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## Homework Statement

A system having ##N## non-degenerate energy eigenstates populated by##N## identical spin-zero particles and ##2N## identical spin-half particles. There are no interaction between any of these particles. If ##N=1000## what is the entropy of the system?

## Homework Equations

*Third law of thermodynamics*$$S=k_B \ln W$$

where ##S## is entropy, ##k_B## is the

*Botlzmann constant*and ##W## is the thermodynamic probability.

Thermodynamic probability, i.e., the total number of independent ways of distributing ##N_i## particles in ##g_i## states is

for ##spin-integral## particles $$W= \prod_{i=1}^{n} {\frac {(N_i +g_i -1)!}{{N_i}! {(g_i -1)!}}}$$

for ##spin-half## particles $$W= \prod_{i=1}^{n} {\frac {{g_i}!}{{N_i}! {(g_i -N_i)!}}}$$

Stirling's formula $$\ln {n!} = n \ln n - n$$

## The Attempt at a Solution

for ##spin-integral## particles $$W={\frac {(N+N -1)!}{{N}! {(N -1)!}}}$$

Simplifying $$W={\frac {(2N -1)!}{{N}! {N!}}}$$

That is approximately $$W={\frac {(2N)!} {{{N}!}^2}}$$

Thus $$S_{spin-0}=k_B \ln{\frac {(2N)!} {{(N!)}^2}}$$

$$\Rightarrow S_{spin-0}=k_B {[\ln {{(2N)!}-\ln {(N!)}^2}]}$$

$$\Rightarrow S_{spin-0}=k_B {[(2N) \ln {(2N)}-(2N)-2\ln {{N}!}]}$$

$$\Rightarrow S_{spin-0}=k_B {[(2N) \ln {(2N)}-(2N)-2N\ln {N} + 2N]}$$

$$\Rightarrow S_{spin-0}=k_B {(2N)[ \ln {(2N)}-\ln {N}]}$$

$$\Rightarrow S_{spin-0}=k_B {(2N) \ln 2}$$

for ##spin-half## particles $$W={\frac {N!}{{(2N)}! {(N -2N)!}}}$$

Which gives a ##negative## factorial at the denominator!

So, does this mean one cannot distribute ##2N## spin-half particles in ##N## states?

Which sounds right, because of the

*Pauli exclusion principle*.

Anyone have any thought?