91Ó°ÊÓ

Each particle can have any $10$states each with zero energy.

So, $E\left(r_i\right)=0for1≤r_i≤10$

Formula for particle function with box having one particle:

$Z_l=\underset{i}{∑}{e}^{-\mathrm{β}E\left(r_i\right)}$

where, $\mathrm{β}=\frac{1}{kT}$$\mathrm{β}=\frac{1}{kT}$

We substitute $E\left(r_i\right)=0$

localid="1647243851142" $$\begin{gathered} Z_l=\underset{1}{\overset{10}{∑}}{e}^{\mathrm{β}\left(0\right)} \\ =\underset{1}{\overset{10}{∑}}1 \\ =10 \end{gathered}$$

So, Partition function of system with box having only one particle is$10$.

For two distinguishable particles partition function is:

$Z={Z_l}^2$

Substitute $Z_l=10$

role="math" localid="1647244178631" $$\begin{gathered} Z={\left(10\right)}^2 \\ =100 \end{gathered}$$

So, the partition function of the system if the box contains two distinguishable particles is$100$.

To put two identical bosons in same single particle state the possibility is $10$.

So, to put two identical bosons in different single particle state the possibility is calculated as:

$$\begin{gathered} {}_2{}^{10}C=\frac{10!}{2!8!} \\ =45 \end{gathered}$$

So, total possible states are:

$$\begin{gathered} Z=10+45 \\ =55 \end{gathered}$$

Hence, the partition function if the box contains two identical bosons is$55$.

There is no possibility to put two identical fermions in same single particle state.

For different single particle state possibilities are calculated by:

$$\begin{gathered} {}_2{}^{10}C=\frac{10!}{2!8!} \\ =45 \end{gathered}$$

So, the partition function if the box contains two identical fermions is$45$.

Partition function for $N$indistinguishable and non interacting particles is given by,

$Z=\frac{{Z_l}^N}{N!}$

Substitute $Z_l=10$and $N=2$

$$\begin{gathered} Z=\frac{{10}^2}{2!} \\ =50 \end{gathered}$$

So, the partition function of the system when box have indistinguishable particles is$50$.

For distinguishable particles number of accessible states are $Z=100$.

For both particles in same single particle state, accessible states are $10$.

So, probability of finding both particle in same single particle state is:

$$\begin{gathered} P=\frac{10}{100} \\ =10\% \end{gathered}$$

So, probability for distinguishable particles is role="math" localid="1647245922911" $10\%$.

For identical bosons number of accessible states are $Z=55$.

For both bosons in same single particle state, accessible states are $10$.

So, probability of finding both bosons in same single particle state is:

$$\begin{gathered} P=\frac{10}{55} \\ =18\% \end{gathered}$$

So, probability for identical bosons is $18\%$.

For identical fermions number of accessible states are $Z=45$.

For both fermions in same single particle state, accessible states is $0$.

So, probability of finding both fermions in same single particle state is:

$$\begin{gathered} P=\frac{0}{45} \\ =0\% \end{gathered}$$

So, probability for identical fermions is $0\%$.