91Ó°ÊÓ

Given material: Ideal two-state electronic paramagnet such as DPPH.

$\mathrm{μ}={\mathrm{μ}}_{\mathrm{B}}$

Magnetic field strength $=2.06T$

Minimum Temperaturerole="math" localid="1647345687440" $=2.2K$

Here, the magnetic moment is equal to the Bohr magneton i.e.

${\mathrm{μ}}_B=\frac{eh}{4\mathrm{π}{m}_e}$

By substituting $9.11×{10}^{-31}\mathrm{kg}$for $m_e$, localid="1647346175321" $6.62×{10}^{-34}\mathrm{J}.\mathrm{s}$for $h$and $1.6×{10}^{-19}\mathrm{C}$for $e$in the above equation, we get,

$$\begin{gathered} {\mathrm{μ}}_B=\frac{\left(1.6×{10}^{−19}\right)\left(6.62×{10}^{−34}\right)}{4×\mathrm{π}\left(9.11×{10}^{−31}\right)} \\ {\mathrm{μ}}_B=9.26×{10}^{−24}{\mathrm{JT}}^{−1} \end{gathered}$$

Since the tanh function has a maximum value of $+1$which is reached asymptotically as its argument goes to $+∞$, the maximum magnetization is reached in the limit of the very high applied field $B$and at a very low temperature.

For an applied field of $2.06\mathrm{T}$and a temperature of $2.2\mathrm{K}$, fractional change in Magnetization is given as:

localid="1647348010263" $\frac{M}{\mathrm{μ}N}=\tanh \left(\frac{\mathrm{μ}B}{kT}\right)..........\left(1\right)$

By substituting $1.38×{10}^{-23}{\mathrm{JK}}^{-1}$for $k$, $9.26×{10}^{-24}{\mathrm{JT}}^{-1}$for ${\mathrm{μ}}_B$, $2.06\mathrm{T}$for $B$in the above equation, we get,

$$\begin{gathered} \frac{M}{\mathrm{μ}N}=\tanh \left(\frac{9.26×{10}^{−24}×2.06}{1.38×{10}^{−23}×2.2}\right) \\ \frac{M}{\mathrm{μ}N}=0.557 \end{gathered}$$

Now, the magnetization can also be written as:

localid="1647346764792" $$\begin{gathered} M=\mathrm{μ}\left(N_{↑}-N_{↓}\right) \\ M=\mathrm{μ}\left(2N_{↑}−N\right) \end{gathered}$$

By dividing both sides by $\mathrm{μ}N$on the above equation, we get,

localid="1647348000233" $\frac{M}{\mathrm{μ}N}=2×\frac{N_{\mathrm{t}}}{N}-1………..\left(2\right)$

On comparing equations (1) and (2),

$$\begin{gathered} \tanh \left(\frac{\mathrm{μ}B}{kT}\right)=(2×\frac{N_{\mathrm{t}}}{N}−1) \\ \text{But}n=\frac{N_{\mathrm{t}}}{N}\text{so,}\tanh \left(\frac{\mathrm{μ}B}{kT}\right)=2n−1 \\ n=\frac{1}{2}[1+\tan \left(\frac{\mathrm{μ}B}{kT}\right)] \end{gathered}$$

By substituting $0.557$for $\tan \left(\frac{\mathrm{μ}B}{kT}\right)$in the above equation, we get,

$n=\frac{1}{2}[1+0.557]=0.78$

The entropy is given as:

$\frac{S}{k}=N\ln \left(N\right)−\left(N_{↑}\right)\ln \left(N_{↑}\right)−(N−N_{↑})\ln (N−N_{↑})$

Replace each $N_{↑}$with $\left(\frac{N_{↑}N}{N}\right)$in the above equation,

$\frac{S}{k}=N\ln \left(N\right)−\left(\frac{N_{\mathrm{t}}N}{N}\right)\ln \left(\frac{N_{\mathrm{t}}N}{N}\right)−\left(N−\left(\frac{N_{\mathrm{t}}N}{N}\right)\right)\ln \left(N−\left(\frac{N_{\mathrm{t}}N}{N}\right)\right)$

