{c}{\lambda\left(1+\frac{\Delta\lambda}{\lambda}\right)}\approx f\left(1-\dfrac{\Delta\lambda}{\lambda}\right)\;\Longrightarrow\;\dfrac{\Delta f}{f}=\dfrac{f-f'}{f}\approx\dfrac{\Delta\lambda}{\lambda}

Result\dfrac{\Delta\lambda}{\lambda}=\dfrac{\Delta f}{f}=\dfrac{v}{c}\qquad(v\ll c)

The larger the red shift, the faster the source is moving away. Measuring the shift of a known spectral line gives the recession speed directly, v=c\,\Delta\lambda/\lambda.

Astronomers see red shift in the light of almost all galaxies. The book explains it as follows: the space between the Earth and the galaxies is expanding. This expansion increases the wavelength of the light from these galaxies, shifting it towards the red end of the spectrum. The more red-shifted the light from a galaxy is, the faster the galaxy is moving away from the Earth.

Example 23.3: speed of a galaxy from a frequency shift

A specific line in the spectrum of light obtained from a source in laboratory has a frequency of 4.570\times10^{14}\,\mathrm{Hz}. The same line in the spectrum of light from a distant galaxy has a frequency of 4.547\times10^{14}\,\mathrm{Hz}. What speed is the distant galaxy moving in relation to the Earth? Is it moving towards or away from the Earth?

Given

f = 4.570\times10^{14}\,\mathrm{Hz} emitted (laboratory) frequency
f' = 4.547\times10^{14}\,\mathrm{Hz} observed frequency
c = 3\times10^{8}\,\mathrm{m\,s^{-1}}

To find

speed v and direction

Shift in frequency (emitted minus observed).
\Delta f=(4.570-4.547)\times10^{14}=2.3\times10^{12}\,\mathrm{Hz}
Use the red shift equation and rearrange for v.
\dfrac{\Delta f}{f}=\dfrac{v}{c}\;\Longrightarrow\;v=\dfrac{c\,\Delta f}{f}
Substitute, using the emitted (laboratory) frequency for f.
v=\dfrac{(3\times10^{8})(2.3\times10^{12})}{4.570\times10^{14}}=1.5\times10^{6}\,\mathrm{m\,s^{-1}}
Decide the direction. The observed frequency is lower than the emitted frequency, so the wavelength is longer: a red shift.

The galaxy is moving away from the Earth (receding) at 1.5\times10^{6}\,\mathrm{m\,s^{-1}}, which is 0.5\% of the speed of light.

v\approx1.5\times10^{6}\,\mathrm{m\,s^{-1}}\ \text{(away from the Earth)}

Speed of a galaxy from a wavelength shift

The hydrogen line of wavelength 656\,\mathrm{nm} measured in the laboratory is observed at 663\,\mathrm{nm} in the light from a galaxy. Find the speed of the galaxy and say whether it is approaching or receding.

Find the shift in wavelength.
\Delta\lambda=663-656=7\,\mathrm{nm}
Use \Delta\lambda/\lambda=v/c. The nanometres cancel in the ratio, so there is no need to convert them.
v=c\,\dfrac{\Delta\lambda}{\lambda}=(3\times10^{8})\,\dfrac{7}{656}=3.2\times10^{6}\,\mathrm{m\,s^{-1}}
The observed wavelength is longer than the laboratory value (red shift), so the galaxy is receding.

v\approx3.2\times10^{6}\,\mathrm{m\,s^{-1}}\ \text{(receding)}

23.4The expanding universeOften examined

After the discovery of the Doppler red shift, astronomers began to realise that almost all the galaxies in the universe are receding, and that the farther away a galaxy is, the larger its red shift. This led to the idea that the space between the Earth and the galaxies, and between all the galaxies, must be expanding: the universe is expanding.

How red shift leads to the idea of an expanding universe

Spectral lines from distant galaxies are shifted to longer wavelengths than their laboratory values (red shift).
By the Doppler effect, a red shift means the galaxy is moving away from us, with speed v=c\,\Delta\lambda/\lambda.
Almost all galaxies, in every direction, show a red shift, so almost all galaxies are moving away from us and from each other.
More distant galaxies show larger red shifts, so they recede faster (v\propto d, Hubble's law).
The simplest explanation is that space itself is stretching and carrying the galaxies apart, so the universe is expanding. The expansion also stretches the light waves on their long journey, shifting them further towards the red.

Activity 23.1 (redrawn): dots on an inflating balloon. As the rubber stretches, every dot moves away from every other dot. In the same time B moves 1.5\,\mathrm{cm} farther from A while C, which started 1.5 times as far away, moves 2.25\,\mathrm{cm}: the farther dot recedes faster, as in Hubble's law.

Activity 23.1 (balloon model). When a dotted balloon is inflated, all the dots move away from each other as the rubber stretches. The dots stand for galaxies and the rubber for space. Just like the dots, the galaxies move away from each other. The model matches the real universe in three ways:

No dot is at the centre: an ant on any dot sees all the other dots moving away from it. In the same way, the fact that all galaxies recede from us does not mean the Earth is at the centre of the universe.
The dots do not grow; the space between them grows. Galaxies are held together by their own gravity and do not expand.
A dot twice as far away moves away twice as fast, which is Hubble's law.

23.5Hubble's lawOften examined

In the 1920s Edwin Hubble investigated the light spectra emitted by a large number of galaxies. He used red shift data to find each galaxy's recession velocity v, and standard candles (Cepheid variables) to find its distance d from the Earth. When he plotted v against d, the points lay close to a straight line through the origin. This relationship is now known as Hubble's law.

Hubble's law: The recession speed of a galaxy is directly proportional to its distance from the Earth.

v\propto d\qquad\Longrightarrow\qquad v=H_0\,d

(23.5)

v: recession speed of the galaxy (\mathrm{m\,s^{-1}}), found from its red shift.
d: distance of the galaxy from the Earth (m), found with standard candles.
H_0: the Hubble constant, the rate of expansion of the universe. Its SI unit is \mathrm{s^{-1}}, because H_0=v/d has the unit \mathrm{m\,s^{-1}}/\mathrm{m}=\mathrm{s^{-1}}.

Hubble's law

Figure 23.6 (redrawn): recession velocity v (in 10^{6}\,\mathrm{m\,s^{-1}}) against distance d (in 10^{24}\,\mathrm{m}). The points (galaxies) lie close to a straight line through the origin, and its gradient is the Hubble constant, here about 2.3\times10^{-18}\,\mathrm{s^{-1}}.

Equation (23.5) shows that:

The farther away a galaxy is, the faster its recession velocity.
The graph of recession velocity against distance is a straight line through the origin, and its gradient equals the Hubble constant: v/d=H_0.

Distance of a galaxy from Hubble's law

A galaxy's hydrogen line (656\,\mathrm{nm} in the laboratory) is observed at 663\,\mathrm{nm}. Taking H_0=2.3\times10^{-18}\,\mathrm{s^{-1}}, find the distance of the galaxy.

