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Stefan Boltzmann Law - Formula, Derivation

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What Is Stefan Boltzmann Law?

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

u/A = σT4 . . . . . . (1)

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

A body that is not a black body absorbs, and hence emits 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 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. ”

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:

\(\begin{array}{l}\frac{dP}{d\lambda }\frac{1}{A}=\frac{2\pi hc^{2}}{\lambda ^{5}\left ( e^{\frac{hc}{\lambda kT}}-1 \right )}\end{array} \)


  • P is the power radiated
  • A is the surface area of a blackbody
  • λ is the wavelength of the 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:

\(\begin{array}{l}\frac{d\left ( \frac{P}{A} \right )}{d\lambda }=\frac{2\pi hc^{2}}{\lambda ^{5}\left ( e^{\frac{hc}{\lambda kT}}-1 \right )}\end{array} \)

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

\(\begin{array}{l}\int_{0}^{\infty }\frac{d\left ( \frac{P}{A} \right )}{d\lambda }=\int_{0}^{\infty }\left [ \frac{2\pi hc^{2}}{\lambda ^{5}\left ( e^{\frac{hc}{\lambda kT}}-1 \right )} \right ]d\lambda\end{array} \)

The integrated power after separating the constants is:

\(\begin{array}{l}\frac{P}{A}=2\pi hc^{2}\int_{0}^{\infty }\left [ \frac{d\lambda }{\lambda ^{5}\left ( e^{\frac{hc}{\lambda kT}}-1 \right )} \right ]….(1)\end{array} \)

This can be solved analytically by substituting:

\(\begin{array}{l}x = \frac{hc}{\lambda kT}\end{array} \)


\(\begin{array}{l}dx=-\frac{hc}{\lambda^{2} kT}d\lambda\end{array} \)
\(\begin{array}{l}\Rightarrow h=\frac{x\lambda kT}{c}\end{array} \)
\(\begin{array}{l}\Rightarrow c=\frac{x\lambda kT}{h}\end{array} \)
\(\begin{array}{l}\Rightarrow d\lambda=-\frac{\lambda^{2} kT}{hc}dx\end{array} \)

As a result of substituting them in equation (1)

\(\begin{array}{l}\Rightarrow \frac{P}{A}=2\pi \left ( \frac{x\lambda kT}{c} \right )\left (\frac{x\lambda kT}{h} \right )^{2}\int_{0}^{\infty }\left [ \frac{\left ( -\frac{\lambda^{2}kT}{hc} \right )dx}{e^{x}-1} \right ]\end{array} \)
\(\begin{array}{l}= 2\pi \left ( \frac{x^{3}\lambda^{5} k^{4}T^{4}}{h^{3}c^{2}\lambda ^{5}} \right )\int_{0}^{\infty }\left [ \frac{dx}{e^{x}-1} \right ]\end{array} \)
\(\begin{array}{l}= \frac{2\pi \left ( kT \right )^{4}}{h^{3}c^{2}}\int_{0}^{\infty }\left [ \frac{x^{3}}{e^{x}-1} \right ]dx\end{array} \)

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

\(\begin{array}{l}\int_{0}^{\infty }\left [ \frac{x^{3}}{e^{x}-1} \right ]dx=\frac{\pi ^{4}}{15}\end{array} \)

Thus, substituting the above result, we get,

\(\begin{array}{l}\frac{P}{A}=\frac{2\pi \left ( kT \right )^{4}}{h^{3}c^{2}}\frac{\pi ^{4}}{15}\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{P}{A}=\left ( \frac{2 k^{4}\pi ^{5}}{15h^{3}c^{2}}\right )T^{4}\end{array} \)

On further simplifying, we get,

⇒ P/A = σ T4

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

⇒ ε = σT4


ε = P/A

\(\begin{array}{l}\sigma =\left ( \frac{2 k^{4}\pi ^{5}}{15h^{3}c^{2}}\right )=\left ( 5.670\times 10^{8}\;\frac{watts}{m^{2}K^{4}} \right )\end{array} \)

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

Problems with Stefan Boltzmann Law

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

Using equation (3);

P = eσA (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 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.


Wein’s displacement law is, λm.T = b

i.e. T∝ [1/ λm]

Here, λm becomes half, and 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:

Frequently Asked Questions – FAQs


What is Stefan-Boltzmann law?

Stefan-Boltzmann law states that the amount of radiation emitted by a black body per unit area is directly proportional to the fourth power of the temperature.

What is the value of Stefan’s constant?

5.67 × 10-8 W/m2 k4 is the value of Stefan’s constant.

What is the range of value of emissivity?

The range of value of emissivity is between 0-1.
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