Chapter 14, Problem 14.32E

Interpretation Introduction

(a)

Interpretation:

The most populated rotational level for a sample of $\text{LiH}$ at $298\text{K}$ is to be determined.

Concept introduction:

An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant. This is known as the Franck-Condon principle.

Answer to Problem 14.32E

The most populated rotational level for a sample of $\text{LiH}$ at $298\text{K}$ is $4$.

Explanation of Solution

The most populated rotational level is calculated by the formula as shown below.

$J_{\text{max}}={\left(\frac{kT}{2B}\right)}^{\frac{1}{2}}$…(1)

Where,

• $k$ is the Boltzmann’s constant.

• $T$ is the temperature.

• $B$ is the rotational constant.

The rotational constant is calculated by the formula as shown below.

$B=\frac{h^2}{8{\pi }^2\mu r^2}$…(2)

Where,

• $r$ is the bond length of $\text{LiH}$.

• $\mu$ is the reduced mass.

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)$.

The reduced mass is calculated by the formula as shown below.

$\mu =\frac{m_{\text{Li}}{m}_{\text{H}}}{m_{\text{Li}}+m_{\text{H}}}$…(3)

Where,

• $m_{\text{Li}}$ is the mass of lithium.

• $m_{\text{H}}$ is the mass of hydrogen.

Substitute the value of mass of lithium and hydrogen in equation (3).

$\begin{array}{rll}\mu & =& \frac{6.941\text{amu}\times 1\text{amu}}{6.941\text{amu}+1\text{amu}}\\ & =& 0.874\text{amu}\end{array}$

Convert $0.874\text{amu}$ to $\text{kg}$.

$\begin{array}{rll}0.874\text{amu}& =& \frac{0.874}{6.022\times {10}^{26}}\text{kg}\\ & =& 1.45\times {\text{10}}^{-27}\text{kg}\end{array}$

Substitute the value of reduced mass, bond length, Planck’s constant in equation (2).

$\begin{array}{rll}B& =& \frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8{\left(3.14\right)}^2\times 1.45\times {\text{10}}^{-27}\text{kg}\times {\left(1.60\times {10}^{-10}\right)}^2}\\ & =& \frac{43.9\times {10}^{-68}}{292.8\times {10}^{-47}}\\ & =& 1.5\times {10}^{-22}\text{J}\end{array}$

Substitute the value of rotational constant, Boltzmann’s constant and $J_{\text{max}}$ equation (1).

$\begin{array}{rll}{J}_{\text{max}}& =& {\left(\frac{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\times 298\text{K}}{2\times 1.5\times {10}^{-22}\text{J}}\right)}^{\frac{1}{2}}\\ & =& \sqrt{\frac{411.24\times {10}^{-23}\text{J}}{3\times {10}^{-22}\text{J}}}\\ & =& \sqrt{13.708}\\ & =& 3.70\end{array}$

Therefore, the most populated rotational level for a sample of $\text{LiH}$ at $298\text{K}$ is $\approx 4$.

Conclusion

the most populated rotational level for a sample of $\text{LiH}$ at $298\text{K}$ is $4$.

Interpretation Introduction

(b)

Interpretation:

The most populated rotational level for a sample of $\text{LiH}$ at $1000\text{K}$ is to be determined.

Concept introduction:

An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant. This is known as the Franck-Condon principle.

Answer to Problem 14.32E

The most populated rotational level for a sample of $\text{LiH}$ at $1000\text{K}$ is $7$.

Explanation of Solution

The most populated rotational level is calculated by the formula as shown below.

$J_{\text{max}}={\left(\frac{kT}{2B}\right)}^{\frac{1}{2}}$…(1)

Where,

• $k$ is the Boltzmann’s constant.

• $T$ is the temperature.

• $B$ is the rotational constant.

The rotational constant is calculated by the formula as shown below.

$B=\frac{h^2}{8{\pi }^2\mu r^2}$…(2)

Where,

• $r$ is the bond length of $\text{LiH}$.

• $\mu$ is the reduced mass.

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)$.

The reduced mass is calculated by the formula as shown below.

$\mu =\frac{m_{\text{Li}}{m}_{\text{H}}}{m_{\text{Li}}+m_{\text{H}}}$…(3)

Where,

• $m_{\text{Li}}$ is the mass of lithium.

• $m_{\text{H}}$ is the mass of hydrogen.

Substitute the value of mass of lithium and hydrogen in equation (3).

$\begin{array}{rll}\mu & =& \frac{6.941\text{amu}\times 1\text{amu}}{6.941\text{amu}+1\text{amu}}\\ & =& 0.874\text{amu}\end{array}$

Convert $0.874\text{amu}$ to $\text{kg}$.

