A nanoparticle containing 6 atoms can be modeled approximately as an Einstein solid of 18 independent oscillators. The evenly spaced energy levels of each oscillator are$\text{4}\times {\text{10}}^{-21}\text{\hspace{0.33em}J}$apart. (a) When the nanoparticle’s energy is in the range$\text{5}\times \text{4}\times {\text{10}}^{-21}\text{\hspace{0.33em}J}$J to,$6\times \text{4}\times {\text{10}}^{-21}\text{\hspace{0.33em}J}$ what is the approximate temperature? (In order to keep precision for calculating the specific heat, give the result to the nearest tenth of a kelvin.) (b) When the nanoparticle’s energy is in the rangerole="math" localid="1657107429075" $8\times \text{4}\times {\text{10}}^{-21}\text{\hspace{0.33em}J}$to,$9\times \text{4}\times {\text{10}}^{-21}\text{\hspace{0.33em}J}$ what is the approximate temperature? (In order to keep precision for calculating the specific heat, give the result to the nearest tenth of a degree.) (c) When the nanoparticle’s energy is in the range$\text{5}\times \text{4}\times {\text{10}}^{-21}\text{\hspace{0.33em}J}$to$9\times \text{4}\times {\text{10}}^{-21}\text{\hspace{0.33em}J}$, what is the approximate heat capacity per atom? Note that between parts (a) and (b) the average energy increased from 5.5 quanta to 8.5 quanta. As a check, compare your result with the high temperature limit of $\mathbf{3}{\mathbf{k}}_B$.

The given data is listed below as:

• The number of independent oscillators in the Einstein solid is,$N=18$
• The energy quanta in the first caseare,$a_1=5$
• The energy quanta in the first caseare,$a_2=6$
• The energy quanta in the second case are,$a_3=8$
• The energy quanta in the second case are,$a_4=9$
• The energy of the nanoparticle in the first case is,$\Delta E_1=6\times 4\times {10}^{-21}\text{\hspace{0.33em}J-5}\times 4\times {10}^{-21}\text{\hspace{0.33em}J}$
• The energy of the nanoparticle in the second case is,$\Delta E_2=9\times 4\times {10}^{-21}\text{\hspace{0.33em}J-8}\times 4\times {10}^{-21}\text{\hspace{0.33em}J}$
• The second energy of the nanoparticle in the first case is, .$\Delta E_3=9\times 4\times {10}^{-21}\text{\hspace{0.33em}J-5}\times 4\times {10}^{-21}\text{\hspace{0.33em}J}$

The temperature is described as a physical quantity that mainly measures the atom’s “average kinetic energy” or the system’s molecules. The temperature is the ratio of the energy and the specific heat of an object.

The equation of the number of ways to distribute the energy quanta initially is expressed as:

${\Omega }_1=\frac{\left(a_1+N-1\right)!}{a_1!\left(N-1\right)!}$

Here,$a_1$is the energy quanta in the first case and$N$is the number of independent oscillators in the Einstein solid.

Substitute the values in the above equation.

$\begin{array}{c}{\Omega }_1=\frac{\left(5+18-1\right)!}{5!\left(18-1\right)!}\\ =\frac{\left(22\right)!}{5!\left(17\right)!}\\ =26334\end{array}$

The equation of the entropy in the first case is expressed as:

$S_1=k_B\ln {\Omega }_1$

Here,$k_B$is the Boltzmann constant with the value $\left(1.38\times {10}^{-23}\text{\hspace{0.33em}J/K}\right)$and${\Omega }_1$is the number of ways to distribute the energy quanta.

Substitute the values in the above equation.

$\begin{array}{c}{S}_1=k_B\ln 26334\\ =10.1786k_B\end{array}$

The equation of the number of ways to distribute the energy quanta finally is expressed as:

${\Omega }_2=\frac{\left(a_2+N-1\right)!}{a_2!\left(N-1\right)!}$

Here, $a_2$is the energy quanta in the first case and$N$is the number of independent oscillators in the Einstein solid.

