Chapter 7, Problem 7A.10P

(a)

Interpretation Introduction

Interpretation:

The dimensions of ${\theta }_{\text{E}}$ is of temperature has to be confirmed.  The criterion for the validity of the high-temperature form of Einstein equation in terms of ${\theta }_{\text{E}}$ has to be stated.  The value of ${\theta }_{\text{E}}$ has to be calculated for diamond.

Concept introduction:

The Einstein temperature is denoted by ${\theta }_{\text{E}}$.  It is characteristic to solids.  The value of heat capacities at low temperature, using the Einstein’s introduction of quantization is calculated by the following formula.

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\text{e}}^{-{\theta }_{\text{E}}/T}\right]$

Answer to Problem 7A.10P

The dimensions of ${\theta }_{\text{E}}$ is of temperature is validated.  The criterion for the validity of the high-temperature form of Einstein equation in terms of ${\theta }_{\text{E}}$ is $T>>{\theta }_{\text{E}}$.  The value of ${\theta }_{\text{E}}$ for diamond is $\underset{\_}{2231.8K}$.

Explanation of Solution

The value of Einstein temperature (${\theta }_{\text{E}}$) is expressed as,

    ${\theta }_{\text{E}}=\frac{hv}{k}$                                                                                                         (1)

Where,

• $h$ is the Planck’s constant.
• $v$ is the frequency.
• $k$ is the Boltzmann’s constant.

The units of $h$ is $\text{J}\text{s}$.

The units of $v$ is ${\text{s}}^{-1}$.

The units of $k$ is $\text{J}{\text{K}}^{-1}$.

Substitute the units of $h$, $v$ and $k$ in equation (1).

    $\begin{array}{c}{\theta }_{\text{E}}=\frac{\text{J}\text{s}\times {\text{s}}^{-1}}{\text{J}{\text{K}}^{-1}}\\ =\text{K}\end{array}$

Hence, the dimensions are of the temperature.

For the calculation of heat capacity Einstein gave a quantitative explanation, which is shown by the formula below.

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\left(\frac{{\text{e}}^{{\theta }_{\text{E}}/2T}}{{\text{e}}^{{\theta }_{\text{E}}/T}-1}\right)}^2\right]$                                                               (2)

Where,

• $C_{\text{V,m}}\left(T\right)$ is the heat capacity at temperature $T$.
• $R$ is the gas constant.
• ${\theta }_{\text{E}}$ is the Einstein’s temperature.

Assume,

  $\frac{{\theta }_{\text{E}}}{2T}=x$ and $\frac{{\theta }_{\text{E}}}{T}=y$

Substitute the values of $x$ and $y$ in equation (2).

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\left(\frac{{\text{e}}^x}{{\text{e}}^y-1}\right)}^2\right]$                                                                   (3)

Expand ${\text{e}}^x$ and ${\text{e}}^y$ as shown below.

    $\begin{array}{l}{\text{e}}^x=1+x+\frac{1}{2}{x}^2+\mathrm{...}\\ {\text{e}}^y=1+y+\frac{1}{2}{y}^2+\mathrm{...}\end{array}$

Substitute the values of ${\text{e}}^x$ and ${\text{e}}^y$ in equation (3).

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\left(\frac{1+x}{\left(1+y\right)-1}\right)}^2\right]$

Now, substitute the values of $x$ and $y$ in the above equation.

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\left(\frac{1+\frac{{\theta }_{\text{E}}}{2T}}{\left(1+\frac{{\theta }_{\text{E}}}{T}\right)-1}\right)}^2\right]$

At high temperature, $T>>{\theta }_{\text{E}}$.  Therefore,

    ${\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2\left(\frac{1+\frac{{\theta }_{\text{E}}}{2T}}{\left(1+\frac{{\theta }_{\text{E}}}{T}\right)-1}\right)\approx 1$

Hence, the heat capacity becomes,

    $\begin{array}{c}{C}_{\text{V,m}}\left(T\right)=3R\times 1\\ =3R\end{array}$

The classical value of heat capacity is $3R$.  Therefore, the Einstein equation behaves classically at high temperatures.

Hence, the criterion for the validity of the high-temperature form of Einstein equation in terms of ${\theta }_{\text{E}}$ is $T>>{\theta }_{\text{E}}$.

The value of $v$ is $46.5\text{THz}$.

The conversion of $\text{THz}$ to ${\text{s}}^{-1}$ is done as,

    $\begin{array}{c}v=46.5\text{THz}\left(\frac{{10}^{12}\text{Hz}}{1\text{}\text{THz}}\right)\left(\frac{1{\text{s}}^{-1}}{1\text{Hz}}\right)\\ =46.5\times {10}^{12}{\text{s}}^{-1}\end{array}$

The value of $h$ is $6.626\times {10}^{-34}\text{kg}{\text{m}}^2/\text{s}$.

The value of $k$ is $1.38\times {10}^{-23}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}$.

The value of $v$ is $46.5\times {10}^{12}{\text{s}}^{-1}$.

Substitute the value of $v$, $h$ $k$ and in equation (1).

    $\begin{array}{c}{\theta }_{\text{E}}=\frac{6.626\times {10}^{-34}\text{J}\text{s}\times 46.5\times {10}^{12}{\text{s}}^{-1}}{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}}\\ =\frac{3.08\times {10}^{-20}}{1.38\times {10}^{-23}}\\ =\underset{\_}{2231.8K}\end{array}$

Hence, the value of ${\theta }_{\text{E}}$ for diamond is $\underset{\_}{2231.8K}$.

