91Ó°ÊÓ

Number of atoms in 2-dimensional box =

$N={∫}_0^{∞}g\left(\mathrm{Î\mu }\right)\frac{1}{e^{(\mathrm{Î\mu }-\mathrm{μ})/kT}-1}d\mathrm{Î\mu }$

Here,

$g\left(\mathrm{Î\mu }\right)$=density of states,

$k=$Boltzmann constant,

$T=$temperature of the gas,

$\mathrm{μ}$=chemical potential,

$\mathrm{Î\mu }$= energy of the particle in a two-dimensional box.

$N=g{∫}_0^{∞}\frac{1}{e^{(\mathrm{Î\mu }-\mathrm{μ})/kT}-1}d\mathrm{Î\mu }$

If we take $\mathrm{μ}=0$

$N=g{∫}_0^{∞}\frac{1}{e^{\mathrm{Ï\mu }/kT}-1}d\mathrm{Î\mu }$

Let's take $x=\frac{\mathrm{Î\mu }}{kT}$

$N=g{∫}_0^{∞}\frac{1}{e^x-1}d\mathrm{Î\mu }$

As we know the value of

${∫}_0^{∞}\frac{1}{e^x-1}dx=∞$

Therefore the number of atoms =

$N=g{∫}_0^{∞}\frac{1}{e^x-1}d\mathrm{Î\mu }≈∞$

From the above result, it's clear that there's infinite number of atoms within the low-lying excited state when the chemical potential is zero. But the number of atoms is finite, So our assumption is not possible.

$N=g{∫}_0^{∞}\frac{1}{e^{(\mathrm{Î\mu }-\mathrm{μ})kT}-1}d\mathrm{Î\mu }$

$=g{∫}_0^{∞}\frac{1}{e^{-\mathrm{μ}/kT}{e}^{\mathrm{Î\mu }/kT}-1}d\mathrm{Î\mu }$

The integral will not be divergent if $e^{-\mathrm{μ}/kT}>1$

Applying the natural logarithm and solving the chemical potential we get,

$-\frac{\mathrm{μ}}{kT}>\ln \left(1\right)$

$\mathrm{μ}<\ln \left(1\right)kT$

$\mathrm{μ}<0$

Thus, the chemical potential is less than zero. Hence the given statement is proved.

The chemical potential should be negative as a result of the integral is finite (or the chemical potential is a smaller amount than zero the least bit temperatures). So, the whole particles can settle into the ground-state energy at very-low temperatures. because the temperature of the system goes to zero, the particles endlessly move to the ground-state, with no abrupt transition.

To get abrupt transition, the integral for the number of atoms in the two-dimensional box must be convergent at its lower limit when $\mathrm{μ}=0$. This condition will possible only when the value of the density of states $g\left(\mathrm{Î\mu }\right)$ goes to zero as $\mathrm{Î\mu }$ goes to zero.