Find the recession speed from the red shift.
v=c\,\dfrac{\Delta\lambda}{\lambda}=(3\times10^{8})\,\dfrac{7}{656}=3.2\times10^{6}\,\mathrm{m\,s^{-1}}
Rearrange Hubble's law for the distance.
d=\dfrac{v}{H_0}=\dfrac{3.2\times10^{6}}{2.3\times10^{-18}}=1.4\times10^{24}\,\mathrm{m}
This is about 150 million light-years.

d\approx1.4\times10^{24}\,\mathrm{m}

23.5The Big Bang theory and the age of the universeOften examined

If the galaxies are moving away from each other today, they were closer together in the past. Running the expansion backwards, like a film played in reverse, there must have been a time when all the matter of the universe was at the same point (Figure 23.7). The universe probably began as an extremely hot, dense point, which then began to expand very quickly. This idea is known as the Big Bang theory.

Big Bang theory: About 13.8 billion years ago the whole universe was a very small, extremely hot and dense region. From this tiny point, the whole universe expanded outwards to what exists today.

Figure 23.7 (redrawn): tracing the expansion back in time leads to a beginning, the Big Bang. The same galaxies are farther apart at each later time, and their paths meet at one point in the past.

Evidence for the Big Bang

Red shift of galaxies: almost all galaxies are moving away from us and from each other.
Expansion of the universe (Hubble's law): recession speed is proportional to distance, which is exactly what a uniform expansion from one starting point produces. The data from Hubble's law can be extrapolated back to the moment the expansion began.
Residual radiation: the whole sky glows faintly with microwaves (the cosmic microwave background), with the spectrum of a black body at about 2.7\,\mathrm{K}. It is the cooled remains of the radiation of the hot early universe.

Age of the universe from Hubble's law

Derivation Age of the universe, T_0=\dfrac{1}{H_0}

Aim. Estimate how long ago all the galaxies were together at one point.

Assumptions.

Each galaxy has moved away at a constant speed since the Big Bang (the expansion has not sped up or slowed down).
All galaxies started from the same point at the same moment.

A galaxy now at distance d has covered this distance at a constant speed v in the time T_0 since the Big Bang.
T_0=\dfrac{\text{distance}}{\text{speed}}=\dfrac{d}{v}
Hubble's law gives the speed of the galaxy.
v=H_0\,d
Substitute; the distance cancels.
T_0=\dfrac{d}{H_0\,d}=\dfrac{1}{H_0}\qquad(23.6)

ResultT_0=\dfrac{1}{H_0}

The age does not depend on which galaxy we use, because d cancels: every galaxy gives the same starting time, which supports the idea that all galaxies started together in a Big Bang. With H_0\approx2.3\times10^{-18}\,\mathrm{s^{-1}}, T_0\approx4.3\times10^{17}\,\mathrm{s}, about 14 billion years.

The presently estimated age of the universe ranges from 13 to 14 billion years (the best value is about 13.8 billion years). There is still some discussion about the exact age, because different methods give slightly different values of H_0. Finding accurate measurements of the Hubble constant is therefore a top priority for cosmologists.

Assignment 23.3: age of the universe

A galaxy is found to be moving away with a speed of 2.1\times10^{7}\,\mathrm{m\,s^{-1}}. The galaxy is at a distance of 9.5\times10^{24}\,\mathrm{m} from Earth. Assuming the speed has remained constant, what is the age of the universe in years?

Given

v = 2.1\times10^{7}\,\mathrm{m\,s^{-1}}
d = 9.5\times10^{24}\,\mathrm{m}
1\ \text{year} = 365\times24\times3600\,\mathrm{s}=3.15\times10^{7}\,\mathrm{s}

To find

age T_0 in years

Find the Hubble constant from Hubble's law.
H_0=\dfrac{v}{d}=\dfrac{2.1\times10^{7}}{9.5\times10^{24}}=2.21\times10^{-18}\,\mathrm{s^{-1}}
Age of the universe.
T_0=\dfrac{1}{H_0}=\dfrac{1}{2.21\times10^{-18}}=4.52\times10^{17}\,\mathrm{s}
Convert seconds to years.
T_0=\dfrac{4.52\times10^{17}}{3.15\times10^{7}}=1.4\times10^{10}\ \text{years}

About 14 billion years, close to the accepted value of 13.8 billion years.

T_0\approx1.4\times10^{10}\ \text{years}

The two space telescopes described in the book.
	Hubble Space Telescope	James Webb Space Telescope
Launched	1990	2021
Orbit	low Earth orbit, about 540\,\mathrm{km} (340 miles) up	about 1.5 million km from the Earth
Observes	ultraviolet, visible and near-infrared light	near-infrared and mid-infrared light
Main work	galaxy evolution, star formation, cosmology, rate of expansion of the universe	the early universe, galaxy formation and evolution, star birth, planetary systems

Formula sheet

Wien's displacement law

\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}

\lambda_{\max}: peak wavelength; T: absolute surface temperature

Units: \lambda_{\max} in m, T in K

Wien's law, ratio form

\dfrac{T_1}{T_2}=\dfrac{\lambda_2}{\lambda_1}

\lambda_1,\lambda_2: peak wavelengths of bodies at temperatures T_1,T_2

any consistent units

Stefan-Boltzmann law

\dfrac{P}{A}=\sigma T^{4}

P/A: power emitted per unit area; \sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}; T: absolute temperature

Unit: \mathrm{W\,m^{-2}}

Luminosity of a black body

L=A\sigma T^{4}

A: surface area; \sigma: Stefan-Boltzmann constant; T: surface temperature

Unit: \mathrm{W}

Luminosity of a star

L=4\pi r^{2}\sigma T^{4}

r: radius of the star; T: surface temperature

Unit: \mathrm{W}

Comparing luminosities

\dfrac{L_A}{L_B}=\left(\dfrac{r_A}{r_B}\right)^{2}\left(\dfrac{T_A}{T_B}\right)^{4}

r: radii and T: surface temperatures of stars A and B

ratio (no unit)

Radiant flux intensity

F=\dfrac{L}{A}

L: luminosity; A: area over which the power is spread

Unit: \mathrm{W\,m^{-2}}

Inverse square law

F=\dfrac{L}{4\pi d^{2}}

d: distance from the star to the observer

Unit: \mathrm{W\,m^{-2}}

Flux at two distances

\dfrac{F_2}{F_1}=\dfrac{d_1^{2}}{d_2^{2}}

F_1,F_2: radiant flux intensities at distances d_1,d_2 from the same source

ratio (no unit)

Distance of a standard candle

d=\sqrt{\dfrac{L}{4\pi F}}

L: known luminosity; F: measured radiant flux intensity

Unit: \mathrm{m}

Radius of a star

r=\sqrt{\dfrac{L}{4\pi\sigma T^{4}}}=\dfrac{\lambda_{\max}^{2}}{b^{2}}\sqrt{\dfrac{L}{4\pi\sigma}}

T=b/\lambda_{\max} from Wien's law, b=2.9\times10^{-3}\,\mathrm{m\,K}

Unit: \mathrm{m}

Red shift (Doppler shift)

\dfrac{\Delta\lambda}{\lambda}=\dfrac{\Delta f}{f}=\dfrac{v}{c}

\Delta\lambda, \Delta f: shifts; \lambda, f: emitted values; v: recession speed; c: speed of light (valid for v\ll c)

ratios have no unit; v in \mathrm{m\,s^{-1}}

Hubble's law

v=H_0\,d

v: recession speed; d: distance of the galaxy; H_0: Hubble constant

Unit: H_0 in \mathrm{s^{-1}}

Age of the universe

T_0=\dfrac{1}{H_0}

H_0: Hubble constant

Unit: \mathrm{s} (divide by 3.15\times10^{7} for years)

