91Ó°ÊÓ

Assume they're two Einstein physics structures, $A$and $B$, with $N_A=N_B=10$and $q_A+q_B=20\text{.}$.

(a) As a metric, the count of macro states is:

$q+1=20+1=21$

because we began with $0$ and switch our attention resolute $20$.

(b) The formula for said length of microstates is:

${\mathrm{Î\copyright }}_{\text{overall}}\left(N_{\text{overall}},q_{\text{overall}}\right)=\left(\begin{array}{c}{q}_{\text{overall}}+N_{\text{overall}}−1\\ q_{\text{overall}}\end{array}\right)=\frac{\left(q_{\text{overall}}+N_{\text{overall}}−1\right)!}{q_{\text{overall}}!\left(N_{\text{overall}}−1\right)!}$

$q_{\text{overall}}=q_A+q_B=20,$

$N_{\text{overall}}=N_A+N_B$

$\begin{array}{c}{\mathrm{Î\copyright }}_{\text{overall}}=\left(\begin{array}{c}20+20−1\\ 20\end{array}\right)\\ \end{array}$

$\begin{array}{c}=\frac{(20+20−1)!}{20!(20−1)!}\\ \end{array}$

$\begin{array}{c}=6.892×{10}^{10}\\ \end{array}$

${\mathrm{Î\copyright }}_{\text{overall}}=6.892×{10}^{10}$

(c)The likelihood that while in equilibrium state, most of the facility is in rigid $A$, resulted in:

$q_A=20$$q_B=0$$N_A=10$$N_B=10$

The cumulative number is calculated based:

$\begin{array}{c}{\mathrm{Î\copyright }}_{\text{total}}={\mathrm{Î\copyright }}_A{\mathrm{Î\copyright }}_B\\ \end{array}$

$\begin{array}{c}\\ {\mathrm{Î\copyright }}_A=\left(\begin{array}{c}{q}_A+N_A−1\\ q_A\end{array}\right)=\frac{\left(q_A+N_A−1\right)!}{q_A!\left(N_A−1\right)!}\end{array}$

$\begin{array}{c}{\mathrm{Î\copyright }}_A=\left(\begin{array}{c}20+10−1\\ 20\end{array}\right)=\frac{(20+10−1)!}{20!(10−1)!}=10015005\\ \\ \end{array}$

$\begin{array}{c}{\mathrm{Î\copyright }}_B=\left(\begin{array}{c}{q}_B+N_B−1\\ q_B\end{array}\right)=\frac{\left(q_B+N_B−1\right)!}{q_{B3}!\left(N_B−1\right)!}\\ \end{array}$

${\mathrm{Î\copyright }}_B=\left(\begin{array}{c}0+10−1\\ 0\end{array}\right)=\frac{(0+10−1)!}{0!(10−1)!}=1$

${\mathrm{Î\copyright }}_{\text{total}}={\mathrm{Î\copyright }}_A{\mathrm{Î\copyright }}_B=10015005×1=10015005$

As an answer, this same chances inside this instance is:

localid="1650298575703" $P=\frac{{\mathrm{Î\copyright }}_{\text{total}}}{{\mathrm{Î\copyright }}_{\text{overall}}}$

localid="1650298609352" $=\frac{10015005}{6.892×{10}^{10}}$

localid="1650298489904" $=1.453×{10}^{−4}$

(d)The likelihood that while in equilibrium state, most of the power is in rigid , resulted in:

$q_A=10$$q_B=10$$N_A=10$$N_B=10$

The cumulative number is calculated based:

$\begin{array}{c}{\mathrm{Î\copyright }}_{\text{total}}={\mathrm{Î\copyright }}_A{\mathrm{Î\copyright }}_B\\ \end{array}$

localid="1650382124449" $\begin{array}{c}\\ \end{array}\begin{array}{c}\\ {\mathrm{Î\copyright }}_A=\left(\begin{array}{c}{q}_A+N_A−1\\ q_A\end{array}\right)=\frac{\left(q_A+N_A−1\right)!}{q_A!\left(N_A−1\right)!}\\ \end{array}$

$\begin{array}{c}{\mathrm{Î\copyright }}_A=\left(\begin{array}{c}20+10−1\\ 20\end{array}\right)=\frac{(10+10−1)!}{10!(10−1)!}=92378\\ \\ \end{array}$

$\begin{array}{c}{\mathrm{Î\copyright }}_B=\left(\begin{array}{c}{q}_B+N_B−1\\ q_B\end{array}\right)=\frac{\left(q_B+N_B−1\right)!}{q_{B3}!\left(N_B−1\right)!}\\ \end{array}$

${\mathrm{Î\copyright }}_B=\left(\begin{array}{c}0+10−1\\ 0\end{array}\right)=\frac{(10+10−1)!}{10!(10−1)!}=92378$

As a solution, this same chances inside this instance is:

$P=\frac{{\mathrm{Î\copyright }}_{\text{total}}}{{\mathrm{Î\copyright }}_{\text{overall}}}$

$=\frac{8.5337×{10}^9}{6.892×{10}^{10}}$

$=0.1238$

(e) it'll be more likely that it photon energy will spread them divided between the $2$ solids, and when that does, it's rare that its position will revert about one solid having similar more quanta than others. That is, an irreversible state is that the one where the quanta are spaced equally.