Chapter 10, Problem 4P

To determine

The sketch of the absorption spectra at temperature $4\text{K}$ and $1\text{K}$ .

Answer to Problem 4P

The sketch of the absorption spectra at temperature $4\text{K}$ and $1\text{K}$ is plotted below.

Explanation of Solution

Write the expression for number of particles.

  $n_i=g_i{f}_{\text{MB}}$        (I)

Here, $n_i$ is the number of particles, $f_{\text{MB}}$ is the Maxwell-Boltzmann probability and $g_i$ is the degeneracy.

Substitute $A{e}^{-E/k_{B}T}$ for $f_{\text{MB}}$ in equation (I).

  $n_i=g_{i}A{e}^{-E/k_{B}T}$

Here, $E$ is the energy of particle, $k_B$ is Boltzmann constant and $T$ is the temperature.

Write the expression for strength of absorption spectra.

  $S\left(E_i\to E_f\right)=n_{i}P\left(i\to f\right)$

Here, $S\left(E_i\to E_f\right)$ is the strength and $P\left(i\to f\right)$ is the probability.

Here, the transition probability is equal for all allowed transitions.

It gives $P\left(0\to 3\right)=P\left(1\to 3\right)=P\left(2\to 3\right)$ .

Compare the strength for level $\left(0\to 3\right)$ and $\left(2\to 3\right)$ .

  $\frac{S\left(0\to 3\right)}{S\left(2\to 3\right)}=\frac{n_0}{n_2}$

Substitute $g_{0}A{e}^{-E_0/k_{B}T}$ for $n_0$ and $g_{2}A{e}^{-E_2/k_{B}T}$ for $n_2$ in above expression and simplify.

  $\frac{S\left(0\to 3\right)}{S\left(2\to 3\right)}=\frac{g_0}{g_2}{e}^{-\left(E_2-E_0\right)/k_{B}T}$        (II)

Similarly the strength for level $\left(1\to 3\right)$ and $\left(2\to 3\right)$

  $\frac{S\left(1\to 3\right)}{S\left(2\to 3\right)}=\frac{g_1}{g_2}{e}^{-\left(E_2-E_1\right)/k_{B}T}$        (III)

Conclusion:

Calculate the value $E_2-E_0$ .

   $\begin{array}{c}{E}_2-E_0=2\Delta \\ =24.82\times {10}^{-5}\text{eV}\end{array}$

Calculate the value $E_2-E_1$ .

   $\begin{array}{c}{E}_2-E_1=\Delta \\ =12.41\times {10}^{-5}\text{eV}\end{array}$

For $T=4\text{K}$:

Substitute $24.82\times {10}^{-5}\text{eV}$ for $E_2-E_0$, $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $K_B$, $1$ for $g_0$, $2$ for $g_2$ and $4\text{K}$ for $T$ in equation (II).

  $\begin{array}{c}\frac{S\left(0\to 3\right)}{S\left(2\to 3\right)}=\frac{1}{2}{e}^{\left(\frac{-\left(24.82\times {10}^{-5}\text{eV}\right)}{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left(4\text{K}\right)}\right)}\\ =0.2434\end{array}$

Substitute $12.41\times {10}^{-5}\text{eV}$ for $E_2-E_1$, $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $K_B$, $2$ for $g_1$, $2$ for $g_2$ and $4\text{K}$ for $T$ in equation (III).

  $\begin{array}{c}\frac{S\left(1\to 3\right)}{S\left(2\to 3\right)}=\frac{2}{2}{e}^{\left(\frac{-\left(12.41\times {10}^{-5}\text{eV}\right)}{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left(4\text{K}\right)}\right)}\\ =0.6977\end{array}$

The sketch of the absorption spectra at temperature $4\text{K}$ is plotted above.

For $T=1\text{K}$:

Substitute $24.82\times {10}^{-5}\text{eV}$ for $E_2-E_0$, $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $K_B$, $1$ for $g_0$, $2$ for $g_2$ and $1\text{K}$ for $T$ in equation (II).

  $\begin{array}{c}\frac{S\left(0\to 3\right)}{S\left(2\to 3\right)}=\frac{1}{2}{e}^{\left(\frac{-\left(24.82\times {10}^{-5}\text{eV}\right)}{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left(1\text{K}\right)}\right)}\\ =0.0281\end{array}$

Substitute $12.41\times {10}^{-5}\text{eV}$ for $E_2-E_1$, $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $K_B$, $2$ for $g_1$, $2$ for $g_2$ and $1\text{K}$ for $T$ in equation (III).

  $\begin{array}{c}\frac{S\left(1\to 3\right)}{S\left(2\to 3\right)}=\frac{2}{2}{e}^{\left(\frac{-\left(12.41\times {10}^{-5}\text{eV}\right)}{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left(1\text{K}\right)}\right)}\\ =0.1185\end{array}$

The sketch of the absorption spectra at temperature $1\text{K}$ is plotted above.