Question

P7A.4 The wavelength λmax at which the Planck distribution is a maximum can be found by solving dp(AT)/d7-0. Differentiate ρα'T) with respect to T and show that the condition for the maximum can be expressed as xe 5(e-1) = 0, where x = hc/AKT. There are no analytical solutions to this equation, but a numerical approach gives x = 4.965 as a solution. Use this result to confirm Wien's law, that λmaxT is a constant, deduce an expression for the constant, and compare it to the value quoted in the text.

Answer 1

The Planck distribution function is given by
  $\rho(\lambda, T)=\frac{8\pi hc}{\lambda^5}\frac{1}{e^{\frac{hc}{kT\lambda}}-1}$
And so, we now calculate
   $\frac{\partial \rho(\lambda, T)}{\partial T}=\frac{\partial }{\partial T}\bigg[\frac{8\pi hc}{\lambda^5}\frac{1}{e^{\frac{hc}{kT\lambda}}-1}\bigg]$
  $\Rightarrow \frac{\partial \rho(\lambda, T)}{\partial T}=8\pi hc\bigg[-\frac{5}{\lambda^6}\frac{1}{e^{\frac{hc}{kT\lambda}}-1}+\frac{1}{\lambda^5}\frac{\partial }{\partial T}\bigg(\frac{1}{e^{\frac{hc}{kT\lambda}}-1}\bigg)\bigg]$
$\Rightarrow \frac{\partial \rho(\lambda, T)}{\partial T}=8\pi hc\bigg[-\frac{5}{\lambda^6}\frac{1}{e^{\frac{hc}{kT\lambda}}-1}-\frac{1}{\lambda^5}\bigg(\frac{1}{e^{\frac{hc}{kT\lambda}}-1}\bigg)^2\frac{\partial }{\partial T}(e^{\frac{hc}{kT\lambda}})\bigg]$
$\Rightarrow \frac{\partial \rho(\lambda, T)}{\partial T}=8\pi hc\bigg[-\frac{5}{\lambda^6}\frac{1}{e^{\frac{hc}{kT\lambda}}-1}-\frac{1}{\lambda^5}\bigg(\frac{1}{e^{\frac{hc}{kT\lambda}}-1}\bigg)^2e^{\frac{hc}{kT\lambda}}\big(-\frac{hc}{kT\lambda^2}\big)\bigg]$
$\Rightarrow \frac{\partial \rho(\lambda, T)}{\partial T}=\frac{8\pi hc}{\lambda^6}\frac{1}{e^{\frac{hc}{kT\lambda}}-1}\bigg[-5+\frac{hc}{kT\lambda}\frac{e^{\frac{hc}{kT\lambda}}}{e^{\frac{hc}{kT\lambda}}-1}\bigg]$
Now for the maximum of the distribution function, we require
   $\frac{\partial \rho(\lambda, T)}{\partial T}=0$
$\Rightarrow \frac{8\pi hc}{\lambda^6}\frac{1}{e^{\frac{hc}{kT\lambda}}-1}\bigg[-5+\frac{hc}{kT\lambda}\frac{e^{\frac{hc}{kT\lambda}}}{e^{\frac{hc}{kT\lambda}}-1}\bigg]=0$
$\Rightarrow -5+\frac{hc}{kT\lambda}\frac{e^{\frac{hc}{kT\lambda}}}{e^{\frac{hc}{kT\lambda}}-1}=0$
And we define
     $\frac{hc}{kT\lambda}\equiv x$
And so, the above equation becomes
  $\Rightarrow -5+\frac{xe^{x}}{e^{x}-1}=0$
$\Rightarrow xe^{x}-5(e^{x}-1)=0$

The numerical solution of this equation is
   $x=4.965$
And by using the definition, we get
  $\frac{hc}{kT\lambda_{max}}=4.965$
  $\Rightarrow T\lambda_{max}=\frac{hc}{4.965\times k}=\text{constant}$
This is the Wien's law.
  The value of the constant is
  $\Rightarrow T\lambda_{max}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.965\times 1.38\times 10^{-23}}=2.90\times 10^{-3}~\text{m.K}$