Constants for this unit

\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}},\quad b=2.9\times10^{-3}\,\mathrm{m\,K},\quad c=3\times10^{8}\,\mathrm{m\,s^{-1}}

\sigma: Stefan-Boltzmann constant; b: Wien's constant; c: speed of light

Unit: as shown

Useful values

1\,\text{ly}=9.46\times10^{15}\,\mathrm{m},\quad 1\,\text{year}=3.15\times10^{7}\,\mathrm{s},\quad L_{\text{Sun}}=3.84\times10^{26}\,\mathrm{W}

ly: light-year, the distance light travels in one year

Unit: as shown

Key terms

Cosmology: The study of the origin, development, structure, history and future of the entire universe.
Black body: An ideal object which is perfectly opaque and non-reflecting; it absorbs all radiation falling on it and is also the best possible emitter.
Black body radiation: Radiation emitted by a black body because of its temperature; its continuous spectrum depends only on the temperature.
Peak wavelength (\lambda_{\max}): The wavelength at which the intensity of a black body's radiation is greatest.
Wien's displacement law: The black body radiation curve peaks at a wavelength inversely proportional to the absolute temperature: \lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}.
Luminosity: The total power output of radiation emitted by a star (energy per unit time); SI unit the watt.
Stefan-Boltzmann law: The total energy emitted by a black body per unit area per second is proportional to the fourth power of its absolute temperature.
Stefan-Boltzmann constant (\sigma): The constant in the Stefan-Boltzmann law, 5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}.
Radiant flux intensity: The luminosity per unit area measured on the surface of the Earth, F=L/4\pi d^{2}; unit \mathrm{W\,m^{-2}}.
Inverse square law: The radiant flux intensity from a point source is inversely proportional to the square of the distance from it.
Standard candle: An astronomical object which has a known luminosity due to a characteristic quality possessed by that class of object.
Cepheid variable: A pulsating star whose luminosity is fixed by its period of pulsation; used as a standard candle.
Type Ia supernova: The explosion of a white dwarf star; all reach nearly the same peak luminosity, so they are standard candles for distant galaxies.
Cosmic distance ladder: The chain of distance methods, each working over a certain range and calibrating the next, from nearby stars to distant galaxies.
Nebula: An interstellar cloud of gas and dust.
Continuous spectrum: A spectrum containing all wavelengths without gaps, emitted by a hot dense body.
Emission spectrum: Bright lines at particular wavelengths on a dark background, emitted by a hot thin gas.
Absorption spectrum: A continuous spectrum crossed by dark lines, formed when light passes through a cooler gas that absorbs certain wavelengths.
Spectral signature: The unique pattern of spectral lines of an element, used to identify it.
Doppler effect: The apparent change in wavelength or frequency of radiation due to the relative motion of the source towards or away from the observer.
Red shift: The fractional increase in wavelength (or decrease in frequency) due to the source and observer receding from each other.
Blue shift: The decrease in wavelength (increase in frequency) of radiation from a source moving towards the observer.
Recession speed: The speed at which a galaxy moves away from the Earth.
Hubble's law: The recession speed of a galaxy is directly proportional to its distance from the Earth, v=H_0d.
Hubble constant (H_0): The constant of proportionality in Hubble's law; the rate of expansion of the universe, with SI unit \mathrm{s^{-1}}.
Expanding universe: The idea, based on the red shift of galaxies, that the space between galaxies is stretching so that galaxies move apart.
Big Bang theory: About 13.8 billion years ago the whole universe was a very small, extremely hot and dense region, from which it expanded to what exists today.
Age of the universe: The time since the Big Bang, estimated as T_0=1/H_0; about 13 to 14 billion years.
Cosmic microwave background: Faint microwave radiation filling the universe, the cooled residual radiation of the Big Bang.
Light-year: The distance light travels in one year, 9.46\times10^{15}\,\mathrm{m}.

Multiple choice questionsp. 181

12026 Annual I, Set D, MCQ 5

SI unit of Luminosity is

AwattsAnswer
Bjoules
Cnewtons/meter
Dcd/m

Why

Luminosity is the total power output of a star, that is, energy emitted per second. Its SI unit is the joule per second, which is the watt. The joule (B) is the unit of energy, not power, and newtons/metre and cd/m are not units of power.

Luminosity is associated with energy produced by

Aobjects
Bcelestial bodiesAnswer
Clight source
Dfluorescent lamp

Why

In astronomy, luminosity is the total electromagnetic energy emitted per unit time by a star or other celestial body. The other options are too general (objects, light sources) or an everyday lamp, which is not what the term describes in this unit.

Luminosity depends on

ASize of the star
BTemperature of the star
CColour of the star
DOption 1 and 2Answer

Why

From the Stefan-Boltzmann law, L=4\pi r^{2}\sigma T^{4}, so luminosity depends on both the size (radius) and the temperature of the star. Colour is not a separate factor: it is decided by the temperature through Wien's law.

Which law states that brightness of a source is inversely proportional to the square of its distance

ALaw of brightness
BDirect Square Law of brightness
CInverse Square Law of brightness
DInverse Square Law of lightAnswer

Why

Brightness (radiant flux intensity) falls as 1/d^{2} because the same power spreads over a sphere of area 4\pi d^{2}: F=L/4\pi d^{2}. The standard name of this law in physics and astronomy is the inverse square law of light. Option B has the wrong dependence, and A is not a named law.

Note. Options C and D describe the same law in different words. D is chosen because 'inverse square law of light' is the standard name used in physics and astronomy texts; some keys may give C.

SI unit of Hubble's constant (H_0) is

A\mathrm{s^{-1}}Answer
B\mathrm{m^{-1}}
C\mathrm{m}
D\mathrm{m\,s^{-1}}

Why

From Hubble's law v=H_0d, H_0=v/d. Its unit is \mathrm{m\,s^{-1}}/\mathrm{m}=\mathrm{s^{-1}}.

Which statement about the Big-Bang theory is not correct:

AThere was a giant explosion known as the big bang.
BThis caused the universe to expand from a single point.
CEach point in the universe expands away from the others.
DThe further away galaxies are the slower they are moving.Answer

Why

Hubble's law says that the farther away a galaxy is, the faster it moves away (v=H_0d). So statement D is not correct. Statements A to C describe the Big Bang theory as given in the book.

Short questionsp. 181–182

13 marks

What is meant by blackbody radiation?

Solution

Black body radiation is the electromagnetic radiation emitted by a black body because of its temperature. A black body is an ideal object that is perfectly opaque and non-reflecting: it absorbs all the radiation falling on it and is also the best possible emitter. A small hole in a hollow, blackened cavity is a good model; stars are the best natural examples.

Black body radiation has a continuous spectrum whose shape depends only on the temperature. The intensity against wavelength curve has a peak at \lambda_{\max}. As the temperature rises, the peak shifts to shorter wavelengths (\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}) and the total power emitted rises as T^{4}.

23 marks

What is the relationship between luminosity and radiant flux intensity?

Solution

Luminosity L is the total power radiated by a star (unit \mathrm{W}). Radiant flux intensity F is the power received per unit area at the observer on Earth (unit \mathrm{W\,m^{-2}}).

