91影视

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,$螖E_1=6脳4脳{10}^{鈭�21}\text{鈥凧-5}脳4脳{10}^{鈭�21}\text{鈥凧}$
• The energy of the nanoparticle in the second case is,$螖E_2=9脳4脳{10}^{鈭�21}\text{鈥凧-8}脳4脳{10}^{鈭�21}\text{鈥凧}$
• The second energy of the nanoparticle in the first case is, .$螖E_3=9脳4脳{10}^{鈭�21}\text{鈥凧-5}脳4脳{10}^{鈭�21}\text{鈥凧}$

The temperature is described as a physical quantity that mainly measures the atom鈥檚 鈥渁verage kinetic energy鈥� or the system鈥檚 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:

${惟}_1=\frac{\left(a_1+N鈭�1\right)!}{a_1!\left(N鈭�1\right)!}$

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.

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

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

$S_1=k_B\ln {惟}_1$

Here,$k_B$is the Boltzmann constant with the value $\left(1.38脳{10}^{鈭�23}\text{鈥凧/碍}\right)$and${惟}_1$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:

${惟}_2=\frac{\left(a_2+N鈭�1\right)!}{a_2!\left(N鈭�1\right)!}$

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

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

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

$S_2=k_B\ln {惟}_2$

Here,$k_B$is the Boltzmann constant and ${惟}_2$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:

$\begin{array}{c}{T}_1=\frac{螖E_1}{螖S}\\ =\frac{螖E_1}{S_2鈭�S_1}\end{array}$

Here, $螖E_1$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.

$\begin{array}{c}{T}_1=\frac{6脳4脳{10}^{鈭�21}\text{鈥凧-5}脳4脳{10}^{鈭�21}\text{鈥凧}}{\left(11.5224鈭�10.1786\right)k_B}\\ =\frac{6脳4脳{10}^{鈭�21}\text{鈥凧-5}脳4脳{10}^{鈭�21}\text{鈥凧}}{\left(11.5224鈭�10.1786\right)\left(1.38脳{10}^{鈭�23}\text{鈥凧/碍}\right)}\\ =\frac{4脳{10}^{鈭�21}\text{鈥凧}}{\left(1.3438\right)\left(1.38脳{10}^{鈭�23}\text{鈥凧/碍}\right)}\\ =\frac{4脳{10}^{鈭�21}\text{鈥凧}}{\left(1.85脳{10}^{鈭�23}\text{鈥凧/碍}\right)}\\ =215.71\text{鈥凨}\end{array}$

Thus, the approximate temperature is .$215.71\text{鈥凨}$

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

${惟}_3=\frac{\left(a_3+N鈭�1\right)!}{a_3!\left(N鈭�1\right)!}$

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.

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

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

$S_3=k_B\ln {惟}_3$

Here,$k_B$is the Boltzmann constant and${惟}_3$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:

${惟}_4=\frac{\left(a_4+N鈭�1\right)!}{a_4!\left(N鈭�1\right)!}$

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.

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

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

$S_4=k_B\ln {惟}_4$

Here,$S_4$is the entropy in the second case, $k$is the Boltzmann constant and ${惟}_4$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:

$\begin{array}{c}{T}_2=\frac{螖E_2}{螖S}\\ =\frac{螖E_2}{S_4鈭�S_3}\end{array}$

Here, $螖E_1$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.

$\begin{array}{c}{T}_2=\frac{9脳4脳{10}^{鈭�21}\text{鈥凧-8}脳4脳{10}^{鈭�21}\text{鈥凧}}{\left(14.9548鈭�13.8939\right)k_B}\\ =\frac{9脳4脳{10}^{鈭�21}\text{鈥凧-8}脳4脳{10}^{鈭�21}\text{鈥凧}}{\left(14.9548鈭�13.8939\right)\left(1.38脳{10}^{鈭�23}\text{鈥凧/碍}\right)}\\ =\frac{4脳{10}^{鈭�21}\text{鈥凧}}{\left(1.0609\right)\left(1.38脳{10}^{鈭�23}\text{鈥凧/碍}\right)}\\ =\frac{4脳{10}^{鈭�21}\text{鈥凧}}{\left(1.46脳{10}^{鈭�23}\text{鈥凧/碍}\right)}\\ =273.22\text{鈥凨}\end{array}$

Thus, the approximate temperature is $273.22\text{鈥凨}$.

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:

$\begin{array}{c}c=\frac{螖E_3}{螖T}\\ =\frac{螖E_3}{T_2鈭�T_1}\end{array}$

Here, $螖E_3$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.

$\begin{array}{c}c=\frac{9脳4脳{10}^{鈭�21}\text{鈥凧-5}脳4脳{10}^{鈭�21}\text{鈥凧}}{\text{273.22鈥凨-215.71鈥凨}}\\ =\frac{\text{4}脳4脳{10}^{鈭�21}\text{鈥凧}}{\text{57.51鈥凨}}\\ =2.78脳{10}^{鈭�22}\text{鈥凧/碍}\end{array}$

Thus, the approximate heat capacity per atom is .$2.78脳{10}^{鈭�22}\text{鈥凧/碍}$

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.

$\begin{array}{c}{c}_1=3脳\left(1.38脳{10}^{鈭�23}\text{鈥凧/碍}\right)\\ =4.14脳{10}^{鈭�23}\text{鈥凧/碍}\end{array}$

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

$c_2=c鈭�c_1$

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.

$\begin{array}{c}{c}_2=2.78脳{10}^{鈭�22}\text{鈥凧/碍}鈭�4.14脳{10}^{鈭�23}\text{鈥凧/碍}\\ =鈭�\text{2.36}脳{\text{10}}^{鈭�22}\text{鈥凧/碍}\end{array}$

Thus, the higher temperature limit of$3k_B$ is lower by $2.36脳{10}^{鈭�22}\text{鈥凧/碍}$from the approximate heat capacity per atom.