Question

The interatomic spring stiffness for tungsten is determined from Young’s modulus measurements to be 90 N. The mass of one mole of tungsten is 0185 kg . If we model a block of tungsten as a collection of atomic “oscillators” (masses on springs), note that since each oscillator is attached to two “springs,” and each “spring” is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above.Use these precise values for the constants: $\mathbf{h}\mathbf{̶}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{\hspace{0.33em}}\mathbf{J}\mathbf{.}\mathbf{s}$(Planck’s constant divided by 2π), Avogadro’s number = $\mathbf{6}\mathbf{.}\mathbf{0221}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{\hspace{0.33em}}\mathbf{molecule}\mathbf{/}\mathbf{mole}$, ${\mathbf{k}}_{\mathbf{B}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3807}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{\hspace{0.33em}}\mathbf{J}\mathbf{/}\mathbf{K}$(the Boltzmann constant). (a) What is one quantum of energy for one of these atomic oscillators? (b) Figure 12.56 contains the number of ways to arrange a given number of quanta of energy in a particular block of tungsten. Fill in the blanks to complete the table, including calculating the temperature of the block. The energy E is measured from the ground state. Nothing goes in the shaded boxes. Be sure to give the temperature to the nearest 0.1 kelvin. (c) There are about 60 atoms in this object. What is the heat capacity on a per-atom basis? (Note that at high temperatures the heat capacity on a per-atom basis approaches the classical limit of 3kB = 4.2×10−23 J/K/atom.)

Short answer

a) the energy of the one quantum of energy is $3.61\times {10}^{-21}\hspace{0.33em}\mathrm{J}$.

b) The table is given below.

c) The specific heat capacity per atom is $2.07\times {10}^{-23}\hspace{0.33em}\mathrm{J}/\mathrm{K}/\mathrm{atom}$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• Young modulus of tungsten is,$\mathrm{Y}=90\hspace{0.33em}\mathrm{N}/\mathrm{m}$.

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

Concept/Significance of energy of the oscillator

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.

Determination of one quantum of energy for one of these atomic oscillators

The energy of the atomic oscillator is given by,

${\mathrm{E}}_{\mathrm{osci}}=\hslash \mathrm{\omega }$

Here,$\hslash$ is the Planck constant whose value is$1.05\times {10}^{-34}\mathrm{J}.\mathrm{s}$ and$\omega$ is the frequency.

Substitute all the values in the above,

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

Thus, the energy of the one quantum of energy is$3.61\times {10}^{-21}\hspace{0.33em}\mathrm{J}$ .

(b) Determination of the content of the table given

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

$E_1=20\times E_{osci}$

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

Substitute all the values in the above,

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

The entropy for the first way is given by,

$S_1=k_{B}ln\Omega$

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

Substitute all the values in the above,

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

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

$E_2=21\times E_{osci}$

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

Substitute all the values in the above,

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

The entropy for the second way is given by,

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

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

Substitute all the values in the above,

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

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

$E_3=22\times E_{osci}$

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

Substitute all the values in the above,

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

The entropy for third way is given by,

$S_3=k_{B}ln\Omega$

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

Substitute all the values in the above,

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

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

$\Delta E=E_2-E_1$

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,

$$\begin{gathered} ∆\mathrm{E}=7.58\times {10}^{-20}\mathrm{J}-7.22\times {10}^{-20}\mathrm{J} \\ =3.61\times {10}^{-21}\mathrm{J} \end{gathered}$$

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

$\Delta S=S_2-S_1$

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

Substitute values in the above,

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

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

$T=\frac{\Delta E}{\Delta S}$

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

Substitute all the values in the above,

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

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

$\Delta E=E_3-E_2$

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,

$$\begin{gathered} ∆\mathrm{E}=7.94\times {10}^{-20}\mathrm{J}-7.58\times {10}^{-20}\mathrm{J} \\ =3.61\times {10}^{-21}\mathrm{J} \end{gathered}$$

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

$\Delta S=S_3-S_2$

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

Substitute values in the above,

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

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

$T=\frac{\Delta E}{\Delta S}$

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

Substitute all the values in the above,

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

Substitute all these values obtained above in the table below.

(c) Determination of the heat capacity on a per-atom

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 $1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$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,

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

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