Prakash Gupta Heat and Thermodynamics

Black Body Radiation

1 Some Useful Terms

1.1 Total Energy Density

Total Energy Density (u) at any point denotes the total radiant

energy for all wavelengths from 0 to ∞per unit volume around

that point. Its unit is Jm−3.

1.2 Spectral energy Density

Spectral Energy Density (uλ) for the wavelength λis a measure

of the energy per unit volume per unit wavelength. Therefore,

uλdλ denotes the energy per unit volume in the wavelength

range between λand λ+dλ. It is related to toal energy density

through the relation

u=Z∞

uλdλ

1.3 Emissivity

Emissivity(e), also known as total emissive power of a body, is

deﬁned as the total radiant energy of all wavelengths from 0 to

∞emitted per second per unit surface area of the body. Its unit

is Jm−2s−1or W m−2.

1.4 Spectral emissive power

Spectral Emissive power (eλ) of a body for the wavelength λ

signiﬁes the radiant energy per second per unit surface area per

unit range of wavelength. Therefore, eλdλ denotes the energy

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per unit area per second in the wavelength range between λand

λ+dλ. It is related to emissivity through the relation

e=Z∞

eλdλ

1.5 Spectral absorptivity

Spectral absorptivity (aλ) is deﬁned as the fraction of incident

energy absorbed per unit surface area per second at wavelength

λ. Suppose that δQλradiation of wavelength between λand

λ+dλ is incident on a unit area of the surface of the body

per second from all possible directions. If aλδQλis the amount

of radiation absorbed, then aλsigniﬁes the absorptivity of the

body for wavelength λ. Note that aλhas no dimensions; it is a

fraction whose maximum value is unity. Such a body is said to

be perfectly blackbody.

2 Kirchoﬀ’s Law:

It states that the ratio of the emissive power (eλ) to the absorp-

tio power (aλ) for a given wavelength at a given temperature

is the same for all bodies and equal to the emissive power of a

perfectly black-body (Eλ) at that temperature.

i.e., eλ

aλ

=Eλ

Proof:

Let us consider a body placed in an isothermal enclosure. Let

dQ be the amount of radiant energy of wavelength lying between

λand λ+dλ, incident on unit surface area per second. If aλ

is the absorptive power of the body for the wavelength λand

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at the temperature of enclosure, then amount of radiant energy

absorbed by unit surface area of the body per second is aλdQ.

The remainder of the incident energy (1−aλ)dQ will be reﬂected

or transmitted.

If eλbe the emmisive power of the body for wavelength λat the

temperature of enclosure, then the total radiation lying between

λand λ+dλ emitted by unit surface area of the body per second

is eλdλ.

As the body is in temperature equillibrium with the enclosure.

(1 −aλ)dQ +eλdλ =dQ

or, eλdλ =aλdQ

or, eλ

aλ

=dQ

dλ ...(1)

But dQ

dλ depends only on temperature, So, eλ

aλis the same for all

substances for a given temperature.

If the body under consideration is perfectly black, the absorp-

tive power aλ= 1 for all wavelength and eλhas maximum value

which we denote by Eλ.

Therefore, for such a body, we have

Eλdλ =dQ

Eλ=dQ

dλ ...(2)

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Comparing equn (1) and (2), we get

eλ

aλ

=Eλ

This gives Kirchoﬀ’s Law. Here, it follows that good absorbers

of radiation are also good emitters of radiations.

3 Pressure of Radiation

Radiation posses the properties of light, so like light it exerts a

small but deﬁnite pressure on the surface on which it is incident.

Maxwell proved on the basis of his electromagnetic theory that

the pressure is equal to the energy density for normal incidence

on a surface.

Let a photon of energy hν move with velocity of light c. Ac-

cording to the theory of relativity thus energy of the photon

corresponds to a mass mgiven by,

mc2=hν ...(1)or, m=hν

Then the momentum of photon = mass ×velocity

=hν

c2×c

=hν

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where, eis the energy of photon.

Now, the momentum incident on the surface PQ per unit area

per second is given by,

p=Xe

c=Pe

c...(2)

Here, Pe= total incident energy on the surface per unit area

per second = E (say)

So, equn(2) becomes

p=E

c....(3)

If uis the energy density, i.e., energy per unit volume, then the

total energy passing through any area s of the surface normal to

the radiation per second = u×sc

Therefore, the energy ﬂux (energy radiation per unit area per

sec) is

E=ucs

s=uc

So, equn (3) becomes

p=uc

c=u

Here, the momentum per unit area per second i.e., pressure of

radiation is equal to the energy density for normal incidence.

4 Pressure of Diﬀuse Radiation

If the photons move in all possible directions i.e., difussed radi-

ation, the pressure of radiations will be one third of the energy

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density.

Let the beam of radiation make an angle θwith the normal to

the surface QR. The energy incident on the surface PQ per unit

area per second = uc.

If S′be the surface area of plane QR, surface area of the plane

PQ = S′cosθ

The energy incident on surface PQ per second = ucS′cosθ

The energy incident on plane PQ per second is equal to the en-

ergy incident on the plane QR per second.