But $\left(n=\frac{N_{\mathrm{t}}}{N}\right)$,

Hence, the above equation can be written as:

$$\begin{gathered} \frac{S}{k}=N\ln \left(N\right)−\left(nN\right)\ln \left(nN\right)−(N−(nN\left)\right)\ln (N−(nN\left)\right) \\ \frac{S}{k}≃N\left[\ln \right(N)−(n\left)\ln \right(nN)−(1−n\left)\ln \right(N−\left(nN\right)\left)\right] \end{gathered}$$

But $\ln \left(ab\right)=\ln a+\ln b$

Hence, the above equation becomes,

$$\begin{gathered} \frac{S}{k}≃N\left[\ln \right(N)-(n\left)\right[\ln \left(n\right)+\ln \left(N\right)]-(1-n\left)\right\{\ln \left(N\right)+\ln (1-n)\left\}\right] \\ \frac{S}{k}=N\left[n\ln \right(n)+(1-n\left)\ln \right(1-n\left)\right] \end{gathered}$$

The maximum entropy occurs at $\left(n=\frac{N_{\mathrm{t}}}{N}=0.5\right)$and it can be given as:

$$\begin{gathered} \frac{S_{\max }}{Nk}=-[0.5×\ln (0.5)+(1-0.5\left)\ln \right(1-0.5\left)\right] \\ \frac{S_{\max }}{Nk}=0.693 \end{gathered}$$

The entropy for an applied field of $2.06\mathrm{T}$and a temperature of $2.2\mathrm{K}$

Hence,

localid="1647348762642" $$\begin{gathered} \frac{S}{Nk}=-[0.778×\ln (0.7785)+(1-0.7785\left)\ln \right(1-0.7785\left)\right] \\ \frac{S}{Nk}=0.5288 \end{gathered}$$

Fractional entropy is therefore given as:

localid="1647348833396" $$\begin{gathered} \frac{S}{Nk}×\frac{Nk}{S_{\max }}=\frac{0.5288}{0.693} \\ \frac{S}{S_{\max }}=0.763 \end{gathered}$$

The energy is given as:

localid="1647348937658" $$\begin{gathered} U=\mathrm{μ}B\left(N-2N_{↑}\right) \\ U=\mathrm{μ}BN\left(1-2\frac{N_{\mathrm{t}}}{N}\right) \\ U=\mathrm{μ}BN(1-2n) \end{gathered}$$

That is maximum when $n=0$and it has value of $U_{\max }=\mathrm{μ}BN(1-2n)$

So the fractional change in energy is given as:

localid="1647349059847" $$\begin{gathered} \frac{U}{U_{\max }}=\frac{\mathrm{μ}BN}{\mathrm{μ}BN}(1-2n)=(1-2n) \\ \frac{U}{U_{\max }}=(1-2n) \end{gathered}$$

By substituting the value of $n$as $0.7785$in the above equation, we get,

$$\begin{gathered} \frac{U}{U_{\max }}=(1-2×0.7785) \\ \frac{U}{U_{\max }}=-0.557 \end{gathered}$$

To achieve $99\%$magnetization, from equation (1), we get,

$$\begin{gathered} \frac{M}{\mathrm{μ}N}=\tanh \left(\frac{\mathrm{μ}B}{kT}\right)=0.99 \\ \frac{\mathrm{μ}B}{kT}={\tanh }^{-1}(0.99) \\ \frac{\mathrm{μ}B}{kT}=2.647 \end{gathered}$$

Now, substitute $1.38×{10}^{-23}{\mathrm{JK}}^{-1}$for $k$and $9.26×{10}^{-24}{\mathrm{JT}}^{-1}$for $\mathrm{μ}$in the above equation, we get,

role="math" localid="1647349781971" $$\begin{gathered} \frac{\mathrm{μ}B}{kT}=2.647 \\ \frac{B}{T}=\frac{k}{\mathrm{μ}}×2.647 \\ \frac{B}{T}=\left(\frac{1.38×{10}^{−23}}{9.26×{10}^{−24}}\right)×2.647 \\ \frac{B}{T}=3.945{\mathrm{TK}}^{−1} \end{gathered}$$

Hence, the required values are calculated as:

$\frac{M}{\mathrm{μ}N}=0.557$

$\frac{S}{S_{max}}=0.763$

$\frac{U}{U_{max}}=−0.557$

$\frac{B}{T}=3.945{\mathrm{TK}}^{−1}$