At a distance d from the star, its power is spread evenly over a sphere of area 4\pi d^{2}, so

F=\dfrac{L}{A}=\dfrac{L}{4\pi d^{2}}

Thus F is directly proportional to L and inversely proportional to the square of the distance. L is a property of the star, while F also depends on how far away the observer is.

32026 Annual I, Set B, Q.6 (a)3 marks

What is the inverse square law for radiant flux intensity?

Solution

Inverse square law: the radiant flux intensity received from a point source is inversely proportional to the square of the distance from the source.

F=\dfrac{L}{4\pi d^{2}}\qquad\Longrightarrow\qquad F\propto\dfrac{1}{d^{2}}

Reason: the light leaving a star spreads out uniformly over a sphere of area 4\pi d^{2}. If the distance is doubled, the same luminosity is spread over four times the area, so F falls to one-fourth; at three times the distance it falls to one-ninth. That is why distant stars look fainter.

43 marks

What is meant by standard candles? Give any two examples.

Solution

A standard candle is an astronomical object which has a known luminosity due to a characteristic quality possessed by that class of object.

Because its luminosity L is known, measuring its radiant flux intensity F on Earth gives its distance from the inverse square law, d=\sqrt{L/4\pi F}. Standard candles are therefore used to find the distances of galaxies.

Examples: (1) Cepheid variable stars, whose luminosity is fixed by their period of pulsation; (2) Type Ia supernovae, which all reach nearly the same peak luminosity.

52026 Annual I, Set D, Q.2 (x) OR2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ii) OR, old syllabus3 marks

What is Wien's displacement law?

Solution

Wien's displacement law: the black body radiation curve for different temperatures peaks at a wavelength which is inversely proportional to the absolute temperature.

\lambda_{\max}\propto\dfrac{1}{T}\qquad\Longrightarrow\qquad\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}

Here \lambda_{\max} is the peak wavelength (where the intensity is greatest) and T is the surface temperature in kelvin. As a body gets hotter its peak shifts to shorter wavelengths, so hot stars look blue-white and cool stars look red. Measuring \lambda_{\max} for a star gives its surface temperature, T=b/\lambda_{\max}.

63 marks

Define Stefan-Boltzmann law.

Solution

Stefan-Boltzmann law: the total energy emitted by a black body per unit area per second is directly proportional to the fourth power of its absolute temperature.

\dfrac{P}{A}=\sigma T^{4},\qquad\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}

For a star of radius r (surface area 4\pi r^{2}), the total power emitted, its luminosity, is L=4\pi r^{2}\sigma T^{4}. Because of the fourth power, doubling the temperature makes a star 16 times more luminous.

73 marks

How can we estimate the radius of a star by applying Wien's displacement law and the Stefan-Boltzmann law?

Solution

A star behaves almost like a black body, so its radius is found in three steps:

Measure the peak wavelength of its spectrum and find the surface temperature from Wien's law: T=\dfrac{2.9\times10^{-3}\,\mathrm{m\,K}}{\lambda_{\max}}.
Find the luminosity from the measured radiant flux intensity F and the known distance d (inverse square law): L=4\pi d^{2}F.
Apply the Stefan-Boltzmann law L=4\pi r^{2}\sigma T^{4} and solve for the radius:

r=\sqrt{\dfrac{L}{4\pi\sigma T^{4}}}

For example, this method gives the Sun's radius as about 7\times10^{8}\,\mathrm{m}.

82026 Annual I, Set B, Q.2 (x) OR3 marks

How redshift leads to the idea that the Universe is expanding?

Solution

Spectral lines in the light of distant galaxies are found at longer wavelengths than their laboratory values: they are red-shifted. By the Doppler effect this means the galaxies are moving away from us, with speed v=c\,\Delta\lambda/\lambda.

Almost all galaxies, in every direction, show red shift, and the farther the galaxy the larger its red shift (Hubble's law, v=H_0d). So galaxies are moving away from each other, which means the space between them is stretching. Therefore the universe is expanding, like dots on an inflating balloon.

93 marks

How does the expansion of the universe support the Big Bang theory?

Solution

If the galaxies are moving apart now, they must have been closer together in the past. Tracing the expansion back in time, all the matter of the universe must once have been in one extremely small, hot and dense region. The expansion from this state is the Big Bang, about 13.8 billion years ago.

Hubble's law supports this. Since v=H_0d, every galaxy took the same time T_0=d/v=1/H_0 to reach its present distance, so all galaxies started together. This gives an age of about 13 to 14 billion years.

103 marks

How do different types of spectra (emission, absorption, continuous) provide insights into the composition and properties of celestial objects?

Solution

Continuous spectrum (all wavelengths, from a hot dense surface): its peak wavelength gives the temperature by Wien's law, T=b/\lambda_{\max}; with the total power it gives the luminosity and radius.
Emission spectrum (bright lines, from a hot thin gas such as a nebula): each element has its own set of lines, so the lines show which elements the gas contains.
Absorption spectrum (dark lines on a continuous background, from cooler gas in a star's atmosphere): the lines are at the same wavelengths as the element's emission lines, so they reveal the star's composition, mainly hydrogen and helium.
A shift of the whole line pattern to longer wavelengths (red shift) shows that the object is moving away, and gives its speed, v=c\,\Delta\lambda/\lambda.

Comprehensive questionsp. 182

12026 Annual I, Set D, Q.6 (a)2026 Annual I, Set B, Q.6 (a)6 marks

Explain the term luminosity and radiant flux intensity.

Solution

Luminosity

Definition. Luminosity L is the total amount of electromagnetic energy emitted per unit time by a star or other celestial object. It is the total power output of the star's radiation. SI unit: joule per second, which is the watt (\mathrm{W}). For example, the Sun's luminosity is 3.84\times10^{26}\,\mathrm{W}.

Luminosity is a property of the star itself and is the same wherever the observer is. It depends on the star's surface temperature and its size, through the Stefan-Boltzmann law: the energy emitted by a black body per unit area per second is proportional to the fourth power of its absolute temperature.

Derivation Luminosity of a star

Aim. Relate the luminosity of a star to its radius r and surface temperature T.

Treat the star as a black body. By the Stefan-Boltzmann law, each square metre of its surface emits power \sigma T^{4}.
\dfrac{P}{A}=\sigma T^{4}
Luminosity is the power from the whole surface, so multiply by the surface area.
L=A\sigma T^{4}
The surface of a spherical star has area 4\pi r^{2}.
L=4\pi r^{2}\sigma T^{4}

ResultL=4\pi r^{2}\sigma T^{4}

A star that is twice as hot is 2^{4}=16 times more luminous; a star with twice the radius and the same temperature is 2^{2}=4 times more luminous.

Radiant flux intensity

Definition. Radiant flux intensity F is the luminosity per unit area measured on the surface of the Earth, that is, the power of the star's radiation received per square metre facing the star. SI unit: \mathrm{W\,m^{-2}}. It tells us how bright the star appears.

Derivation Inverse square law

Aim. Find the radiant flux intensity at distance d from a star of luminosity L.

The same power passes through areas A, 4A and 9A at distances r, 2r and 3r, so F falls to F/4 and F/9.

The star radiates equally in all directions. At distance d its power is spread over a sphere of radius d.
A=4\pi d^{2}
Radiant flux intensity is power per unit area.
F=\dfrac{L}{A}=\dfrac{L}{4\pi d^{2}}
L and 4\pi are constant for a given star.
F\propto\dfrac{1}{d^{2}}

ResultF=\dfrac{L}{4\pi d^{2}}

Doubling the distance spreads the same power over 4 times the area, so F falls to one-fourth: the inverse square law.

