Wein’s Displacement Law

PhysicsThermal Properties Of MatterFor JEE aspirants

Wien's displacement law states that the wavelength at which a black body radiates most strongly is inversely proportional to its absolute temperature: $λ_{m} T = b = 2.9 \times 1 0^{- 3} m K$ . Hotter bodies glow bluer. This page builds the whole theory of thermal radiation around it: black bodies, absorptive and emissive power, Kirchhoff's law, the Stefan-Boltzmann law and the black-body spectrum. Wien's displacement law and Stefan's law are regular one-mark questions in NEET and JEE Main.

On this page1Thermal radiation2Prevost's theory3a + r + t = 14Black body5Emissive power6Kirchhoff's law7Stefan-Boltzmann law8Black-body spectrum9Wien's law10Sun and greenhouse

Key Formulas - Quick Reference

★ Must learnWien's displacement law: $λ_{m} T = b$ , $b = 2.9 \times 1 0^{- 3} m K$ ( $0.29 cm K$ )
★ Must learnStefan-Boltzmann law: $E = σ T^{4}$ (black body); $P = e σ A T^{4}$ ; $σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$
★ Must learnNet loss to surroundings at $T_{0}$ : $P_{net} = e σ A (T^{4} - T_{0}^{4})$
$a + r + t = 1$ ; black body $a = 1$ ; emissivity $e = \frac{E}{E _{black}}$ ( $0 \leq e \leq 1$ )
Kirchhoff's law: $\frac{E _{λ}}{a _{λ}}$ is the same for all bodies at a given $T$ $= E_{λ}$ (black); so $e = a$
Total emissive power $E = \int_{0}^{\infty} E_{λ} d λ$ ; peak height $E_{λ, max} \propto T^{5}$
Rate of cooling by radiation: $- \frac{d T}{d t} = \frac{e σ A}{m s} (T^{4} - T_{0}^{4})$
Solar flux at distance $R$ from a star of radius $r$ : $S = σ T^{4} \frac{r ^{2}}{R ^{2}}$

1. Thermal Radiation

Radiation is the transfer of heat from one place to another without heating the medium in between. The radiation emitted by a body because of its temperature is thermal radiation. It is electromagnetic: oscillating electric and magnetic fields perpendicular to each other, produced by vibrating charged particles in atoms and molecules.

Figure 1: Thermal radiation is electromagnetic radiation emitted because of temperature. Bodies near room temperature emit mainly infrared; only above about

800 K

does enough visible light appear for a body to glow.

Every body above $0 K$ emits thermal radiation and absorbs some of what falls on it.
It travels in straight lines at the speed of light, needs no medium and crosses vacuum (this is how sunlight reaches us).
Like light, it can be reflected, refracted, diffracted and polarised; from a point source its intensity falls as $\frac{1}{r ^{2}}$ .
A hotter body emits more, and its most intense wavelength moves from long (infrared) towards short (visible, then ultraviolet). A heated iron rod glows dull red, then orange, then white.

1.1 Prevost's theory of exchange

All bodies radiate at all temperatures, and at the same time absorb radiation from their surroundings. If a body emits more than it absorbs, it cools; if less, it warms; at the same temperature as its surroundings it emits and absorbs at equal rates, so its temperature stays constant. Equilibrium is dynamic, not a stop.

Figure 2: Every body radiates all the time (Prevost). A body hotter than its surroundings emits more than it absorbs and cools; at equal temperatures emission and absorption balance, so there is no net exchange.

2. Absorption, Reflection and Transmission

Radiation energy $Q$ falling on a body is partly reflected ( $Q_{r}$ ), partly absorbed ( $Q_{a}$ ) and partly transmitted ( $Q_{t}$ ):

Q = Q_{r} + Q_{a} + Q_{t} \Rightarrow r + a + t = 1

Figure 3: Energy conservation for incident radiation: reflectance

r

, absorptance (absorptive power)

a

and transmittance

t

satisfy

r + a + t = 1

. A perfect black body has

a = 1

r = t = 0

Body	$r$	$a$	$t$
Perfect black body	0	1	0
Perfect reflector	1	0	0
Perfectly transparent (diathermanous)	0	0	1
Opaque body	$r$	$1 - r$	0

Absorptive power (absorptance) $a$ of a surface: the fraction of the incident radiation it absorbs, $a = \frac{energy absorbed}{energy incident}$ . It has no unit. For a black body $a = 1$ .