$\begin{array}{rll}0.874\text{amu}& =& \frac{0.874}{6.022\times {10}^{26}}\text{kg}\\ & =& 1.45\times {\text{10}}^{-27}\text{kg}\end{array}$

Substitute the value of reduced mass, bond length, Planck’s constant in equation (2).

$\begin{array}{rll}B& =& \frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8{\left(3.14\right)}^2\times 1.45\times {\text{10}}^{-27}\text{kg}\times {\left(1.60\times {10}^{-10}\right)}^2}\\ & =& \frac{43.9\times {10}^{-68}}{292.8\times {10}^{-47}}\\ & =& 1.5\times {10}^{-22}\text{J}\end{array}$

Substitute the value of rotational constant, Boltzmann’s constant and $J_{\text{max}}$ equation (1).

$\begin{array}{rll}{J}_{\text{max}}& =& {\left(\frac{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\times 1000\text{K}}{2\times 1.5\times {10}^{-22}\text{J}}\right)}^{\frac{1}{2}}\\ & =& \sqrt{\frac{1.38\times {10}^{-20}\text{J}}{3\times {10}^{-22}\text{J}}}\\ & =& \sqrt{46}\\ & =& 6.78\end{array}$

Therefore, the most populated rotational level for a sample of $\text{LiH}$ at $1000\text{K}$ is $\approx 7$.

Conclusion

The most populated rotational level for a sample of $\text{LiH}$ at $1000\text{K}$ is $7$.

Interpretation Introduction

(c)

Interpretation:

The most populated rotational level for a sample of $\text{LiH}$ at $5000\text{K}$ is to be determined.

Concept introduction:

An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant. This is known as the Franck-Condon principle.

Answer to Problem 14.32E

The most populated rotational level for a sample of $\text{LiH}$ at $5000\text{K}$ is $15$.

Explanation of Solution

The most populated rotational level is calculated by the formula as shown below.

$J_{\text{max}}={\left(\frac{kT}{2B}\right)}^{\frac{1}{2}}$…(1)

Where,

• $k$ is the Boltzmann’s constant.

• $T$ is the temperature.

• $B$ is the rotational constant.

The rotational constant is calculated by the formula as shown below.

$B=\frac{h^2}{8{\pi }^2\mu r^2}$…(2)

Where,

• $r$ is the bond length of $\text{LiH}$.

• $\mu$ is the reduced mass.

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)$.

The reduced mass is calculated by the formula as shown below.

$\mu =\frac{m_{\text{Li}}{m}_{\text{H}}}{m_{\text{Li}}+m_{\text{H}}}$…(3)

Where,

• $m_{\text{Li}}$ is the mass of lithium.

• $m_{\text{H}}$ is the mass of hydrogen.

Substitute the value of mass of lithium and hydrogen in equation (3).

$\begin{array}{rll}\mu & =& \frac{6.941\text{amu}\times 1\text{amu}}{6.941\text{amu}+1\text{amu}}\\ & =& 0.874\text{amu}\end{array}$

Convert $0.874\text{amu}$ to $\text{kg}$.

$\begin{array}{rll}0.874\text{amu}& =& \frac{0.874}{6.022\times {10}^{26}}\text{kg}\\ & =& 1.45\times {\text{10}}^{-27}\text{kg}\end{array}$

Substitute the value of reduced mass, bond length, Planck’s constant in equation (2).

$\begin{array}{rll}B& =& \frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8{\left(3.14\right)}^2\times 1.45\times {\text{10}}^{-27}\text{kg}\times {\left(1.60\times {10}^{-10}\right)}^2}\\ & =& \frac{43.9\times {10}^{-68}}{292.8\times {10}^{-47}}\\ & =& 1.5\times {10}^{-22}\text{J}\end{array}$

Substitute the value of rotational constant, Boltzmann’s constant and $J_{\text{max}}$ equation (1).

$\begin{array}{rll}{J}_{\text{max}}& =& {\left(\frac{1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\times 5000\text{K}}{2\times 1.5\times {10}^{-22}\text{J}}\right)}^{\frac{1}{2}}\\ & =& \sqrt{\frac{6.9\times {10}^{-20}\text{J}}{3\times {10}^{-22}\text{J}}}\\ & =& \sqrt{230}\\ & =& 15.16\end{array}$

Therefore, the most populated rotational level for a sample of $\text{LiH}$ at $5000\text{K}$ is $\approx 15$.

Conclusion

The most populated rotational level for a sample of $\text{LiH}$ at $5000\text{K}$ is $15$.