So radiant flux intensity follows the inverse square law. At three times the distance it falls to one-ninth. This is why distant stars look faint even when they are very luminous.

Comparison

	Luminosity L	Radiant flux intensity F
What it is	total power emitted by the star	power received per unit area on Earth
Depends on	radius and temperature	luminosity and distance
Unit	\mathrm{W}	\mathrm{W\,m^{-2}}

Because F depends on both L and d, measuring F for a star of known luminosity (a standard candle) gives its distance, d=\sqrt{L/4\pi F}. For two stars of the same luminosity, the one with the greater radiant flux intensity is closer to the Earth.

26 marks

Discuss the use of standard candles to determine distances to galaxies.

Solution

The problem

Distances to stars and galaxies are enormous and cannot be measured directly. A geometric method (parallax) works only for nearby stars. For galaxies, which are millions of light-years away, astronomers use an indirect method based on the inverse square law: standard candles.

Standard candle

A standard candle is an astronomical object which has a known luminosity due to a characteristic quality possessed by that class of object. Its luminosity is known because every member of the class has the same, or a predictable, luminosity.

Method

Identify a standard candle (for example a Cepheid variable) in the galaxy. Its class, or its period of pulsation, gives its luminosity L.
Measure its radiant flux intensity F on Earth, that is, how bright it appears.
Use the inverse square law, rearranged for the distance.
F=\dfrac{L}{4\pi d^{2}}\quad\Longrightarrow\quad d=\sqrt{\dfrac{L}{4\pi F}}
The distance of the standard candle is the distance of the galaxy that contains it.

Examples

Cepheid variable stars: very luminous stars whose brightness rises and falls regularly. The longer the period, the greater the luminosity, so measuring the period gives L. They are used for our own and nearby galaxies.
Type Ia supernovae: exploding white dwarf stars that all reach nearly the same peak luminosity, several billion times the Sun's. They are bright enough to be seen in very distant galaxies.

Illustration

A Cepheid in a nearby galaxy has a luminosity of 1.0\times10^{30}\,\mathrm{W} (from its period), and its measured radiant flux intensity is 1.0\times10^{-14}\,\mathrm{W\,m^{-2}}. Then

d=\sqrt{\dfrac{1.0\times10^{30}}{4\pi(1.0\times10^{-14})}}=2.8\times10^{21}\,\mathrm{m}\approx3\times10^{5}\ \text{light-years}

The cosmic distance ladder

Each method works only within a certain range of distances. Nearer methods calibrate the next, more distant one, like the rungs of a ladder: parallax for nearby stars, Cepheids for nearby galaxies, Type Ia supernovae for distant galaxies, and finally Hubble's law. This is the cosmic distance ladder.

The cosmic distance ladder.

Importance

Edwin Hubble used Cepheid variables to find the distances of galaxies. Combined with their red shifts, these distances gave Hubble's law, the value of the Hubble constant and hence the age of the universe.

36 marks

Explain that how the radius of a star can be estimated by applying Wien's displacement law and the Stefan-Boltzmann law.

Solution

Idea

Stars are too far away for their size to be measured directly. But a star radiates almost like a black body, so two black body laws can be combined: Wien's law gives the surface temperature from the star's colour, and the Stefan-Boltzmann law connects temperature, luminosity and radius.

Step 1: temperature from Wien's law

The black body curve peaks at a wavelength inversely proportional to the absolute temperature, \lambda_{\max}T=b with b=2.9\times10^{-3}\,\mathrm{m\,K}. Measure the peak wavelength of the star's spectrum and calculate

T=\dfrac{b}{\lambda_{\max}}

Step 2: luminosity from the inverse square law

Measure the radiant flux intensity F on Earth. If the distance d is known (from parallax or a standard candle), the luminosity is

F=\dfrac{L}{4\pi d^{2}}\quad\Longrightarrow\quad L=4\pi d^{2}F

Step 3: radius from the Stefan-Boltzmann law

Derivation Radius of a star

Aim. Find the radius r from L and T.

The luminosity of a spherical black body of radius r is given by the Stefan-Boltzmann law.
L=4\pi r^{2}\sigma T^{4}
Make r^{2} the subject.
r^{2}=\dfrac{L}{4\pi\sigma T^{4}}
Take the square root.
r=\sqrt{\dfrac{L}{4\pi\sigma T^{4}}}
Using T=b/\lambda_{\max} from step 1, the radius can also be written directly in terms of the measured peak wavelength.
r=\dfrac{\lambda_{\max}^{2}}{b^{2}}\sqrt{\dfrac{L}{4\pi\sigma}}

Resultr=\sqrt{\dfrac{L}{4\pi\sigma T^{4}}}

For the same luminosity, a cooler star must have a larger surface, so it is bigger.

The three steps for estimating the radius of a star.

Example: the Sun

The Sun's spectrum peaks at 500\,\mathrm{nm}.
T=\dfrac{2.9\times10^{-3}}{5.0\times10^{-7}}=5800\,\mathrm{K}
F=1.36\times10^{3}\,\mathrm{W\,m^{-2}} at d=1.5\times10^{11}\,\mathrm{m}.
L=4\pi(1.5\times10^{11})^{2}(1.36\times10^{3})=3.85\times10^{26}\,\mathrm{W}
Radius:
r=\sqrt{\dfrac{3.85\times10^{26}}{4\pi(5.67\times10^{-8})(5800)^{4}}}=6.9\times10^{8}\,\mathrm{m}

Conclusion

The radius follows from r=\sqrt{L/4\pi\sigma T^{4}}. The method gives the Sun's radius correctly, and it shows that luminous but cool red stars such as Betelgeuse are giants, about 1000 times the Sun's radius.

42026 Annual I, Set B, Q.2 (x) OR6 marks

Why redshift leads to the idea that the Universe is expanding? Discuss.

Solution

Red shift

When the light of a distant galaxy is spread into a spectrum, its lines have the same pattern as the lines of known elements in the laboratory, but every line is at a longer wavelength. The lines are shifted towards the red end of the spectrum. Red shift is the fractional increase in wavelength (or decrease in frequency) due to the source and the observer receding from each other.

Hydrogen lines from a laboratory source (a) and from a distant galaxy (b): the same pattern, shifted towards red.

Doppler effect and speed of recession

The Doppler effect is the apparent change in wavelength or frequency of radiation due to the motion of the source towards or away from the observer. A source moving away shows a red shift; a source moving towards us shows a blue shift. For speeds much less than c:

\dfrac{\Delta\lambda}{\lambda}=\dfrac{\Delta f}{f}=\dfrac{v}{c}

So the red shift of a galaxy tells us that it is receding, and how fast.

What the observations show

Almost all galaxies show red shift, in whichever direction we look. Almost all galaxies are moving away from us.
The farther away a galaxy is, the larger its red shift, so the faster it recedes: v=H_0d (Hubble's law).

Conclusion: the universe is expanding

If every galaxy moves away from every other, with a speed proportional to its distance, the galaxies are not flying through space away from one centre. Instead, the space between them is stretching. The universe is expanding. This expansion also stretches the light waves as they travel through space, which is the red shift we observe.

Balloon model: as the balloon is inflated, every dot moves away from every other dot, and farther dots move away faster.