3. The Perfectly Black Body

★ Must learn

A perfectly black body absorbs all radiation of every wavelength falling on it; it neither reflects nor transmits any. When heated it emits the maximum possible radiation at every wavelength for its temperature: it is the ideal emitter as well as the ideal absorber.

No real surface is perfectly black: lamp black and platinum black absorb about 99%. Fery's black body is a double-walled hollow sphere, blackened inside, with a small hole and a conical projection opposite the hole. Radiation entering the hole is reflected many times inside and absorbed almost entirely, so the hole acts as a black body. When the sphere is heated, radiation coming out of the hole is black-body radiation. A small opening into a furnace behaves the same way.

Figure 4: Fery's black body. Radiation entering the hole suffers many reflections (drawn with the true law of reflection) and is absorbed at each, so almost none escapes: the hole is a near-perfect absorber and, when heated, a perfect emitter.

4. Emissive Power, Spectral Emissive Power and Emissivity

Quantity	Definition	Unit
Emissive power $E$	Energy radiated per unit time per unit area of the surface (over all wavelengths)	$W m^{- 2}$
Spectral emissive power $E_{λ}$	Emissive power per unit wavelength range at $λ$ : $E = \int_{0}^{\infty} E_{λ} d λ$ , $E_{λ} = \frac{d E}{d λ}$	$W m^{- 3}$
Emissivity $e$	$\frac{E of the body}{E of a black body at the same T}$ ; $0 \leq e \leq 1$	none
Absorptive power $a$	Fraction of incident radiation absorbed	none

The booklet's definition per unit solid angle, $E = \frac{Δ U}{Δ A Δ ω Δ t}$ (along the normal), is the radiance; in exam problems "emissive power" means the total power per unit area, $σ T^{4}$ for a black body.

5. Kirchhoff's Law of Radiation

★ Must learn

Kirchhoff's law: at a given temperature, the ratio of spectral emissive power to spectral absorptive power is the same for all bodies and equals the spectral emissive power of a black body at that temperature:

\frac{E _{λ}}{a _{λ}} = E_{λ} (black body) \Rightarrow e = a

Good absorbers are good emitters; good reflectors are poor emitters.

Figure 5: Kirchhoff's law. At a given temperature the ratio

\frac{E _{λ}}{a _{λ}}

is the same for all bodies and equals

E_{λ}

of a black body. A good absorber is a good emitter; a good reflector is a poor emitter (vacuum flasks and fire suits are shiny).

A black spot on a white china plate looks dark when cold but glows brightest when the plate is heated in a furnace.
Vacuum-flask walls, fire-fighters' suits and emergency blankets are shiny: poor emitters and poor absorbers.
Cooking pots are blackened at the bottom (good absorbers); radiator fins are often painted black (good emitters).
Dark Fraunhofer lines in the solar spectrum: cooler gases in the Sun's atmosphere absorb exactly the wavelengths they would emit.

Quick Recall: tap to check

What are r, a and t for a perfectly black body?

r = 0, a = 1, t = 0.

Why is the hole in Fery's black body, not the sphere, the black body?

Radiation entering the hole is almost completely absorbed after many internal reflections.

A body has absorptive power 0.6. What is its emissivity at the same temperature?

0.6 (Kirchhoff's law: e = a).

6. Stefan-Boltzmann Law

★ Must learn

The total energy radiated per second per unit area by a black body is proportional to the fourth power of its absolute temperature:

E = σ T^{4}, σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}

For a real body of emissivity $e$ and area $A$ : $P = e σ A T^{4}$ .

Figure 6: Stefan-Boltzmann law

E = σ T^{4}

(

T

in kelvin). Doubling the absolute temperature multiplies the radiated power by

2^{4} = 16

; a log-log plot is a straight line of slope 4.

