91影视

The given data can be listed below,

• Mass of the tungsten is,$\mathrm{M}=0.185\mathrm{kg}$.

The source of resonance for an oscillator is quantized energy levels in atoms or molecules. Its premise is based on the notion that every atom is a perfect pendulum whose oscillations may be counted to determine a time interval.

The energy of the atomic oscillator is given by,

${\mathrm{E}}_{\mathrm{osci}}=鈩�\mathrm{蝇}$

Here,$鈩�$ is the Planck constant whose value is$1.05脳{10}^{-34}\mathrm{J}.\mathrm{s}$ and$蝇$ is the frequency.

Substitute all the values in the above,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{osci}}=鈩�\sqrt{\frac{{\mathrm{k}}_{\mathrm{eff}}{\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}} \\ =\left(1.05脳{10}^{-34}\mathrm{J}.\mathrm{s}\right)\sqrt{\frac{4\left(40\mathrm{N}/\mathrm{m}\right)\left(6.0221脳{10}^{23}\mathrm{molecule}/\mathrm{mole}\right)}{0.185\mathrm{kg}}} \\ =3.61脳{10}^{-21}\mathrm{J} \end{gathered}$$

Thus, the energy of the one quantum of energy is$3.61脳{10}^{-21}鈥�\mathrm{J}$ .

The energy of the first way for 20 quanta is given by,

$E_1=20脳E_{osci}$

Here,$E_{osci}$is the energy of one quanta.

Substitute all the values in the above,

$$\begin{gathered} \mathrm{E}=20脳3.61脳{10}^{-21}\mathrm{J} \\ =7.22脳{10}^{-20}\mathrm{J} \end{gathered}$$

The entropy for the first way is given by,

$S_1=k_{B}ln惟$

Here,$K_B$is the Boltzmann constant and$惟$is the number of ways,

Substitute all the values in the above,

$$\begin{gathered} S_1=\left(1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}\right)\mathrm{In}\left(4.91脳{10}^{26}\mathrm{J}/\mathrm{K}\right) \\ =8.49脳{10}^{-22}\mathrm{J}/\mathrm{K} \end{gathered}$$

The energy of the second way for 21 quanta is given by,

$E_2=21脳E_{osci}$

Here, $E_{osci}$is the energy of one quanta.

Substitute all the values in the above,

$$\begin{gathered} {\mathrm{E}}_2=21脳3.61脳{10}^{-21}\mathrm{J} \\ =7.51脳{10}^{-20}\mathrm{J} \end{gathered}$$

The entropy for the second way is given by,

${\mathrm{S}}_2={\mathrm{k}}_{\mathrm{B}}\ln \mathrm{惟}$

Here,$K_B$is the Boltzmann constant and $惟$is the number of ways,

Substitute all the values in the above,

$$\begin{gathered} S_2=\left(1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}\right)\mathrm{In}\left(4.44脳{10}^{27}\mathrm{J}/\mathrm{K}\right) \\ =8.79脳{10}^{-22}\mathrm{J}/\mathrm{K} \end{gathered}$$

The energy of the third way for 22 quanta is given by,

$E_3=22脳E_{osci}$

Here, $E_{osci}$is the energy of one quanta.

Substitute all the values in the above,

$$\begin{gathered} {\mathrm{E}}_3=22脳3.61脳{10}^{-21}\mathrm{J} \\ =7.942脳{10}^{-20}\mathrm{J} \end{gathered}$$

The entropy for third way is given by,

$S_3=k_{B}ln惟$

Here, $K_B$is the Boltzmann constant and$惟$is the number of ways.

Substitute all the values in the above,

$$\begin{gathered} S_3=\left(1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}\right)\mathrm{In}\left(3.85脳{10}^{28}\mathrm{J}/\mathrm{K}\right) \\ =9.1脳{10}^{-22}\mathrm{J}/\mathrm{K} \end{gathered}$$

The change in energy for the first way of arrangement is given by,

$螖E=E_2-E_1$

Here, $E_2$is the energy of second way and $E_1$is the energy of first way.