The balloon model (Activity 23.1) shows the idea. Dots drawn on a balloon move apart as it is inflated, although no dot moves over the rubber. From any dot, all the others recede, and a dot twice as far away recedes twice as fast. The dots are like galaxies and the rubber is like space.

Consequence

Running the expansion backwards leads to the Big Bang theory: about 13.8 billion years ago the universe was a very small, extremely hot and dense region, and it has been expanding ever since.

56 marks

State and explain Hubble's law and how it leads to the Big Bang theory.

Solution

Statement

Hubble's law: the recession speed of a galaxy is directly proportional to its distance from the Earth.

v\propto d\qquad\Longrightarrow\qquad v=H_0\,d

Here v is the recession speed, d the distance of the galaxy and H_0 the Hubble constant, the rate of expansion of the universe. Its SI unit is \mathrm{s^{-1}} (from H_0=v/d), and its value is about 2.3\times10^{-18}\,\mathrm{s^{-1}} (70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}).

Explanation

Edwin Hubble measured the red shifts of many galaxies to find their recession speeds (v=c\,\Delta\lambda/\lambda), and used standard candles (Cepheid variables) to find their distances. A graph of v against d is a straight line through the origin whose gradient is H_0.

Hubble's law

Recession velocity v (in 10^{6}\,\mathrm{m\,s^{-1}}) against distance d (in 10^{24}\,\mathrm{m}): a straight line through the origin with gradient H_0.

The farther a galaxy is, the faster it moves away.
This is exactly the pattern produced by a uniform expansion of space: as on an inflating balloon, a galaxy twice as far away recedes twice as fast.
From every galaxy the same law would be seen, so we are not at the centre of the universe.

How it leads to the Big Bang theory

Since the galaxies are moving apart now, they were closer together in the past. Tracing Hubble's law back in time, all galaxies were at the same point at one moment: the universe began from an extremely small, hot and dense state and has expanded ever since. This is the Big Bang theory.

Derivation Age of the universe

Aim. Find the time since all galaxies were together.

Assume a galaxy at distance d has moved at a constant speed v since the beginning. The time taken is distance divided by speed.
T_0=\dfrac{d}{v}
Use Hubble's law for the speed.
v=H_0d
Substitute; d cancels, so every galaxy gives the same time.
T_0=\dfrac{d}{H_0d}=\dfrac{1}{H_0}

ResultT_0=\dfrac{1}{H_0}

All galaxies left the same point at the same time: this is the Big Bang. With H_0\approx2.3\times10^{-18}\,\mathrm{s^{-1}}, T_0\approx4.3\times10^{17}\,\mathrm{s}\approx1.4\times10^{10} years.

Tracing the expansion back in time leads to the Big Bang.

The present estimate of the age of the universe is 13 to 14 billion years (about 13.8 billion years). Further support for the Big Bang comes from the faint microwave radiation that fills the universe, the cooled residual radiation of the hot early universe.

Numerical problemsp. 182

There are two stars, let named A and B. Which star has the greater temperature? Given that: L_A=L_B=10^{4}\,\mathrm{J/s}; R_A=10^{4}\,\mathrm{m}; R_B=10^{5}\,\mathrm{m}.

Solution

Given

L_A=L_B = 10^{4}\,\mathrm{J\,s^{-1}}=10^{4}\,\mathrm{W}
R_A = 10^{4}\,\mathrm{m}
R_B = 10^{5}\,\mathrm{m}
\sigma = 5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}

To find

T_A and T_B; which is hotter

Rearrange the Stefan-Boltzmann relation for the temperature.
L=4\pi R^{2}\sigma T^{4}\;\Longrightarrow\;T=\left(\dfrac{L}{4\pi R^{2}\sigma}\right)^{1/4}
Star A:
T_A=\left(\dfrac{10^{4}}{4\pi(10^{4})^{2}(5.67\times10^{-8})}\right)^{1/4}=\left(\dfrac{10^{4}}{71.3}\right)^{1/4}=(140)^{1/4}=3.44\,\mathrm{K}
Star B:
T_B=\left(\dfrac{10^{4}}{4\pi(10^{5})^{2}(5.67\times10^{-8})}\right)^{1/4}=\left(\dfrac{10^{4}}{7.13\times10^{3}}\right)^{1/4}=(1.40)^{1/4}=1.09\,\mathrm{K}
Quick check with a ratio: equal luminosities mean R^{2}T^{4} is the same for both stars.
R_A^{2}T_A^{4}=R_B^{2}T_B^{4}\;\Longrightarrow\;\dfrac{T_A}{T_B}=\sqrt{\dfrac{R_B}{R_A}}=\sqrt{10}=3.16

Star A is smaller. To give out the same power from a smaller surface, each square metre must emit more, so star A must be hotter.

T_A\approx3.4\,\mathrm{K}>T_B\approx1.1\,\mathrm{K}:\ \text{star A is hotter}

Book answer. star A

Note. The data are not realistic for real stars (real stars have radii near 10^{9}\,\mathrm{m} and luminosities near 10^{26}\,\mathrm{W}), which is why the temperatures come out tiny. The question tests the method and the comparison.

The Sun has a surface temperature of 6000\,\mathrm{K} produces peak radiation of 420\,\mathrm{nm}. Find out the temperature of the Sirius if peak radiation of Sirius is 72\,\mathrm{nm}.

Solution

Given

T_S = 6000\,\mathrm{K} Sun
\lambda_S = 420\,\mathrm{nm} Sun's peak wavelength
\lambda_{Si} = 72\,\mathrm{nm} peak wavelength of Sirius

To find

T_{Si}

By Wien's law, \lambda_{\max}T has the same value for every black body, so write it for both stars and equate.
\lambda_S T_S=\lambda_{Si}T_{Si}
Rearrange for the temperature of Sirius.
T_{Si}=T_S\,\dfrac{\lambda_S}{\lambda_{Si}}
Substitute. The wavelengths are both in nm, so their units cancel.
T_{Si}=6000\times\dfrac{420}{72}=3.5\times10^{4}\,\mathrm{K}

Sirius peaks at a wavelength about 6 times shorter than the Sun, so it is about 6 times hotter.

T_{\text{Sirius}}=3.5\times10^{4}\,\mathrm{K}\ (35\,000\,\mathrm{K})

Book answer. 35\,000\,\mathrm{K}

Note. The question's data do not fit the usual Wien constant: 6000\times420\times10^{-9}=2.52\times10^{-3}\,\mathrm{m\,K}, not 2.9\times10^{-3}\,\mathrm{m\,K}. Using T=2.9\times10^{-3}/(72\times10^{-9}) directly gives 4.0\times10^{4}\,\mathrm{K}. The ratio method uses only the data given and matches the book's answer. (The real Sirius has a surface temperature of about 9900\,\mathrm{K} and peaks near 290\,\mathrm{nm}.)

If a source has a luminosity of 3.826\times10^{26}\,\mathrm{W} and is at a distance of 3 lightyear, what is the radiant flux intensity?

