Chapter 8, Problem 8.1P

To determine

The proof that at room temperature thermal energy is $kT\approx 1/140\text{eV}$, the temperature for $1\text{eV}$ and $13.6\text{eV}$.

Answer to Problem 8.1P

It is proved that at room temperature thermal energy is $kT\approx 1/140\text{eV}$, the temperature for $1\text{eV}$ and $13.6\text{eV}$ thermal energy are $11604.4\text{K}$ and $157820.22\text{K}$ respectively.

Explanation of Solution

Write the expression for the thermal energy.

    $\text{Thermal}\text{energy}=kT$        (1)

Here, $k$ is the Boltzmann constant and $T$ is the temperature.

Conclusion:

Substitute $8.6174\times {10}^{-5}\text{eV}/\text{K}$ for $k$ and $300\text{K}$ for $T$ in equation (1).

    $\begin{array}{c}\text{Thermal}\text{energy}=\left(8.6174\times {10}^{-5}\text{eV}/\text{K}\right)\left(300\text{K}\right)\\ =0.0253\text{eV}\\ =\frac{1}{40}\text{eV}\end{array}$

Substitute $8.6174\times {10}^{-5}\text{eV}/\text{K}$ for $k$ and $1\text{eV}$ for $\text{Thermal}\text{energy}$ in equation (1).

    $\begin{array}{c}\left(1\text{eV}\right)=\left(8.6174\times {10}^{-5}\text{eV}/\text{K}\right)T\\ T=\left(\frac{1\text{eV}}{8.6174\times {10}^{-5}\text{eV}/\text{K}}\right)\\ =11604.4\text{K}\end{array}$

Substitute $8.6174\times {10}^{-5}\text{eV}/\text{K}$ for $k$ and $13.6\text{eV}$ for $\text{Thermal}\text{energy}$ in equation (1).

    $\begin{array}{c}\left(13.6\text{eV}\right)=\left(8.6174\times {10}^{-5}\text{eV}/\text{K}\right)T\\ T=\left(\frac{13.6\text{eV}}{8.6174\times {10}^{-5}\text{eV}/\text{K}}\right)\\ =157820.22\text{K}\end{array}$

Thus, it is proved that at room temperature thermal energy is $kT\approx 1/140\text{eV}$, the temperature for $1\text{eV}$ and $13.6\text{eV}$ thermal energy are $11604.4\text{K}$ and $157820.22\text{K}$ respectively.