A body at $T$ in surroundings at $T_{0}$ emits $e σ A T^{4}$ and absorbs $e σ A T_{0}^{4}$ (by Kirchhoff's law its absorptance equals $e$ ), so the net rate of heat loss and the rate of cooling are

P_{net} = e σ A (T^{4} - T_{0}^{4}), - \frac{d T}{d t} = \frac{e σ A}{m s} (T^{4} - T_{0}^{4})

Exam Trick

Ratios, not numbers. $P \propto A T^{4} \propto r^{2} T^{4}$ for spheres. Rate of cooling $\propto \frac{A}{m s} \propto \frac{r ^{2}}{ρ r ^{3} s} \propto \frac{1}{r}$ at the same temperature: a smaller sphere of the same material cools faster. Always convert to kelvin: $2 7^{\circ} C \to 300 K$ , $32 7^{\circ} C \to 600 K$ gives $P$ ratio $2^{4} = 16$ .

7. Black-Body Spectrum and Wien's Displacement Law

The energy radiated by a black body is spread over all wavelengths but not evenly. Measuring $E_{λ}$ against $λ$ at several temperatures (Lummer and Pringsheim) gives curves with these features:

At each temperature the energy is small at very short and very long wavelengths and has a maximum at one wavelength $λ_{m}$ .
The area under a curve is the total emissive power; it grows as $T^{4}$ (Stefan's law).
A hotter curve lies above a cooler one at every wavelength.
As $T$ rises, $λ_{m}$ shifts to shorter wavelengths, and the peak height grows as $T^{5}$ .

Figure 7: Black-body spectra computed from Planck's law. As

T

rises: (i)

E_{λ}

increases at every wavelength; (ii) the total area

E = \int E_{λ} d λ \propto T^{4}

(Stefan); (iii) the peak shifts to shorter wavelength,

λ_{m} \propto \frac{1}{T}

(Wien), and its height grows as

T^{5}

★ Must learn

Wien's displacement law: the wavelength of maximum emission is inversely proportional to the absolute temperature,

λ_{m} \propto \frac{1}{T} \Rightarrow λ_{m} T = b = 2.9 \times 1 0^{- 3} m K

$b$ is Wien's constant, $2.898 \times 1 0^{- 3} m K = 0.2898 cm K$ .

Figure 8:

λ_{m} = \frac{b}{T}

is a rectangular hyperbola. Cool stars peak in the red or infrared and look red; hot stars peak in the ultraviolet and look blue-white. The Sun peaks in the visible (about

500 nm

Body	T	$λ_{m}$	Region
Human body	$310 K$	$9.4 μ m$	far infrared (thermal cameras)
Red-hot iron	$1000 K$	$2.9 μ m$	infrared (dull red glow)
Tungsten filament	$2800 K$	$1.0 μ m$	near infrared (bulbs waste heat)
Sun's surface	$5800 K$	$0.50 μ m$	visible (green-yellow)
Blue star	$12000 K$	$0.24 μ m$	ultraviolet

Stefan-Boltzmann law

How much is radiated: $E = σ T^{4}$ (area under the curve). Doubling $T$ gives 16 times the power.

Wien's displacement law

Where the peak is: $λ_{m} = \frac{b}{T}$ . Doubling $T$ halves $λ_{m}$ (colour shifts towards blue).

JEE Advanced

Both laws follow from Planck's radiation law, $E_{λ} = \frac{2 π h c ^{2}}{λ ^{5}} \cdot \frac{1}{e ^{h c / λk T} - 1}$ (per unit area, all directions). Setting $\frac{d E _{λ}}{d λ} = 0$ gives $\frac{h c}{λ _{m} k T} = 4.965$ , i.e. Wien's law with $b = \frac{h c}{4.965 k}$ ; integrating over $λ$ gives $σ = \frac{2 π ^{5} k ^{4}}{15 h ^{3} c ^{2}}$ . Combining the two laws: $E \propto T^{4} \propto \frac{1}{λ _{m}^{4}}$ .

8. Solar Radiation and the Greenhouse Effect

The Sun radiates almost as a black body. Its power spreads over a sphere of radius $R$ (Sun-Earth distance), so the energy received per second per unit area at the Earth, normal to the rays (the solar constant), is $S = σ T^{4} (\frac{r _{s}}{R})^{2} \approx 1.36 kW m^{- 2}$ . Measuring $S$ and the Sun's angular size gives the Sun's surface temperature, about $5800 K$ .

