Answer to Question #309158 in Molecular Physics | Thermodynamics for Abhishek

Question

derive an expression for planck's law of blackbody radiation using this law of wien's displacement law and stefan boltzmann law

Answer 1

Solution;

The probability that a single mode has energy $E_n=nhv$

is given by the usual Boltzmann factor;

$P(n)=\frac{e^{\frac{-E_n}{kT}}}{\displaystyle\sum_{n=0}^{\infin}e^{\frac{E_n}{kT}}}$

where the denominator ensures that the total probability is unity, the usual normalisation procedure.

In the language of photons, this is the probability

that the state contains n photons of frequency ν.The mean energy of the mode of frequency ν is therefore;

$\bar{E_n}=\displaystyle\sum_{n=0}^{\infin}E_np(n)=\frac{\displaystyle\sum_{n=0}^{\infin}E_ne^{\frac{-E_n}{kT}}}{\displaystyle\sum{n=0}^{\infin}e^{\frac{-E_n}{kT}}}$

Which is;

$\bar{E_n}=\frac{\displaystyle\sum_{n=0}^{\infin}nhv e^{\frac{-nhv}{kT}}}{\displaystyle\sum_{n=0}^{\infin}e^{\frac{-nhv}{kT}}}$

To simplify the calculation we use ;

$e^{\frac{-nhv}{kT}}=x$

We have;

$\bar{E_n}=\frac{hv \displaystyle \sum_{n=0}^{\infin}nx^n}{\displaystyle\sum _{n=0}^{\infin}x^n}=hv\frac{x+2x^2+3x^3+...}{1+x+x^2+...}=hvx\frac{1+2x+3x^2+...}{1+x+x^2+...}$

Now remember the following series expansion;

$\frac{1}{1-x}=1+x+x^2+x^3+...$

$\frac{1}{(1-x)^2}=1+2x+3x^2+...$

Hence, the mean energy of the mode is

$E=\frac{hvx}{1-x}=\frac {hv}{x^{-1}-1}=\frac{hv}{e^{\frac{hv}{kT}}-1}$

To find the classical limit, we allow the energy quanta hν to tend to zero. Expanding $e^{\frac{hv}{kT}}$ ​ for small values of $\frac{hv}{kT}$

$e^{\frac{hv}{kT}}-1=1+\frac{hv}{kT}+\frac{1}{2!}(\frac{hv}{kT})^2+...-1$

For small values of $\frac{hv}{kT}$ ;

$e^{\frac{hv}{kT}}-1=\frac{hv}{kT}$

And so;

$\bar E=\frac{hv}{e^{\frac{hv}{kT}}-1}=\frac{hv}{\frac {\epsilon}{kT}}=kT$

We can now complete the determination of Planck’s radiation formula. We know that the number of modes in the frequency interval ν to v+dv is $\frac{8πv^2}{c^3}dv$ per unit volume.

The energy density of radiation in the frequency range is;

$u(v)dv=\frac{8πv^2}{c^3}\bar{E_v}dv$

$u(v)dv=\frac{8πv^2}{c^3}\frac{1}{e^{\frac{hv}{kT}}-1}$

Which is the Planck's distribution function.