Question

Verify every entry in the third line of Table 3.2 (starting with $\left.N_{\uparrow }=98\right)$.

Short answer

All the entries in the third line are verified.

Step-by-step solution

Given Information

In a magnetic field $B$, a system of $N$magnetic dipoles arrange themselves so that their magnetic moment $\mu$points either parallel or anti parallel to the field.

Below is a table with the information provided.

Calculation

The expression for total energy is given as:

$U=\mu B\left(N_{\downarrow }-N_{\uparrow }\right)\dots \dots ..\left(1\right)$

Where,

$N_{\uparrow }$= the number of dipoles pointing up

$N_{\downarrow }$= the number of dipoles pointing down

$N=\left(N_{\downarrow }+N_{\uparrow }\right)$= the total number of dipoles

The net magnetization is given as:

$M=\mu \left(N_{\downarrow }-N_{\uparrow }\right)=-\frac{U}{B}\dots \dots ...\left(2\right)$

Where,

$B$= the magnetic field

$U$= the total energy

The multiplicity of states is given as:

$\mathrm{\Omega }=\frac{N!}{N_{\downarrow }!(N-N_{\downarrow })!}=\frac{N!}{N_{!}!N_{\downarrow }!}\dots \dots \dots .\left(3\right)$

For a small system, entropy is given as:

$\frac{S}{k}=\ln \left(\Omega \right)\dots \dots ..\left(4\right)$

By using these formulae, let's verify the table:

For $N_{\uparrow }=98,N_{\downarrow }=2$and $N=100$,

By substituting these values in the equation (1), we get,

$$\begin{gathered} U=\mu B(100-2\times 98) \\ \frac{U}{\mu B}=-96\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=-\frac{U}{B}=96\mu \\ \frac{M}{N\mu }=\frac{96}{N}=\frac{96}{100}=0.96\begin{array}{}\end{array} \end{gathered}$$

Now, substitute the values in equation (3) to get $\Omega$

$\Omega =\frac{100!}{98!2!}=4950$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\Omega \right)=\ln \left(4950\right)=8.5$

For $N_{\uparrow }=97,N_{\downarrow }=3$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mu B(100-2\times 97) \\ \frac{U}{\mu B}=-94\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=-\frac{U}{B}=94\mu \\ \frac{M}{N\mu }=\frac{94}{N}=\frac{94}{100}=0.94\begin{array}{}\end{array} \end{gathered}$$

Substitute the values in equation (3) to get $\Omega$

$\Omega =\frac{100!}{97!3!}=161700$

Now, by substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\Omega \right)=\ln \left(161700\right)=12$

For $N_{\uparrow }=52,N_{\downarrow }=48$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mu B(100-2\times 52) \\ \frac{U}{\mu B}=-4\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=-\frac{U}{B}=4\mu \\ \frac{M}{N\mu }=\frac{4}{N}=\frac{4}{100}=0.04\begin{array}{}\end{array} \end{gathered}$$

Now, substitute the values in equation (3) to get $\Omega$

$\Omega =\frac{100!}{52!48!}=9.32\times {10}^{28}$

By substituting this value in equation (4), we get,

$\begin{array}{rl}\frac{S}{k}& =\ln \left(\mathrm{\Omega }\right)=\ln \left(161700\right)=12\end{array}$

For $\begin{array}{rl}{N}_{\uparrow }& =52,N_{\downarrow }=48\end{array}$and $N=100,$

by substituting these values in equation (1), we get,

$$\begin{gathered} U=\mu B(100-2\times 51) \\ \frac{U}{\mu B}=-2\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=-\frac{U}{B}=2\mu \\ \frac{M}{N\mu }=\frac{2}{N}=\frac{2}{100}=0.02\begin{array}{}\end{array} \end{gathered}$$

Now, substitute the values in equation (3) to get $\Omega$

$\Omega =\frac{100!}{51!49!}=9.9\times {10}^{28}$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\Omega \right)=\ln \left(9.9\times {10}^{28}\right)=66.765$

For $N_{\uparrow }=50,N_{\downarrow }=50$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mu B(100-2\times 50) \\ \frac{U}{\mu B}=0\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=-\frac{U}{B}=0 \\ \frac{M}{N\mu }=\frac{0}{N}=\frac{0}{100}=0\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\Omega$

$\Omega =\frac{100!}{50!50!}=1.009\times {10}^{29}$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\Omega \right)=\ln \left(1.009\times {10}^{29}\right)=66.784$

For $N_{\uparrow }=49,N_{\downarrow }=51$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mu B(100-2\times 49) \\ \frac{U}{\mu B}=2\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=-\frac{U}{B}=-2\mu \\ \frac{M}{N\mu }=\frac{2}{N}=\frac{2}{100}=-0.02\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\Omega$

$\Omega =\frac{100!}{49!51!}=9.9\times {10}^{28}$

By substituting this value in equation (4), we get,

$\begin{array}{rl}\frac{S}{k}& =\ln \left(\mathrm{\Omega }\right)=\ln (9.9\times {10}^{28})=66.765\end{array}$

For $\begin{array}{rl}{N}_{\uparrow }& =48,N_{\downarrow }=52\end{array}$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mu B(100-2\times 48) \\ \frac{U}{\mu B}=4\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=-\frac{U}{B}=-4\mu \\ \frac{M}{N\mu }=-\frac{4}{N}=-\frac{4}{100}=-0.04\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\Omega$

$\Omega =\frac{100!}{48!52!}=9.32\times {10}^{28}$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\Omega \right)=\ln \left(9.32\times {10}^{28}\right)=66.7$

For $N_{\uparrow }=1,N_{\downarrow }=99$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mu B(100-2\times 1) \\ \frac{U}{\mu B}=98\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=-\frac{U}{B}=-98\mu \\ \frac{M}{N\mu }=-\frac{98}{N}=-\frac{98}{100}=-0.98\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\Omega$

$\Omega =\frac{100!}{1!99!}=100$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\Omega \right)=\ln \left(100\right)=4.6$

For $N_{\uparrow }=0,N_{\downarrow }=100$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mu B(100-2\times 0) \\ \frac{U}{\mu B}=100\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=-\frac{U}{B}=-100\mu \\ \frac{M}{N\mu }=-\frac{100}{N}=-\frac{100}{100}=-1\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\Omega$

$\Omega =\frac{100!}{0!100!}=1$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\Omega \right)=\ln \left(1\right)=0$

Final answer

Hence, all the table values are verified.