Figure 9: The Sun's total power

σ T^{4} \cdot 4 π (D /2)^{2}

spreads over a sphere of radius

R

, so the solar flux at Earth is

S = σ T^{4} (\frac{D}{2 R})^{2}

. With

\frac{D}{R} = 0.5 3^{\circ}

in radians this gives

T

(drawing not to scale).

Greenhouse effect. The Sun's radiation, peaking in the visible, passes through glass and the atmosphere. The warmed ground re-radiates at about $300 K$ , i.e. in the far infrared near $10 μ m$ , which glass, carbon dioxide and water vapour absorb and partly send back down. Heat is trapped, keeping the Earth's average surface temperature near $1 5^{\circ} C$ instead of about $- 1 8^{\circ} C$ ; extra $C O_{2}$ strengthens the effect (global warming).

Figure 10: Greenhouse effect. By Wien's law the Sun (about

5800 K

) emits mostly visible light, which passes through; the ground (about

300 K

) emits infrared near

10 μ m

, which glass and greenhouse gases block.

Figure 11: Choosing between Wien's law and the Stefan-Boltzmann law. Every radiation formula needs absolute temperature.

Figure 12: Mind map of this concept for quick revision.

9. Solved Examples

Solved Example 1

The Earth receives solar radiation at

8.2 J cm^{- 2} min^{- 1}

. Treating the Sun as a black body that subtends an angle of

0.5 3^{\circ}

at the Earth, find the Sun's surface temperature. (

σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}

)

Solution:

Let the Sun's diameter be $D$ and its distance $R$ : $\frac{D}{R} \approx 0.53 \times \frac{π}{180} = 9.25 \times 1 0^{- 3}$ .

Power emitted $= σ T^{4} \cdot 4 π (\frac{D}{2})^{2} = π D^{2} σ T^{4}$ . At distance $R$ it spreads over $4 π R^{2}$ , so the flux is $\frac{σ T ^{4}}{4} (\frac{D}{R})^{2}$ .

Flux $= \frac{8.2}{1 0 ^{- 4} \times 60} = 1.37 \times 1 0^{3} W m^{- 2}$ . So $T^{4} = \frac{4 \times 1.37 \times 1 0 ^{3}}{( 5.67 \times 1 0 ^{- 8} ) ( 9.25 \times 1 0 ^{- 3} ) ^{2}} = 1.13 \times 1 0^{15}$ .

Answer: $T \approx 5.8 \times 1 0^{3} K$ (about $5790 K$ ).

Solved Example 2

Find the wavelength of maximum emission for (a) the Sun,

T = 5800 K

, and (b) the human body,

T = 310 K

. (

b = 2.9 \times 1 0^{- 3} m K

)

Solution:

(a) $λ_{m} = \frac{2.9 \times 1 0 ^{- 3}}{5800} = 5.0 \times 1 0^{- 7} m$ . (b) $λ_{m} = \frac{2.9 \times 1 0 ^{- 3}}{310} = 9.4 \times 1 0^{- 6} m$ .

Answer: (a) $500 nm$ (visible); (b) $9.4 μ m$ (infrared), which is why thermal cameras work in the infrared.

Solved Example 3

The spectrum of a star peaks at

350 nm

. Estimate its surface temperature.

Solution:

$T = \frac{b}{λ _{m}} = \frac{2.9 \times 1 0 ^{- 3}}{350 \times 1 0 ^{- 9}}$ .

Answer: $T \approx 8.3 \times 1 0^{3} K$ (a bluish-white star).

Solved Example 4

The peak wavelength of a black body shifts from

700 nm

350 nm

when it is heated. By what factor does the power it radiates change?

Solution:

$λ_{m} \propto \frac{1}{T}$ : halving $λ_{m}$ doubles $T$ . Then $P \propto T^{4}$ grows by $2^{4}$ .

Answer: 16 times.

