Q. 7.65 - An Introduction to Thermal Physics

Question

Evaluate the integral in equation $N=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{\sqrt{ϵ}dϵ}{e^{ϵ/kT}-1}$ numerically, to confirm the value quoted in the text.

Step-by-Step Solution

Answer

The integral in equation$N=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{\sqrt{ϵ}dϵ}{e^{ϵ/kT}-1}$ is evaluated in simpler form.

Step 1. Given information

The total number of atoms, $N$ in Bose-Einstein distribution over all the states is given as:

$N=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V{\int }_0^{\infty }\frac{\sqrt{\epsilon }d\epsilon }{e^{\frac{\epsilon }{kT}}-1}$

$=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}V{\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx$

Where, 

 $h$ = Planck's constant,

 $k$= Boltzmann's constant,

 $V$= volume of the box,

 $\epsilon$= energy of the atom for higher energy level,

 $T$ = temperature,

  $m$ = mass of the atom,

  $x=\frac{\epsilon }{kT}$ is a new variable

Step 2. Calculating the integral ∫ 0 ∞ x d x e x - 1

Now, 

${\int }_0^{\infty }\frac{\sqrt{x}dx}{e^x-1}={\int }_0^{\infty }\frac{\sqrt{x}{e}^{-x}}{1-e^{-x}}dx$

$={\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-x}{\left(1-e^{-x}\right)}^{-1}$

$={\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-x}\left(1+e^{-x}+e^{-2x}+e^{-3x}+\dots \dots \right)dx$

$={\int }_0^{\infty }{x}^{\frac{1}{2}}\left(e^{-x}+e^{-2x}+e^{-3x}+e^{-4x}+\dots \dots \right)dx$

$={\int }_0^{\infty }{x}^{\frac{1}{2}}\sum _{k=0}^{\infty }{e}^{-(k+1)x}dx$

$=\sum _{k=0}^{\infty }{\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-(k+1)x}dx$

Step 3. Solving the integral ∫ 0 ∞ x 1 2 e - ( k + 1 ) x d x using the formula ∫ 0 ∞ x n e - a x d x = ( n ! ) a - ( n + 1 )

Therefore, 

${\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-(k+1)x}=\left(\frac{1}{2}\right)!(k+1{)}^{-\left(\frac{1}{2}+1\right)}$

As we know $\left(\frac{1}{2}\right)!=\Gamma \left(\frac{1}{2}+1\right)$,  and also $\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$

Now, 

$\Gamma \left(\frac{1}{2}+1\right)=\frac{1}{2}\Gamma \left(\frac{1}{2}\right)$

$=\frac{1}{2}\sqrt{\pi }$

Substituting the value of  $\left(\frac{1}{2}\right)!=\frac{1}{2}\sqrt{\pi }$in the equation  ${\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-(k+1)x}=\left(\frac{1}{2}\right)!(k+1{)}^{-\left(\frac{1}{2}+1\right)}$ we get,

${\int }_0^{\infty }{x}^{\frac{1}{2}}{e}^{-(k+1)x}=\frac{\sqrt{\pi }}{2}(k+1{)}^{\frac{-3}{2}}$

Step 4. Substituting the value of π 2 ( k + 1 ) - 3 2 = ∫ 0 1 2 x - ( k + 1 ) x in the equation

we get,

${\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx=\sum _{k=0}^{\infty }\frac{\sqrt{\pi }}{2}(k+1{)}^{\frac{-3}{2}}$

$=\frac{\sqrt{\pi }}{2}\sum _{k=0}^{\infty }(k+1{)}^{\frac{-3}{2}}$

$=\frac{\sqrt{\pi }}{2}\left[1+\frac{1}{2^{\frac{3}{2}}}+\frac{1}{3^{\frac{3}{2}}}+\frac{1}{4^{\frac{3}{2}}}+\dots ..\right]$

$=\frac{\sqrt{\pi }}{2}\zeta \left(\frac{3}{2}\right)$

Step 5. Substitute the value of ζ 3 2 = ∑ k = 1 ∞ 1 k 3 2 = 2 . 612 in the equation

we get,

${\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx=\frac{\sqrt{\pi }}{2}(2.612)$

Step 6. Substituting the value of π 2 ( 2 . 612 ) = ∫ 0 ∞ x 1 2 e x - 1 d x in the equation N = 2 π 2 π m k T h 2 3 2 V ∫ 0 ∞ x e x - 1 d x

$N=\left(\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}V\right)\left(\frac{\sqrt{\pi }}{2}(2.612)\right)$

$=2.612{\left(\frac{2\pi mkT}{h^2}\right)}^{\frac{3}{2}}V$

The equation $N=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{\sqrt{ϵ}dϵ}{e^{ϵ/kT}-1}$ is evaluated in simpler form.