Substitute all the values in the above,

$$\begin{gathered} 鈭�\mathrm{E}=7.58脳{10}^{-20}\mathrm{J}-7.22脳{10}^{-20}\mathrm{J} \\ =3.61脳{10}^{-21}\mathrm{J} \end{gathered}$$

The change in entropy of the first way is given by,

$螖S=S_2-S_1$

Here, $S_1$is the entropy for first way and$S_2$is the entropy for second way

Substitute values in the above,

$$\begin{gathered} \mathrm{螖厂}=8.79脳{10}^{-22}\mathrm{J}/\mathrm{K}-8.49脳{10}^{-22}\mathrm{J}/\mathrm{K} \\ =3脳{10}^{-23}\mathrm{J}/\mathrm{K} \end{gathered}$$

The temperature for the first way of arranging energy quanta is given by,

$T=\frac{螖E}{螖S}$

Here,$鈭�E$is the change in energy and $鈭�S$is the change in entropy.

Substitute all the values in the above,

$$\begin{gathered} \mathrm{T}=\frac{3.61脳{10}^{-21}\mathrm{J}}{3.1脳{10}^{-23}\mathrm{J}/\mathrm{K}} \\ =120\mathrm{K} \end{gathered}$$

The change in energy for the second way of arrangement is given by,

$螖E=E_3-E_2$

Here, $E_3$is the energy of 22 quanta of energy and$E_2$is the energy for 21 quanta of energy.

Substitute all the values in the above,

$$\begin{gathered} 鈭�\mathrm{E}=7.94脳{10}^{-20}\mathrm{J}-7.58脳{10}^{-20}\mathrm{J} \\ =3.61脳{10}^{-21}\mathrm{J} \end{gathered}$$

The change in entropy of the second way is given by,

$螖S=S_3-S_2$

Here,$S_3$is the entropy for third way and$S_2$is the entropy for second way

Substitute values in the above,

$$\begin{gathered} \mathrm{螖厂}=9.1脳{10}^{-22}\mathrm{J}/\mathrm{K}-8.79脳{10}^{-22}\mathrm{J}/\mathrm{K} \\ =3.1脳{10}^{-23}\mathrm{J}/\mathrm{K} \end{gathered}$$

The temperature for the second way of arranging energy quanta is given by,

$T=\frac{螖E}{螖S}$

Here,$螖E$is the change in energy and$螖S$is the change in entropy.

Substitute all the values in the above,

$$\begin{gathered} \mathrm{T}=\frac{3.61脳{10}^{-21}\mathrm{J}}{3.1脳{10}^{-23}\mathrm{J}/\mathrm{K}} \\ =116\mathrm{K} \end{gathered}$$

Substitute all these values obtained above in the table below.

The internal energy of the atoms is given by,

$\mathrm{U}=\frac{3}{2}{\mathrm{nk}}_{\mathrm{B}}\mathrm{T}$

Here, n is the number of atoms,$k_B$is the Boltzmann constant whose value is $1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}$and T is the temperature.

The specific heat capacity per atom is given by,

$$\begin{gathered} \frac{{\mathrm{c}}_{\mathrm{v}}}{\mathrm{n}}=\frac{\mathrm{dU}}{\mathrm{dT}} \\ =\frac{3}{4}{\mathrm{k}}_{\mathrm{B}} \end{gathered}$$

Substitute values in the above,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{v}}=\frac{3}{2}\left(1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}\right) \\ =2.07脳{10}^{-23}\mathrm{J}/\mathrm{K}/\mathrm{atom} \end{gathered}$$

Thus, the specific heat capacity per atom is role="math" localid="1657884716241" $2.07脳{10}^{-23}\mathrm{J}/\mathrm{K}/\mathrm{atom}$.