Solution

Given

L = 3.826\times10^{26}\,\mathrm{W}
d = 3\,\text{ly}=3\times9.46\times10^{15}\,\mathrm{m}=2.84\times10^{16}\,\mathrm{m}

To find

radiant flux intensity F

Convert the distance to metres: one light-year is the distance light travels in one year, 9.46\times10^{15}\,\mathrm{m}.
d=3\times9.46\times10^{15}=2.838\times10^{16}\,\mathrm{m}
Use the inverse square law.
F=\dfrac{L}{4\pi d^{2}}
Substitute.
F=\dfrac{3.826\times10^{26}}{4\pi(2.838\times10^{16})^{2}}=\dfrac{3.826\times10^{26}}{1.012\times10^{34}}
Evaluate.
F=3.78\times10^{-8}\,\mathrm{W\,m^{-2}}

F\approx3.78\times10^{-8}\,\mathrm{W\,m^{-2}}

Book answer. 3.78\times10^{-8}\,\mathrm{W\,m^{-2}}

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 31 mark

Two stars have the same luminosity. If one appears dimmer, it is because:

AIt is farther from the observerAnswer
BIt has lower temperature
CIt has smaller size
DIt emits less energy

Why

How bright a star appears is its radiant flux intensity, F=L/4\pi d^{2}. With the same L, a smaller F can only come from a larger distance d. Options B, C and D would each change the luminosity itself (L=4\pi r^{2}\sigma T^{4}), but the question says the luminosities are equal.

2026 Annual I, Set B, MCQ 141 mark

If the distance from a star (luminosity L) is doubled, the radiant flux intensity becomes:

AOne-fourthAnswer
BHalf
CDouble
DFour times

Why

F=L/4\pi d^{2}, so F\propto1/d^{2}. Doubling d multiplies F by (1/2)^{2}=1/4. 'Half' would be true only if F were proportional to 1/d.

2026 Annual I, Set B, MCQ 161 mark

If a star is receding from the earth it exhibits:

ARed shiftAnswer
BBlue shift
CGreen shift
DYellow light

Why

A receding source's waves are stretched (Doppler effect), so its spectral lines move to longer wavelengths, towards the red end of the spectrum: a red shift. A blue shift is seen when a source approaches; 'green shift' and 'yellow light' are not Doppler effects.

2026 Annual I, Set B, Q.2 (x) OR1+2 = 3 marks

To which idea does a red-shift exhibited by a star leads and why?

Solution

Idea (1 mark): red shift leads to the idea that the universe is expanding: galaxies, with their stars, are moving away from us and from each other.

Why (2 marks): red shift means the spectral lines of the star or galaxy are at longer wavelengths than their laboratory values. By the Doppler effect, this happens when the source moves away from the observer, with speed given by

\dfrac{\Delta\lambda}{\lambda}=\dfrac{\Delta f}{f}=\dfrac{v}{c}

Almost all galaxies, in every direction, show red shift, and the farther they are the greater the shift (v=H_0d). The simple explanation is that the space between the galaxies is stretching, like the rubber of an inflating balloon carrying its dots apart. So the universe is expanding; running this backwards leads to the Big Bang.

2026 Annual I, Set B, Q.63+3 = 6 marks

Answer both parts (a) and (b).

Solution

(a) What is meant by Radiant Flux Intensity? How it follows inverse square law?

Radiant flux intensity (1 mark): the luminosity per unit area measured on the surface of the Earth, that is, the power of a star's radiation received per square metre, F=L/A. SI unit: \mathrm{W\,m^{-2}}.

Inverse square law (2 marks): light leaving a star of luminosity L spreads out uniformly in all directions. At a distance d it is spread over the surface of a sphere of area 4\pi d^{2}, so

F=\dfrac{L}{4\pi d^{2}}\qquad\Longrightarrow\qquad F\propto\dfrac{1}{d^{2}}

If the distance is doubled, the same power covers 4 times the area, so F falls to one-fourth; at three times the distance it falls to one-ninth. Hence radiant flux intensity is inversely proportional to the square of the distance: it follows the inverse square law.

The same power passes through areas A, 4A and 9A at distances r, 2r and 3r.

(b) A star has a known luminosity of 9.7\times10^{27}\,\mathrm{W}. Observations of the star show that the radiant flux intensity of light received on Earth from the star is 114\,\mathrm{nW\,m^{-2}}. Determine the distance of the star from Earth.

Given

L = 9.7\times10^{27}\,\mathrm{W}
F = 114\,\mathrm{nW\,m^{-2}}=114\times10^{-9}\,\mathrm{W\,m^{-2}}

To find

distance d

Formula (1 mark): the inverse square law, rearranged for d.
F=\dfrac{L}{4\pi d^{2}}\;\Longrightarrow\;d=\sqrt{\dfrac{L}{4\pi F}}
Substitution (1 mark):
d=\sqrt{\dfrac{9.7\times10^{27}}{4\pi(114\times10^{-9})}}=\sqrt{\dfrac{9.7\times10^{27}}{1.43\times10^{-6}}}=\sqrt{6.77\times10^{33}}
Answer with unit (1 mark):
d=8.2\times10^{16}\,\mathrm{m}\ (\approx8.7\ \text{light-years})

d\approx8.2\times10^{16}\,\mathrm{m}

Note. Part (b) is Example 23.1 of the book, word for word; part (a) matches short question 3.

2026 Annual I, Set D, MCQ 51 mark

The SI unit of luminosity 'L' is:

AWatt \mathrm{m^{-2}}
BKelvin
CWattAnswer
DJoule

Why

Luminosity is power (energy emitted per second), so its unit is \mathrm{J\,s^{-1}}, the watt. \mathrm{W\,m^{-2}} (A) is the unit of radiant flux intensity, the kelvin (B) is the unit of temperature, and the joule (D) is energy, not power.

2026 Annual I, Set D, MCQ 161 mark

According to the inverse square law, radiant flux intensity is inversely proportional to:

ATemperature
BLuminosity
CDistance
DSquare of distanceAnswer

Why

F=L/4\pi d^{2}: the radiant flux intensity is inversely proportional to the square of the distance. It is directly (not inversely) proportional to the luminosity, and 'distance' alone misses the square.

2026 Annual I, Set D, Q.2 (x) OR2+1 = 3 marks

How does Wien's displacement law be used to determine the temperature of a star? Also write its formula.

Solution

Use (2 marks): Wien's displacement law states that the black body radiation curve peaks at a wavelength which is inversely proportional to the absolute temperature. A star radiates almost like a black body, so:

Record the star's spectrum and find the peak wavelength \lambda_{\max}, where the intensity is greatest.
Convert it to metres and divide Wien's constant by it: T=\dfrac{2.9\times10^{-3}\,\mathrm{m\,K}}{\lambda_{\max}}.
A shorter peak wavelength means a hotter star, so blue-white stars are hotter than red stars.

Formula (1 mark):

\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}

For example, Rigel peaks at 263\,\mathrm{nm}, so T=2.9\times10^{-3}/(263\times10^{-9})\approx1.1\times10^{4}\,\mathrm{K}.

2026 Annual I, Set D, Q.63+3 = 6 marks

Answer both parts (a) and (b).

Solution

(a) What is meant by luminosity of a star? Write the relation between luminosity, radius and temperature.

Luminosity (1 mark): the total amount of electromagnetic energy emitted per unit time by a star, that is, the total power output of its radiation. SI unit: watt (\mathrm{J\,s^{-1}}). The Sun's luminosity is 3.84\times10^{26}\,\mathrm{W}.