So, the energy incident on surface QR per second = ucS′cosθ

and, the energy incident on plane QR per unit area per second

=ucS′cosθ

S′=uccosθ

Therefore, the momentum incident on the surface QR per unit

area per second is = uccosθ

c=ucosθ

The component of this momentum in the direction normal to

surface QR = (ucosθ)cosθ

This gives pressure of radiation on the surface QR

i.e., p=ucos2θ

To ﬁnd the pressure of diﬀuse radiation, we must take average

value of cos2θin all direction.

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Hence, p=u¯

cos2θ

But, ¯

cos2θ=1

∴p=u

i.e., Pressure of diﬀused radiation = 1

3×energy density.

5 Stefan’s Boltzman’s law

Q. State stefan’s law of radiation and deduce it from

thermodynamical considerations

Stefan;s law states that the rate of emission of radiant energy

by unit area of a perfectly black body is directly proportional to

he fourth power of its absolute temperature

i.e.,

E∝T4

or, E=σT 4

where σis a constant called Stefan’s constant.

This law refers to the emission only. It can be extended to rep-

resent the net loss of heat in respect to surrounding & may be

enunciated as follows:

A black body at absolute temperature T surrounded by other

blackbody at absolute temperature To, then the amount of heat

lost by the former per unit area per unit time is given by

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E=σ(T4−T4

This law is known as stefan Boltzmann’s law. Thermodynamic

proof:

Let us consider a cylinder enclose ABCD of uniform cross-section

with perfectly reﬂecting walls and provided with piston P.

Let it be ﬁll with diﬀuse radiation of density uat temperature

Then, total internal energy of radiation inside the enclosure is

U=uV (1)

where V = volume of enclosure

Let a small amount of heat dQ ﬂow in the enclosure from out-

side and piston moves out so that the volume changes by a small

amount dV . If dU be change in internal energy and dW is the

external workdone due to change in volume dV , then from 1st

law of thermodynamics

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dQ =dU +dW

or, dQ =d(uV ) + pDV (where p= pressure of radiation)

or, dQ =d(uV ) + u

3dV

or, dQ =uDv +V du +1

3udV

or, dQ =V du +4

3udV (2)

From 2nd law of thermodynamics, change in entropy of radia-

tion is

dS =dQ

Tdu +4

3u

TdV ......(3)

∴s=f(u, V )

or, dS =∂S

∂udu +∂S

∂V dV ....(4)

comparing (3) and (4), we get

∂S

∂u =V

∂S

∂V =4u

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As dS is a perfect diﬀerential, we have

∂2S

∂u∂V =∂2S

∂V ∂u

∂

∂u ∂S

∂V =∂

∂V ∂S

∂u

∂

∂u 4u

3T=∂

∂V V

T

since T is independent of V and function of u, we get

3T−4u

3T2

∂T

∂u =1

3T=4u

3T2

∂T

∂u

u= 4∂T

Integrating, we get

log u= 4 log T+ log A(log Ais constant of integration)

log u= log AT 4

or, u =AT 4

Now, total rate of emission of radiant energy per unit area is

related to energy density as

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E=1

4uc, (where cis the speed of light)

∴E=1

4AcT 4

or, E =σT 4

where σis a constant called stefan’s constant having value σ=

5.672 ×10−8Jm−2s−1K−4

6 Spectrum of Black body radiation

The experimental arrangement for investigating distribution of

energy in the spectrum of Black body Radiation is shown in ﬁg-

ure below.

The radiation from black body pass through the slit S1and fall

on the reﬂector M1. After being reﬂected, the parallel beam

of radiation fall on a ﬂuorspar prism ABC placed on the turn

table of the spectrometer. The emergent light is focused by the

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reﬂector M2on a bolometer placed behind the slit S2.

The turn table is rotated slowly so that diﬀerent parts of radia-

tion spectrum successively fall on the bolometer and the corre-

sponding deﬂection in galvanmeter connected in the bolometer

circuit is noted.

The intensity of each line is proportional to the deﬂection in

the galvanometer. A curve is drawn between intensity and the

wavelength at diﬀerent temperatures.

Results:

1) The energy is not uniformly distributed in the radiation spec-

trum of a black body

2) At a given temperature, the intensity of radiation increases

with wavelength becoms maximum at a particular wavelength

and then decreases.

3) An increase in temperature causes decrease in wavelength of

maximum energy λm.

Thus λmis inversely proportional to absolute temperature

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i.e., λm∝1

or, λmT=constant

This is Wein’s Displacement law.

4) An increase in temperature causes increase in energy emission

for all wavelength

5) The area under the curve represents total energy emitted by

black body at a particula temperature. This area increases with

increase in temperature. It is found that

E∝T4(Stefan’s Law)

Wein’s Displacement law:

It states that the product of the wavlength corresponding to

maximum energy λmand the absolute temperature Tis con-

stant i.e.,

λmT=constant

The constant is called Wein’s displacement constant having value

0.2896 ×10−2mK.