(b)

Interpretation Introduction

Interpretation:

The dimensions of ${\theta }_{\text{E}}$ is of temperature has to be confirmed.  The criterion for the validity of the high-temperature form of Einstein equation in terms of ${\theta }_{\text{E}}$ has to be stated.  The value of ${\theta }_{\text{E}}$ has to be calculated for copper.

Concept introduction:

The Einstein temperature is denoted by ${\theta }_{\text{E}}$.  It is characteristic to solids.  The value of heat capacities at low temperature, using the Einstein’s introduction of quantization is calculated by the following formula.

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\text{e}}^{-{\theta }_{\text{E}}/T}\right]$

Answer to Problem 7A.10P

The dimensions of ${\theta }_{\text{E}}$ is of temperature is validated.  The criterion for the validity of the high-temperature form of Einstein equation in terms of ${\theta }_{\text{E}}$ is $T>>{\theta }_{\text{E}}$.  The value of ${\theta }_{\text{E}}$ for copper is $\underset{\_}{343.3K}$.

Explanation of Solution

The value of ${\theta }_{\text{E}}$ is,

    ${\theta }_{\text{E}}=\frac{hv}{k}$                                                                                                         (1)

Where,

• $h$ is the Planck’s constant.
• $v$ is the frequency.
• $k$ is the Boltzmann’s constant.

The units of $h$ is $\text{J}\text{s}$.

The units of $v$ is ${\text{s}}^{-1}$.

The units of $k$ is $\text{J}{\text{K}}^{-1}$.

Substitute the units of $h$, $v$ and $k$ in equation (2).

    $\begin{array}{c}{\theta }_{\text{E}}=\frac{\text{J}\text{s}\times {\text{s}}^{-1}}{\text{J}{\text{K}}^{-1}}\\ =\text{K}\end{array}$

Hence, the dimensions are of the temperature.

For the calculation of heat capacity Einstein gave a quantitative explanation, which is shown by the formula below.

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\left(\frac{{\text{e}}^{{\theta }_{\text{E}}/2T}}{{\text{e}}^{{\theta }_{\text{E}}/T}-1}\right)}^2\right]$                                                               (2)

Where,

• $C_{\text{V,m}}\left(T\right)$ is the heat capacity at temperature $T$.
• $R$ is the gas constant.
• ${\theta }_{\text{E}}$ is the Einstein’s temperature.

Assume,

  $x=\frac{{\theta }_{\text{E}}}{2T}$ and $y=\frac{{\theta }_{\text{E}}}{T}$

Substitute the value of $x$ and $y$ in equation (1).

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\left(\frac{{\text{e}}^x}{{\text{e}}^y-1}\right)}^2\right]$                                                                   (3)

Expand ${\text{e}}^x$ and ${\text{e}}^y$ as shown below.

    $\begin{array}{l}{\text{e}}^x=1+x+\frac{1}{2}{x}^2+\mathrm{...}\\ {\text{e}}^y=1+y+\frac{1}{2}{y}^2+\mathrm{...}\end{array}$

Substitute the values of ${\text{e}}^x$ and ${\text{e}}^y$ in equation (3).

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\left(\frac{1+x}{\left(1+y\right)-1}\right)}^2\right]$

Now, substitute the values of $x$ and $y$ in the above equation.

    $C_{\text{V,m}}\left(T\right)=3R\left[{\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2{\left(\frac{1+\frac{{\theta }_{\text{E}}}{2T}}{\left(1+\frac{{\theta }_{\text{E}}}{T}\right)-1}\right)}^2\right]$

At high temperature, $T>>{\theta }_{\text{E}}$.  Therefore,

    ${\left(\frac{{\theta }_{\text{E}}}{T}\right)}^2\left(\frac{1+\frac{{\theta }_{\text{E}}}{2T}}{\left(1+\frac{{\theta }_{\text{E}}}{T}\right)-1}\right)\approx 1$

Hence, the heat capacity becomes,

    $\begin{array}{c}{C}_{\text{V,m}}\left(T\right)=3R\times 1\\ =3R\end{array}$

The classical value of heat capacity is $3R$.  Therefore, the Einstein equation behaves classically at high temperatures.

Hence, the criterion for the validity of the high-temperature form of Einstein equation in terms of ${\theta }_{\text{E}}$ is $T>>{\theta }_{\text{E}}$.

The value of $v$ is $7.15\text{THz}$.

The conversion of $\text{THz}$ to ${\text{s}}^{-1}$ is done as,

    $\begin{array}{c}v=7.15\text{THz}\left(\frac{{10}^{12}\text{Hz}}{1\text{}\text{THz}}\right)\left(\frac{1{\text{s}}^{-1}}{1\text{Hz}}\right)\\ =7.15\times {10}^{12}{\text{s}}^{-1}\end{array}$

The value of $v$ is $7.15\times {10}^{12}{\text{s}}^{-1}$.

The value of $h$ is $6.626\times {10}^{-34}\text{kg}{\text{m}}^2/\text{s}$.

The value of $k$ is $1.38\times {10}^{-23}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}$.

Substitute the value of $v$, $h$ $k$ and in equation (1).

    $\begin{array}{c}{\theta }_{\text{E}}=\frac{6.626\times {10}^{-34}\text{J}\text{s}\times 7.15\times {10}^{12}{\text{s}}^{-1}}{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}}\\ =\frac{4.73\times {10}^{-21}}{1.38\times {10}^{-23}}\\ =\underset{\_}{343.3K}\end{array}$

Hence, the value of ${\theta }_{\text{E}}$ for copper is $\underset{\_}{343.3K}$.