Chapter 42, Problem 62P

(a)

To determine

The equilibrium ratio of population of $E_4^{\ast }$ and $E_3$.

Answer to Problem 62P

The equilibrium ratio of population of $E_4^{\ast }$ and $E_3$ is $\underset{\_}{1.26\times {10}^{-33}}$.

Explanation of Solution

Write the expression for the number of atoms in a state of energy $E_n$.

    $N=N_g{e}^{-\frac{\left(E_n-E_g\right)}{k_{\text{B}}T}}$                                                                                                   (I)

Here, $N$ is the number of atoms in $E_n$ state, $N_g$ is the number of atoms in the ground state, $E_g$ is the energy of ground state, $E_n$ is the energy in the $n^{th}$ state, $k_{\text{B}}$ is the Boltzmann constant, and $T$ is the temperature.

Write the expression for population of atoms in $E_4^{\ast }$ using expression (I).

    $N_4^{\ast }=N_g{e}^{-\frac{E_3}{k_{\text{B}}T}}$                                                                                                     (II)

Here, $N_4^{\ast }$ is the number of atoms in $E_4^{\ast }$.

Write the expression for population of atoms in $E_3$ using expression (I).

    $N_3=N_g{e}^{-\frac{E_2}{k_{\text{B}}T}}$                                                                                                     (III)

Here, $N_3$ is the number of atoms in $E_3$.

Divide expressions (II) by (III).

    $\begin{array}{c}\frac{N_4^{\ast }}{N_3}=\frac{N_g{e}^{\frac{-E_3}{k_{\text{B}}T}}}{N_g{e}^{\frac{-E_2}{k_{\text{B}}T}}}\\ =e^{\frac{-\left(E_3-E_2\right)}{k_{\text{B}}T}}\\ =e^{\frac{-\Delta E}{k_{\text{B}}T}}\end{array}$                                                                                          (IV)

Write the expression to convert temperature from Celsius scale to Kelvin scale.

    $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$                                                                                     (V)

Here, $T\left(\text{K}\right)$ is the temperature in Kelvin scale, and $T\left(°\text{C}\right)$ is the temperature in Celsius scale.

Write the expression for energy difference between energies $E_4^{\ast }$ and $E_3$.

    $\Delta E=E_4^{\ast }-E_3$                                                                                                   (VI)

Conclusion:

Substitute $27.0°\text{C}$ for $T°\text{C}$ in equation (V) to find $T°\text{K}$.

    $\begin{array}{c}T°\text{K}=27.0°\text{C}+273.15\\ =300.15\text{K}\\ \approx \text{300}\text{.2}\text{K}\end{array}$

Substitute $20.66\text{eV}$ for $E_4^{\ast }$, and $18.70\text{eV}$ for $E_3$ in equation (VI) to find $\Delta E$.

    $\begin{array}{c}\Delta E=20.66\text{eV}-18.70\text{eV }\\ =1.96\text{eV}\end{array}$

Substitute $1.96\text{eV}$ for $\Delta E$, $1.381\times {10}^{-23}\text{J/K}$ for $k_{\text{B}}$, and $300.2\text{K}$ for $T$ in equation (VI) to find $\frac{N_4^{\ast }}{N_3}$.

    $\begin{array}{c}\frac{N_4^{\ast }}{N_3}=e^{-\frac{\left(1.96\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}{\left(1.38\times {10}^{-23}\text{J/K}\right)\left(300.2\text{K}\right)}}\\ =1.26\times {10}^{-33}\end{array}$

Therefore, the equilibrium ratio of population of $E_4^{\ast }$ and $E_3$ is $\underset{\_}{1.26\times {10}^{-33}}$.

(b)

To determine

The temperature for which the distribution shows a $2.00\%$ population inversion.

Answer to Problem 62P

The temperature for which the distribution shows a $2.00\%$ population inversion is $\underset{\_}{-1.15\times {10}^6\text{K}}$

Explanation of Solution

It is given that the number of atoms in the upper state is $2.00\%$ more than the number of atoms in lower state that is the ratio of population increases to $1.02$.

Rewrite expression for $\frac{N_4^{\ast }}{N_3}$ from (IV).

    $\frac{N_4^{\ast }}{N_3}=e^{\frac{-\left(E_3-E_2\right)}{k_{\text{B}}T}}$

Take logarithm of above expression.

    $\begin{array}{c}\log \frac{N_4^{\ast }}{N_3}=-\frac{\left(E_3-E_2\right)}{k_{\text{B}}T}\\ =-\frac{\left(\Delta E\right)}{k_{\text{B}}T}\end{array}$                                                                                              (VII)

Use expression (VII) to find $T$.

    $T=-\frac{\Delta E}{\log \left(\frac{N_4^{\ast }}{N_3}\right)k_{\text{B}}}$                                                                                              (VIII)

Conclusion:

Substitute $1.96\text{eV}$ for $\Delta E$, $1.02$ for $\frac{N_4^{\ast }}{N_3}$, and $1.38\times {10}^{-23}\text{J/K}$ for $k_{\text{B}}$ in equation (VIII) to find $T$.

    $\begin{array}{c}T=-\frac{\left(1.96\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}{\ln \left(1.02\right)\left(1.38\times {10}^{-23}\text{J/K}\right)}\\ =-1.15\times {10}^6\text{K}\end{array}$

Therefore, the temperature for which the distribution shows a $2.00\%$ population inversion is $\underset{\_}{-1.15\times {10}^6\text{K}}$.

(c)

To determine

The reason why $2.00\%$ population inversion does not occur.

Answer to Problem 62P

For $2.00\%$ population inversion, the temperature should become negative. Temperature can never be negative.

Explanation of Solution

The energy difference between states $E_4^{\ast }$ and $E_3$ is always positive. Also the temperature of the system is always positive. Temperature cannot be negative.

Rewrite expression for $\frac{N_4^{\ast }}{N_3}$ from (IV).

    $\frac{N_4^{\ast }}{N_3}=e^{\frac{-\left(\Delta E\right)}{k_{\text{B}}T}}$

In the above equation, temperature is always positive and hence the resultant population ratio will be less than one or it will be a smaller value. So a population cannot happen in thermal equilibrium.

Conclusion:

For $2.00\%$ population inversion, the temperature should become negative. Temperature can never be negative. So this does not occur naturally.