Substitute the values in the above equation

$\begin{array}{c}{\Omega }_2=\frac{\left(6+18-1\right)!}{6!\left(18-1\right)!}\\ =\frac{\left(23\right)!}{6!\left(17\right)!}\\ =100947\end{array}$

The equation of the entropy in the second case is expressed as:

$S_2=k_B\ln {\Omega }_2$

Here,$k_B$is the Boltzmann constant and ${\Omega }_2$is the number of ways to distribute the energy quanta finally.

Substitute the values in the above equation.

$\begin{array}{c}{S}_2=k_B\ln 100947\\ =11.5224k_B\end{array}$

The equation of the approximate temperature is expressed as:

$\begin{array}{c}{T}_1=\frac{\Delta E_1}{\Delta S}\\ =\frac{\Delta E_1}{S_2-S_1}\end{array}$

Here, $\Delta E_1$in the energy in the first case, $S_2$is the entropy in the second case and $S_1$is the entropy in the first case.

Substitute the values in the above equation.

$\begin{array}{c}{T}_1=\frac{6\times 4\times {10}^{-21}\text{\hspace{0.33em}J-5}\times 4\times {10}^{-21}\text{\hspace{0.33em}J}}{\left(11.5224-10.1786\right)k_B}\\ =\frac{6\times 4\times {10}^{-21}\text{\hspace{0.33em}J-5}\times 4\times {10}^{-21}\text{\hspace{0.33em}J}}{\left(11.5224-10.1786\right)\left(1.38\times {10}^{-23}\text{\hspace{0.33em}J/K}\right)}\\ =\frac{4\times {10}^{-21}\text{\hspace{0.33em}J}}{\left(1.3438\right)\left(1.38\times {10}^{-23}\text{\hspace{0.33em}J/K}\right)}\\ =\frac{4\times {10}^{-21}\text{\hspace{0.33em}J}}{\left(1.85\times {10}^{-23}\text{\hspace{0.33em}J/K}\right)}\\ =215.71\text{\hspace{0.33em}K}\end{array}$

Thus, the approximate temperature is .$215.71\text{\hspace{0.33em}K}$

The equation of the number of ways to distribute the energy quanta finally is expressed as:

${\Omega }_3=\frac{\left(a_3+N-1\right)!}{a_3!\left(N-1\right)!}$

Here,$a_3$is the energy quanta in the second case and$N$is the number of independent oscillators in the Einstein solid.

Substitute the values in the above equation.

$\begin{array}{c}{\Omega }_3=\frac{\left(8+18-1\right)!}{8!\left(18-1\right)!}\\ =\frac{\left(25\right)!}{8!\left(17\right)!}\\ =1081575\end{array}$

The equation of the entropy in the first case is expressed as:

$S_3=k_B\ln {\Omega }_3$

Here,$k_B$is the Boltzmann constant and${\Omega }_3$is the number of ways to distribute the energy quanta.

Substitute the values in the above equation.

$\begin{array}{c}{S}_3=k_B\ln 1081575\\ =13.8939k_B\end{array}$

The equation of the number of ways to distribute the energy quanta finally is expressed as:

${\Omega }_4=\frac{\left(a_4+N-1\right)!}{a_4!\left(N-1\right)!}$

Here,$a_4$is the energy quanta in the second case and $N$is the number of independent oscillators in the Einstein solid.

Substitute the values in the above equation.

$\begin{array}{c}{\Omega }_4=\frac{\left(9+18-1\right)!}{9!\left(18-1\right)!}\\ =\frac{\left(26\right)!}{9!\left(17\right)!}\\ =3124550\end{array}$

The equation of the entropy in the second case is expressed as:

$S_4=k_B\ln {\Omega }_4$

Here,$S_4$is the entropy in the second case, $k$is the Boltzmann constant and ${\Omega }_4$is the number of ways to distribute the energy quanta finally.

Substitute the values in the above equation.

