Question

How did Stefan Boltzmann find the constant to the Stefan Boltzmann law?

Answer 1

Hint: First of all we need to know the Stefan Boltzmann law statement and from that we can define what we understand by Stefan Boltzmann’s Constant. Then we need to know Stefan Boltzmann law. And from that we can write equations and on further solving by integrating the equation and a few more steps. Finally we find the value of the Stefan Boltzmann constant.

Compete answer:
Stefan Boltzmann law:
According to this law the amount of radiation emitted from a body per unit time from a surface area of blackbody let say A at absolute temperature let say T is directly proportional to the fourth power of the temperature, T.
Now the total power that radiates from a body per unit area over all wavelengths of blackbody can be obtained by integrating Planck's radiation formula. Hence the radiated power per unit area as a function of wavelength can be represented as,
${\displaystyle \frac{dP}{d\lambda }}{\displaystyle \frac{1}{A}}={\displaystyle \frac{2\pi h{c}^2}{{\lambda }^5\left(e^{{\displaystyle \frac{hc}{\lambda kT}}}-1\right)}}$
Where,
P is the power of radiation.
A is the surface area of the black body that radiates the light.
$\lambda$ is the wavelength of emitted radiation.
$h$ is Planck's constant.
$c$ is the speed of light which is also a const.
$k$ is the Boltzmann constant.
T is the absolute temperature.
On further simplifying the Stefan Boltzmann equation we will get,
${\displaystyle \frac{d\left({\displaystyle \frac{P}{A}}\right)}{d\lambda }}={\displaystyle \frac{2\pi h{c}^2}{{\lambda }^5\left(e^{{\displaystyle \frac{hc}{\lambda kT}}}-1\right)}}$
Now in integrating both side with respect to the wavelength $\lambda$ and the limit is from $0$ to $\mathrm{\infty }$ we will get,
$\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{d\left({\displaystyle \frac{P}{A}}\right)}{d\lambda }}d\lambda =\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{2\pi h{c}^2}{{\lambda }^5\left(e^{{\displaystyle \frac{hc}{\lambda kT}}}-1\right)}}d\lambda$
Now on further solving we will get,
${\displaystyle \frac{P}{A}}=2\pi h{c}^2\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{d\lambda }{{\lambda }^5\left(e^{{\displaystyle \frac{hc}{\lambda kT}}}-1\right)}}$
Now substituting,
${\displaystyle \frac{hc}{\lambda kT}}=x\dots \dots \left(2\right)$
Now differentiating x with respect to $\lambda$ we will get,
$-{\displaystyle \frac{hc}{{\lambda }^{2}kT}}d\lambda =dx$
Or, $d\lambda =-{\displaystyle \frac{{\lambda }^{2}kT}{hc}}dx$
Now rearranging equation $\left(2\right)$ we will get,
$h={\displaystyle \frac{x\lambda kT}{c}}$
Or, $c={\displaystyle \frac{x\lambda kT}{h}}$
Now putting all these in equation $\left(1\right)$ we will get,
${\displaystyle \frac{P}{A}}=2\pi \left({\displaystyle \frac{x\lambda kT}{c}}\right){\left({\displaystyle \frac{x\lambda kT}{h}}\right)}^2\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\left(-{\displaystyle \frac{{\lambda }^{2}kT}{hc}}\right)dx}{{\lambda }^5\left(e^x-1\right)}}$
$\Rightarrow {\displaystyle \frac{P}{A}}=2\pi \left({\displaystyle \frac{x^3{\lambda }^5{k}^4{T}^4}{h^3{c}^2}}\right)\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{dx}{{\lambda }^5\left(e^x-1\right)}}$
Cancelling and rearranging we will get,
${\displaystyle \frac{P}{A}}=2\pi \left({\displaystyle \frac{k^4{T}^4}{h^3{c}^2}}\right)\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{x^{3}dx}{\left(e^x-1\right)}}$
Now we know that,
$\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{x^{3}dx}{\left(e^x-1\right)}}={\displaystyle \frac{{\pi }^4}{15}}$
Hence,
${\displaystyle \frac{P}{A}}=2\pi \left({\displaystyle \frac{k^4{T}^4}{h^3{c}^2}}\right){\displaystyle \frac{{\pi }^4}{15}}\Rightarrow {\displaystyle \frac{P}{A}}=\left({\displaystyle \frac{2k^4{\pi }^5}{15h^3{c}^2}}\right)T^4$
Now we can write,
${\displaystyle \frac{P}{A}}=\sigma T^4\Rightarrow \epsilon =\sigma T^4$
Where,
$\sigma =\left({\displaystyle \frac{2k^4{\pi }^5}{15h^3{c}^2}}\right)$
Putting all the constant value we will get,
$\sigma =5.670\times {10}^{8}W{m}^{-2}{K}^4$

Note:
If a body us not a black body abortion then it will emit less radiation because and the equation is given as $u=e\sigma A{T}^4$ where, e is the emissivity which lies in between $0$ to $1$. Remember that $c=3\times {10}^{8}m{s}^{-1}$, $\pi =3.14$, $k=1.4\times {10}^{-23}J{K}^{-1}$ and $h=6.626\times {10}^{-34}{m}^{2}kg{s}^{-1}$.