Solved Example 5

A sphere of radius

10 cm

and emissivity

0.8

is at

52 7^{\circ} C

in a room at

2 7^{\circ} C

. Find (a) the power it emits and (b) its net rate of heat loss.

Solution:

$T = 800 K$ , $T_{0} = 300 K$ , $A = 4 π (0.1)^{2} = 0.126 m^{2}$ .

(a) $P = e σ A T^{4} = 0.8 (5.67 \times 1 0^{- 8}) (0.126) (800)^{4} = 2.33 \times 1 0^{3} W$ .

(b) $P_{net} = e σ A (T^{4} - T_{0}^{4}) = 0.8 (5.67 \times 1 0^{- 8}) (0.126) (4.096 \times 1 0^{11} - 8.1 \times 1 0^{9})$ .

Answer: (a) $\approx 2.3 kW$ ; (b) $\approx 2.3 kW$ ( $2.29 kW$ ).

Solved Example 6

Two spheres of the same material have radii

r

and

2 r

and temperatures

2 T

and

T

respectively (both in K). Find the ratio of the powers they radiate,

P_{1} : P_{2}

Solution:

$P \propto r^{2} T^{4}$ : $\frac{P _{1}}{P _{2}} = \frac{r ^{2} ( 2 T ) ^{4}}{( 2 r ) ^{2} T ^{4}} = \frac{16}{4}$ .

Answer: $4 : 1$ .

Solved Example 7

Two solid spheres of the same material, radii

r

and

2 r

, are at the same temperature in the same surroundings. Compare their initial rates of cooling.

Solution:

$- \frac{d T}{d t} = \frac{e σ A ( T ^{4} - T _{0}^{4} )}{m s}$ with $\frac{A}{m} = \frac{4 π r ^{2}}{\frac{4}{3} π r ^{3} ρ} = \frac{3}{r ρ}$ , so the rate $\propto \frac{1}{r}$ .

Answer: $2 : 1$ : the smaller sphere cools twice as fast.

Solved Example 8

The filament of a

100 W

bulb runs at

2500 K

with emissivity

0.35

. Find its surface area (neglect absorption from the room).

Solution:

$A = \frac{P}{e σ T ^{4}} = \frac{100}{0.35 ( 5.67 \times 1 0 ^{- 8} ) ( 2500 ) ^{4}} = \frac{100}{775}$ .

Answer: $A \approx 1.3 \times 1 0^{- 4} m^{2} = 1.3 cm^{2}$ .

Solved Example 9

A glass plate reflects

10%

and absorbs

20%

of the radiation falling on it. What fraction does it transmit, and what is its emissivity?

Solution:

$t = 1 - r - a = 1 - 0.1 - 0.2$ . By Kirchhoff's law $e = a$ .

Answer: $t = 0.7$ ; $e = 0.2$ .

Solved Example 10

Taking the solar constant as

1366 W m^{- 2}

and treating the Earth as a black body in equilibrium (no atmosphere, no reflection), estimate the Earth's mean temperature.

Solution:

Absorbed: $S π R_{E}^{2}$ (the disc facing the Sun). Emitted: $σ T^{4} \cdot 4 π R_{E}^{2}$ . Equating, $T^{4} = \frac{S}{4 σ} = \frac{1366}{4 ( 5.67 \times 1 0 ^{- 8} )}$ .

Answer: $T \approx 279 K$ ( $\approx 6^{\circ} C$ ). Reflection lowers this to about $255 K$ ; the greenhouse effect raises the real value to about $288 K$ .

Solved Example 11

If the temperature of a black body is increased so that its

λ_{m}

decreases by

20%

, its total emissive power increases by about
(A)

59%

(B)

144%

(C)

80%

(D)

244%

Solution:

Answer: (B). $λ_{m}^{'} = 0.8 λ_{m} \Rightarrow T^{'} = \frac{T}{0.8} = 1.25 T$ . $E^{'} = (1.25)^{4} E = 2.44 E$ , an increase of $144%$ .