7 Planck’s Radiation Law

Q. Write down the physical basis on which Planck’s radiation

law is obtained. Hence derive a relation for energy density within

the wavelength range λand λ+dλ, in the spectrum of blackbody

radiation.

Planck found an empirical formula to explain the experimentally

observed distribution of energy in the spectrum of a blackbody.

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This formula may be deduced using the following postulates:

1) A black body radiation chamber is ﬁlled up not pnly with

radiation, but also with simple harmonic oscillator or resonators

of the molecular dimensions which can have value of energy only

given by

E=nhν

where νis the frequency of oscillator, his Planck’s constant and

n is an integer. i.e., n = 0,1,2,3 ....

2) The oscillator cannot radiate or absorb energy continously but

an oscillator of frequency νcan only radiate or absorb energy in

units or quanta of magnitude hν.

i.e., the exchange of energy between radiation and matter cannot

take place continously, but are limited to discrete set of value

0, hν, 2hν, .....nhν .

The average energy of a Planck’s Oscillator is given by

E=hν

ehν/kT −1

The no. of resonators per unit volume in the frequency range ν

and ν+dν is given by

Ndν =8πν2

c3dν, where, c is the speed of light

Thus, the energy density i.e., total energy per unit volume be-

longing to the range is given by

Eνdν =8πν2

c3.hν

ehν/kT −1dν

or, Eνdν =8πhν3

c3.dν

ehν/kT −1...(1)

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This is Planck’s radiation law

Since ν=c

λ, hence |dν|=

−c

λ2dλ



i.e.,

Eλdλ =8πh

c3c

λ3

ehc/λkT −1−c

λ2dλ

∴Eλdλ =8πhc

λ5.1

ehc/λkT −1dλ

which gives the enrgy density for wavelength range λand λ+dλ

in the spectrum of a black body.

8 Rayleigh - Jean’s Law of spectral distribution of energy

The no. of modes of vibration per unit volume in the frequency

range νand ν+dν is

Nνdν =8πν2

c3dν

Rayleigh and Jeans assumed that the average energy of an os-

cillator as

E=kT

Thus, the energy density within the frequency range νand ν+dν

is given by

Uνdν = no. of modes of vibration per unit vol. ×avg. energy per mode of vibration

or, Uνdν =8πν2

c3dν ×kT

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This is Rayleigh-Jean’s Law in terms of frequency.

In terms of wavelength

Uνdν =8π

c3c

λ2−c

λdλkT

=8πkT

λ4dλ

This law explains the experimental measurement of the energy

distribution at long wavelength fairly well, but it fails for short

wavelengths.

9 Q. Deduce Wein’s Displacement Law, Rayleigh-Jean’s law and Stefan’s

Law from Planck’s radiation formula

Planck’s formula is given by

Eλdλ =8πhc

λ5.1

ehc/λkT −1dλ ...(1)

diﬀerentiating partially with respect to λ, we get

∂Eλ

∂λ =8πhc

ehc/λkT −1.−5

λ6+8πhc

λ5.

λ2kT ehc/λkT

(ehc/λkT −1)2

=−40πhc

ehc/λkT −1.1

λ6+8πh2c2

λ7kT .ehc/λkT

(ehc/λkT −1)2

For maximum energy of emission ∂Eλ

∂λ = 0

or, −40πhc

ehc/λkT −1.1

λ6+8πh2c2

λ7kT .ehc/λkT

(ehc/λkT −1)2= 0

or, 8πhc

(ehc/λkT −1)λ6−5 + hc

λkT .ehc/λkT

(ehc/λkT −1)= 0

or, −5 + hc

λkT .ehc/λkT

(ehc/λkT −1) = 0

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Put hc

lambdakT =x, then above equation become

−5 + xex

ex−1= 0

or, xex

ex−1= 5

Solving by the method of approximation, we get

x=hc

λkT = 4.965

The energy is maximum when λ=λm

∴hc

λmkT = 4.965

or, λmT=hc

4.965k=constant

This is Wein’s Displacement law.

Again, from equation (1)

Eλdλ =8πhc

λ5.1

ehc/λkT −1dλ

For longer wavelength,

ehc/λkT =1 + hc

λkT 

Hence, Planck’s law reduces to

Eλdλ =8πhc

λ5.1

1 + hc

λkT −1dλ

=8πhc

λ5.1

λkT

dλ

or, Eλdλ =8πkT

λ4dλ

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This is Rayleigh Jean’s Law

Also, Planck’s Law in terms of frequency is

Eνdν =8πν2

c3.hν

ehν/kT −1dν

Then, total radiant energy overall frequencies will be

E=Z∞

Eνdν =8πh

c3Z∞

ν3

ehν/kT −1dν

Put hν

kT =x, so that ν=kT

hxand dν =kT

hdx

Then E=8πk4T4

c3h3Z∞

x3dx

ex−1

or, E=8πk4T4

c3h3fracπ415

or, E=8π5k4

15c3h3T4

∴E=AT 4

where A=8π5k4

15c3h3

But Ac

4=σ(Stefan’s Constant)

∴E=σT 4

This is Stefan’s Law