Relation (2 marks): treating the star as a black body, the Stefan-Boltzmann law says each square metre of its surface emits power \sigma T^{4}. A spherical star of radius r has surface area 4\pi r^{2}, so

L=A\sigma T^{4}=4\pi r^{2}\sigma T^{4}

where \sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}} and T is the surface temperature in kelvin. So L\propto r^{2} and L\propto T^{4}: doubling the radius gives 4 times the luminosity, and doubling the temperature gives 16 times.

(b) What are the relative luminosities of the objects A and B if T_A=100\,\mathrm{K}, T_B=200\,\mathrm{K}, R_A=10\,\mathrm{m} and R_B=5\,\mathrm{m}

Formula (1 mark): write L=4\pi R^{2}\sigma T^{4} for each object and divide; 4\pi\sigma cancels.
\dfrac{L_A}{L_B}=\left(\dfrac{R_A}{R_B}\right)^{2}\left(\dfrac{T_A}{T_B}\right)^{4}
Substitution (1 mark):
\dfrac{L_A}{L_B}=\left(\dfrac{10}{5}\right)^{2}\left(\dfrac{100}{200}\right)^{4}=4\times\dfrac{1}{16}
Result (1 mark):
\dfrac{L_A}{L_B}=\dfrac{1}{4}\quad\Longrightarrow\quad L_B=4L_A

Check: L_A=4\pi(10)^{2}(5.67\times10^{-8})(100)^{4}=7.1\times10^{3}\,\mathrm{W} and L_B=4\pi(5)^{2}(5.67\times10^{-8})(200)^{4}=2.9\times10^{4}\,\mathrm{W}. B is smaller but twice as hot, and the factor 2^{4}=16 from temperature outweighs the factor 2^{2}=4 from size.

L_A:L_B=1:4

Note. Part (b) is Assignment 23.1 of the book; part (a) is covered by comprehensive question 1.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ii) OR, old syllabus3 marks

What is Wein's displacement law? Explain briefly.

Solution

Wien's displacement law: the black body radiation curve for different temperatures peaks at a wavelength which is inversely proportional to the absolute temperature of the body.

\lambda_{\max}\propto\dfrac{1}{T}\qquad\Longrightarrow\qquad\lambda_{\max}T=b=2.9\times10^{-3}\,\mathrm{m\,K}

Explanation: a hot body emits all wavelengths, but not equally; the intensity is greatest at the peak wavelength \lambda_{\max}. As the temperature rises the peak is displaced to shorter wavelengths, and the intensity rises at every wavelength. That is why a heated iron rod glows red, then yellow, then white, and why blue stars are hotter than red stars. Measuring \lambda_{\max} gives a star's surface temperature, T=b/\lambda_{\max}.

Black body radiation curves

Black body curves: as T rises, the peak \lambda_{\max} moves to shorter wavelengths (\lambda in \mu\mathrm{m}, intensity I in relative units).

Note. The paper spells the name 'Wein'; the correct spelling is Wien (Wilhelm Wien).

Exam focus

Define luminosity and write (or derive) L=4\pi r^{2}\sigma T^{4}; compare the luminosities of two stars. (long)
Set D Q.6 (3+3) in 2026, with Assignment 23.1 as the numerical part.
Go to answer
What is radiant flux intensity? Show that it follows the inverse square law, F=L/4\pi d^{2}. (long)
Set B Q.6 (a) in 2026, plus two MCQs on F\propto1/d^{2} (Set B MCQ 14, Set D MCQ 16).
Go to answer
Find the distance of a star from its luminosity and radiant flux intensity (Example 23.1). (numerical)
Came word for word as Set B Q.6 (b) in 2026.
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State Wien's displacement law and use it to find the surface temperature of a star. (short)
Asked in both the new (Set D Q.2 (x) OR) and old-syllabus (Q.2 (ii) OR) 2026 papers; numerical problem 2 uses it.
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Explain how the radius of a star is estimated using Wien's law and the Stefan-Boltzmann law. (long)
Book comprehensive question 3 and short question 7; a three-step method that is easy to score on.
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How does red shift lead to the idea that the universe is expanding? (short)
Set B Q.2 (x) OR in 2026; also book short question 8 and comprehensive question 4.
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Use \dfrac{\Delta\lambda}{\lambda}=\dfrac{\Delta f}{f}=\dfrac{v}{c} to find the speed of a galaxy (Example 23.3). (numerical)
The only formula for red shift problems; put the emitted (laboratory) value in the denominator.
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State Hubble's law, give the unit of H_0 and show that the age of the universe is T_0=1/H_0. (long)
Book comprehensive question 5 and MCQ 5; Assignment 23.3 turns it into a numerical.
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What are standard candles? Give two examples and explain how they give distances. (short)
Book short question 4 and comprehensive question 2; links the inverse square law to Hubble's law.
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Quick facts for MCQs: L in W, F in \mathrm{W\,m^{-2}}, H_0 in \mathrm{s^{-1}}; a receding star shows red shift; equal L but dimmer means farther. (mcq)
Five 2026 MCQs came from this unit (Set B MCQ 3, 14, 16; Set D MCQ 5, 16).
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Emission, absorption and continuous spectra, and what each tells us about a star. (short)
Book short question 10; a common 'differentiate' question.
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23Cosmology and the role of radiation

23.1Black body radiationOften examined

Properties of a black body

A model of a black body

Black body radiation curves

23.1Wien's displacement lawOften examined

What the law tells us

Using Wien's law to find the temperature of a star

Derivation Ratio form of Wien's law

23.2Luminosity and the Stefan-Boltzmann lawOften examined

Derivation Luminosity of a star, L=4πr2σT4

Comparing two stars

23.2Radiant flux intensity and the inverse square lawOften examined

Derivation Inverse square law, F=\dfrac{L}{4\pi d^{2}}

23.3Standard candles: a distance indicatorOften examined

How a standard candle gives a distance

Examples of standard candles

The cosmic distance ladder

23.3Estimating the radius of a starOften examined

Derivation Radius of a star from measured quantities

23.4Spectra of light

Three types of spectra

Spectral signatures

23.4Red shift of light from galaxiesOften examined

The Doppler effect for light

Derivation Doppler red shift, \dfrac{\Delta\lambda}{\lambda}=\dfrac{v}{c}

23.4The expanding universeOften examined

How red shift leads to the idea of an expanding universe

23.5Hubble's lawOften examined

23.5The Big Bang theory and the age of the universeOften examined

Evidence for the Big Bang

Age of the universe from Hubble's law

Derivation Age of the universe, T_0=\dfrac{1}{H_0}

Formula sheet

Key terms

Multiple choice questionsp. 181

Short questionsp. 181–182

Comprehensive questionsp. 182

Luminosity

Derivation Luminosity of a star

Radiant flux intensity

Derivation Inverse square law

Comparison

The problem

Standard candle

Method

Examples

Illustration

The cosmic distance ladder

Importance

Idea

Step 1: temperature from Wien's law

Step 2: luminosity from the inverse square law

Step 3: radius from the Stefan-Boltzmann law

Derivation Radius of a star

Example: the Sun

Conclusion

Red shift

Doppler effect and speed of recession

What the observations show

Conclusion: the universe is expanding

Consequence

Statement

Explanation

How it leads to the Big Bang theory

Derivation Age of the universe

Numerical problemsp. 182

FBISE 2026 questions

Exam focus

Derivation Luminosity of a star, $L = 4 π r^{2} σ T^{4}$