Practice Questions

At what temperature does a black body's peak lie at $1 μ m$ ? ( $b = 2.9 \times 1 0^{- 3} m K$ )Answer: $2900 K$
The temperature of a black body rises from $2 7^{\circ} C$ to $32 7^{\circ} C$ . By what factor does its radiated power increase?Answer: 16
A black body at $1000 K$ has area $1 cm^{2}$ . How much energy does it radiate per minute?Answer: about $340 J$
Why does a piece of red glass look red in daylight but glow green when heated in a furnace?Answer: It absorbs green strongly (so it looks red); by Kirchhoff's law it emits green strongly when hot
Two stars appear red and blue. Which is hotter and why?Answer: The blue star: $λ_{m}$ is shorter, so $T = \frac{b}{λ _{m}}$ is higher
A body's surface temperature is doubled. By what factor does $λ_{m}$ change, and the peak height $E_{λ, max}$ ?Answer: halves; 32 times
Why do white clothes keep you cooler in summer than black clothes?Answer: White reflects most radiation (small $a$ ); black absorbs most

Common Mistakes to Avoid

Watch out

Using temperatures in $^{\circ} C$ in $T^{4}$ or in $λ_{m} T = b$ . Always use kelvin.
Taking Wien's constant as $0.282 cm K$ ; the correct value is $0.2898 cm K$ ( $2.9 \times 1 0^{- 3} m K$ ).
Using $P = e σ A T^{4}$ for the net loss. Net loss is $e σ A (T^{4} - T_{0}^{4})$ , not $e σ A (T - T_{0})^{4}$ .
Thinking a black body looks black when hot. A hot black body is the brightest emitter; the Sun is nearly a black body.
Saying $λ_{m}$ of a body increases with temperature. It decreases: hotter means bluer.
Confusing Stefan's law (how much energy, area under the curve) with Wien's law (where the peak is).
Forgetting that the Sun-Earth power spreads over $4 π R^{2}$ : the flux falls as $\frac{1}{R ^{2}}$ .
Assuming good reflectors are good emitters. By Kirchhoff's law they are poor emitters.

Frequently Asked Questions

What is Wien's displacement law?

Wien's displacement law says that the wavelength at which a black body emits most strongly is inversely proportional to its absolute temperature: $λ_{m} T = b$ , with $b \approx 2.9 \times 1 0^{- 3} m K$ . As a body gets hotter, its peak moves to shorter wavelengths and its colour shifts towards blue.

What is the Stefan-Boltzmann law?

The Stefan-Boltzmann law says the total energy radiated per second per unit area by a black body is $σ T^{4}$ , where $T$ is the absolute temperature and $σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ . A real body of area $A$ and emissivity $e$ radiates $P = e σ A T^{4}$ .

What is a perfectly black body?

A perfectly black body absorbs all the radiation of every wavelength that falls on it, reflecting and transmitting none. When heated it is also the best possible emitter. A small hole in a blackened hollow enclosure, such as Fery's black body, behaves almost exactly like one.

What is Kirchhoff's law of radiation?

Kirchhoff's law says that at a given temperature the ratio of emissive power to absorptive power is the same for all bodies and equals the emissive power of a black body. So a good absorber of a wavelength is also a good emitter of it, and emissivity equals absorptivity.

Why do hot objects change colour as they get hotter?

By Wien's law the peak of emission moves to shorter wavelengths as temperature rises. A heated body first emits only infrared, then glows dull red, orange, yellow and finally white or bluish-white as more of the visible spectrum, including shorter wavelengths, is emitted strongly.

How is the temperature of the Sun or a star found?

Either from Wien's law, by measuring the wavelength of peak emission and using $T = \frac{b}{λ _{m}}$ , or from Stefan's law, by measuring the energy received per unit area at the Earth and the angle the Sun subtends. Both give about $5800 K$ for the Sun.

How is Wien's law asked in NEET?

NEET asks direct questions on Wien's law and Stefan's law: finding the peak wavelength or temperature, the ratio of powers when temperature changes, colour of stars, and reading black-body radiation curves for the temperature order and area under the curve.

What radiation questions come in JEE Main?

JEE Main combines Stefan's law with Wien's law, for example power change when the peak wavelength shifts, net radiation loss to surroundings, rates of cooling of spheres of different sizes, the Sun's temperature from the solar constant, and Kirchhoff's law with absorptivity.

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