$\begin{array}{c}{S}_4=k_B\ln 3124550\\ =14.9548k_B\end{array}$

The equation of the approximate temperature is expressed as:

$\begin{array}{c}{T}_2=\frac{\Delta E_2}{\Delta S}\\ =\frac{\Delta E_2}{S_4-S_3}\end{array}$

Here, $\Delta E_1$in the energy in the second case,$S_2$is the entropy in the second case and $S_1$is the entropy in the first case.

Substitute the values in the above equation.

$\begin{array}{c}{T}_2=\frac{9\times 4\times {10}^{-21}\text{\hspace{0.33em}J-8}\times 4\times {10}^{-21}\text{\hspace{0.33em}J}}{\left(14.9548-13.8939\right)k_B}\\ =\frac{9\times 4\times {10}^{-21}\text{\hspace{0.33em}J-8}\times 4\times {10}^{-21}\text{\hspace{0.33em}J}}{\left(14.9548-13.8939\right)\left(1.38\times {10}^{-23}\text{\hspace{0.33em}J/K}\right)}\\ =\frac{4\times {10}^{-21}\text{\hspace{0.33em}J}}{\left(1.0609\right)\left(1.38\times {10}^{-23}\text{\hspace{0.33em}J/K}\right)}\\ =\frac{4\times {10}^{-21}\text{\hspace{0.33em}J}}{\left(1.46\times {10}^{-23}\text{\hspace{0.33em}J/K}\right)}\\ =273.22\text{\hspace{0.33em}K}\end{array}$

Thus, the approximate temperature is $273.22\text{\hspace{0.33em}K}$.

The graph of the entropy versus the energy quanta has been provide below-

From the above graph, it is observed that the difference between point a and point b and the corresponding temperature change gives the heat capacity.

The equation of the heat capacity is expressed as:

$\begin{array}{c}c=\frac{\Delta E_3}{\Delta T}\\ =\frac{\Delta E_3}{T_2-T_1}\end{array}$

Here, $\Delta E_3$in the second energy in the first case, $T_1$is the approximate temperature in the first case and$T_2$ is the approximate temperature in the second case.

Substitute the values in the above equation.

$\begin{array}{c}c=\frac{9\times 4\times {10}^{-21}\text{\hspace{0.33em}J-5}\times 4\times {10}^{-21}\text{\hspace{0.33em}J}}{\text{273.22\hspace{0.33em}K-215.71\hspace{0.33em}K}}\\ =\frac{\text{4}\times 4\times {10}^{-21}\text{\hspace{0.33em}J}}{\text{57.51\hspace{0.33em}K}}\\ =2.78\times {10}^{-22}\text{\hspace{0.33em}J/K}\end{array}$

Thus, the approximate heat capacity per atom is .$2.78\times {10}^{-22}\text{\hspace{0.33em}J/K}$

The equation of the high temperature limit is expressed as:

$c_1=3k_B$

Here, $c_1$is the heat capacity and$k_B$is the Boltzmann constant.

Substitute the values in the above equation.

$\begin{array}{c}{c}_1=3\times \left(1.38\times {10}^{-23}\text{\hspace{0.33em}J/K}\right)\\ =4.14\times {10}^{-23}\text{\hspace{0.33em}J/K}\end{array}$

The equation of the difference in the heat capacity is expressed as:

$c_2=c-c_1$

Here,$C_2$is the difference in the heat capacity, $C$is the heat capacity per atom and$C_1$is the heat capacity of the higher temperature limit.

Substitute the values in the above equation.

$\begin{array}{c}{c}_2=2.78\times {10}^{-22}\text{\hspace{0.33em}J/K}-4.14\times {10}^{-23}\text{\hspace{0.33em}J/K}\\ =-\text{2.36}\times {\text{10}}^{-22}\text{\hspace{0.33em}J/K}\end{array}$

Thus, the higher temperature limit of$3k_B$ is lower by $2.36\times {10}^{-22}\text{\hspace{0.33em}J/K}$from the approximate heat capacity per atom.