1. JEE
  2. IIT JEE Study Material
  3. Stefan Boltzmann Law

Stefan Boltzmann law - Formula, Derivation

What is Stefan Boltzmann Law?

According to Stefan Boltzmann law, the amount of radiation emitted per unit time from an area A of a black body at absolute temperature T is directly proportional to the fourth power of the temperature.

u = sAT4 . . . . . . (1)

where s is Stefan’s constant = 5.67 × 10-8 W/m2 k4

A body which is not a black body absorbs and hence emit less radiation, given by equation (1)

For such a body, u = e σ AT4 . . . . . . . (2)

where e = emissivity (which is equal to absorptive power) which lies between 0 to 1.

With the surroundings of temperature T0, net energy radiated by an area A per unit time.

Δu = u – uo = eσA [T4 – T04] . . . . . . (3)

⇒ Also Read

Stefan Boltzmann Law relates the temperature of the blackbody to the amount of the power it emits per unit area. The law states that;

“The total energy emitted/radiated per unit surface area of a blackbody across all wavelengths per unit time is directly proportional to the fourth power of the black body’s thermodynamic temperature. ”

⇒ ε = σT4

Derivation of Stefan Boltzmann Law

The total power radiated per unit area over all wavelengths of a black body can be obtained by integrating Plank’s radiation formula. Thus, the radiated power per unit area as a function of wavelength is:

dPdλ1A=2πhc2λ5(ehcλkT−1)\frac{dP}{d\lambda }\frac{1}{A}=\frac{2\pi hc^{2}}{\lambda ^{5}\left ( e^{\frac{hc}{\lambda kT}}-1 \right )}

Where,

  • P is Power radiated.
  • A is the surface area of a blackbody.
  • λ is the wavelength of emitted radiation.
  • h is Planck’s constant
  • c is the velocity of light
  • k is Boltzmann’s constant
  • T is temperature.

On simplifying Stefan Boltzmann equation, we get:

d(PA)dλ=2πhc2λ5(ehcλkT−1)\frac{d\left ( \frac{P}{A} \right )}{d\lambda }=\frac{2\pi hc^{2}}{\lambda ^{5}\left ( e^{\frac{hc}{\lambda kT}}-1 \right )}

On integrating both the sides with respect to λ and applying the limits we get;

∫0∞d(PA)dλ=∫0∞[2πhc2λ5(ehcλkT−1)]dλ\int_{0}^{\infty }\frac{d\left ( \frac{P}{A} \right )}{d\lambda }=\int_{0}^{\infty }\left [ \frac{2\pi hc^{2}}{\lambda ^{5}\left ( e^{\frac{hc}{\lambda kT}}-1 \right )} \right ]d\lambda

The integrated power after separating the constants is:

PA=2πhc2∫0∞[dλλ5(ehcλkT−1)]\frac{P}{A}=2\pi hc^{2}\int_{0}^{\infty }\left [ \frac{d\lambda }{\lambda ^{5}\left ( e^{\frac{hc}{\lambda kT}}-1 \right )} \right ] —(1)

This can be solves analytically by substituting:

x = hcλkT\frac{hc}{\lambda kT}

Therefore, dx=−hcλ2kTdλdx=-\frac{hc}{\lambda^{2} kT}d\lambda

⇒ h=xλkTch=\frac{x\lambda kT}{c}

⇒ c=xλkThc=\frac{x\lambda kT}{h}

⇒ dλ=−λ2kThcdxd\lambda=-\frac{\lambda^{2} kT}{hc}dx

As a result of substituting them in equation (1)

⇒PA=2π(xλkTc)(xλkTh)2∫0∞[(−λ2kThc)dxex−1]\Rightarrow \frac{P}{A}=2\pi \left ( \frac{x\lambda kT}{c} \right )\left (\frac{x\lambda kT}{h} \right )^{2}\int_{0}^{\infty }\left [ \frac{\left ( -\frac{\lambda^{2}kT}{hc} \right )dx}{e^{x}-1} \right ]

= 2π(x3λ5k4T4h3c2λ5)∫0∞[dxex−1]2\pi \left ( \frac{x^{3}\lambda^{5} k^{4}T^{4}}{h^{3}c^{2}\lambda ^{5}} \right )\int_{0}^{\infty }\left [ \frac{dx}{e^{x}-1} \right ]

=2π(kT)4h3c2∫0∞[x3ex−1]dx\frac{2\pi \left ( kT \right )^{4}}{h^{3}c^{2}}\int_{0}^{\infty }\left [ \frac{x^{3}}{e^{x}-1} \right ]dx

The above equation can be comparable to the standard form of integral:

∫0∞[x3ex−1]dx=π415\int_{0}^{\infty }\left [ \frac{x^{3}}{e^{x}-1} \right ]dx=\frac{\pi ^{4}}{15}

Thus, substituting the above result we get,

PA=2π(kT)4h3c2π415\frac{P}{A}=\frac{2\pi \left ( kT \right )^{4}}{h^{3}c^{2}}\frac{\pi ^{4}}{15} ⇒PA=(2k4π515h3c2)T4\Rightarrow \frac{P}{A}=\left ( \frac{2 k^{4}\pi ^{5}}{15h^{3}c^{2}}\right )T^{4}

On further simplifying we get,

⇒ P/A = σ T4

Thus, we arrive at a mathematical form of Stephen Boltzmann law:

⇒ ε = σT4

Where,

ε = P/A

σ=(2k4π515h3c2)=(5.670×108  wattsm2K4)\sigma =\left ( \frac{2 k^{4}\pi ^{5}}{15h^{3}c^{2}}\right )=\left ( 5.670\times 10^{8}\;\frac{watts}{m^{2}K^{4}} \right )

This quantum mechanical result could efficiently express the behavior of gases at low temperature, that classical mechanics could not predict!

Problems on Stefan Boltzmann Law

Example: A body of emissivity (e = 0.75), the surface area of 300 cm2 and temperature 227 ºC are kept in a room at temperature 27 ºC. Using the Stephens Boltzmann law, calculate the initial value of net power emitted by the body.

Using equation (3);

P = rsA (T4 – T04)

= (0.75) (5.67 × 10-8 W/m2 – k4) (300 × 10-4 m2) × [(500 K)4 – (300 K)4]

= 69.4 Watts.

 

Example 2: A hot black body emits the energy at the rate of 16 J m-2 s-1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then find the value of energy radiated in Jm-2 s-1.

Solution:

Wein’s displacement law is, lm.T = b

i.e. Tμ [1/lm]

Here, lm becomes half, the Temperature doubles.

Now from Stefan Boltzmann Law, e = sT4

e1/e2 = (T1/T2)4

⇒ e2 = (T2/T1)4 . e1 = (2)4 . 16

= 16.16 = 256 J m-2 s-1

Also Read: