Chapter 14, Problem 14.66E

Interpretation Introduction

Interpretation:

The values of the five transitions for $\text{HCl}$ using equation for the energy of a Morse oscillator are to be calculated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. 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.

Answer to Problem 14.66E

The values of the five transitions for $\text{HCl}$ using equation for the energy of a Morse oscillator are $2885.61{\text{cm}}^{-1}$, $5668.52{\text{cm}}^{-1}$, $8341.69{\text{cm}}^{-1}$, $10919.20{\text{cm}}^{-1}$ and $13385.96{\text{cm}}^{-1}$. The predicted values are very close to the experimental values.

Explanation of Solution

The energy of a Morse oscillator is calculated by the formula shown below.

$E=h{v}_{\text{e}}\left(v+\frac{1}{2}\right)-h{v}_{\text{e}}{x}_{\text{e}}{\left(v+\frac{1}{2}\right)}^2$ …(1)

Where,

• $v$ is the vibrational quantum number.

• $h$ is the Planck’s constant.

• $x_{\text{e}}$ is the anharmonicity constant.

• $v_{\text{e}}$ is the harmonic vibrational frequency.

The value of $v_{\text{e}}$ for $\text{HCl}$ is $2989.74{\text{cm}}^{-1}$.

The value of $x_{\text{e}}{v}_{\text{e}}$ for $\text{HCl}$ is $52.05{\text{cm}}^{-1}$.

Convert $2989.74{\text{cm}}^{-1}$ to ${\text{s}}^{-1}$.

$\begin{array}{rll}2989.74{\text{cm}}^{-1}& =& 2989.74{\text{cm}}^{-1}\times 2.9979\times {10}^{10}\text{cm}{\text{s}}^{-1}\\ & =& 8962.94\times {10}^{10}{\text{s}}^{-1}\end{array}$

Convert $52.05{\text{cm}}^{-1}$ to ${\text{s}}^{-1}$.

$\begin{array}{rll}52.05{\text{cm}}^{-1}& =& 52.05{\text{cm}}^{-1}\times 2.9979\times {10}^{10}\text{cm}{\text{s}}^{-1}\\ & =& 156.04\times {10}^{10}{\text{s}}^{-1}\end{array}$

Substitute $v=0$ in equation (1).

$\begin{array}{rll}E& =& h{v}_{\text{e}}\left(0+\frac{1}{2}\right)-h{v}_{\text{e}}{x}_{\text{e}}{\left(0+\frac{1}{2}\right)}^2\\ & =& \frac{1}{2}h{v}_{\text{e}}-\frac{1}{4}h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(2)

Substitute $v=1$ in equation (1).

$\begin{array}{rll}E& =& h{v}_{\text{e}}\left(1+\frac{1}{2}\right)-h{v}_{\text{e}}{x}_{\text{e}}{\left(1+\frac{1}{2}\right)}^2\\ & =& \frac{3}{2}h{v}_{\text{e}}-\frac{9}{4}h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(3)

Subtract equation (2) from equation (3).

$\begin{array}{rll}\Delta E\left(v=0\to v=1\right)& =& \left(\frac{3}{2}h{v}_{\text{e}}-\frac{9}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)-\left(\frac{1}{2}h{v}_{\text{e}}-\frac{1}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)\\ & =& h{v}_{\text{e}}-2h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(4)

Substitute the value of Planck’s constant, $v_{\text{e}}$ and $x_{\text{e}}{v}_{\text{e}}$ in equation (4).

$\begin{array}{rll}\Delta E\left(v=0\to v=1\right)& =& \left(\begin{array}{l}6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 8962.94\times {10}^{10}{\text{s}}^{-1}\\ -2\times 6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 156.04\times {10}^{10}{\text{s}}^{-1}\end{array}\right)\\ & =& 59388.44\times {10}^{-24}\text{J}-2067.84\times {10}^{-24}\text{J}\\ & =& 5.732\times {10}^{-20}\text{J}\end{array}$

The frequency of a transition in ${\text{cm}}^{-1}$ is calculated by the formula shown below.

$v=\frac{\Delta E}{hc}$ …(5)

Substitute the value of $\Delta E$, Planck’s constant and speed of light in equation (5).

$\begin{array}{rll}v& =& \frac{5.732\times {10}^{-20}\text{J}}{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^{10}\text{cm}{\text{s}}^{-1}}\\ & =& 2885.61{\text{cm}}^{-1}\end{array}$

Substitute $v=2$ in equation (1).

$\begin{array}{rll}E& =& h{v}_{\text{e}}\left(2+\frac{1}{2}\right)-h{v}_{\text{e}}{x}_{\text{e}}{\left(2+\frac{1}{2}\right)}^2\\ & =& \frac{5}{2}h{v}_{\text{e}}-\frac{25}{4}h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(6)

Subtract equation (2) from equation (6).

$\begin{array}{rll}\Delta E\left(v=0\to v=2\right)& =& \left(\frac{5}{2}h{v}_{\text{e}}-\frac{25}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)-\left(\frac{1}{2}h{v}_{\text{e}}-\frac{1}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)\\ & =& 2h{v}_{\text{e}}-6h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(7)

Substitute the value of Planck’s constant, $v_{\text{e}}$ and $x_{\text{e}}{v}_{\text{e}}$ in equation (7).

$\begin{array}{rll}\Delta E\left(v=0\to v=2\right)& =& \left(\begin{array}{l}2\times 6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 8962.94\times {10}^{10}{\text{s}}^{-1}\\ -6\times 6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 156.04\times {10}^{10}{\text{s}}^{-1}\end{array}\right)\\ & =& 118776.88\times {10}^{-24}\text{J}-6203.53\times {10}^{-24}\text{J}\\ & =& 1.126\times {10}^{-19}\text{J}\end{array}$

The frequency of a transition in ${\text{cm}}^{-1}$ is calculated by the formula shown below.

$v=\frac{\Delta E}{hc}$ …(8)

Substitute the value of $\Delta E$, Planck’s constant and speed of light in equation (8).

$\begin{array}{rll}v& =& \frac{1.126\times {10}^{-19}\text{J}}{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^{10}\text{cm}{\text{s}}^{-1}}\\ & =& 5668.52{\text{cm}}^{-1}\end{array}$

Substitute $v=3$ in equation (1).

$\begin{array}{rll}E& =& h{v}_{\text{e}}\left(3+\frac{1}{2}\right)-h{v}_{\text{e}}{x}_{\text{e}}{\left(3+\frac{1}{2}\right)}^2\\ & =& \frac{7}{2}h{v}_{\text{e}}-\frac{49}{4}h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(9)

Subtract equation (2) from equation (9).

$\begin{array}{rll}\Delta E\left(v=0\to v=3\right)& =& \left(\frac{7}{2}h{v}_{\text{e}}-\frac{49}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)-\left(\frac{1}{2}h{v}_{\text{e}}-\frac{1}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)\\ & =& 3h{v}_{\text{e}}-12h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(10)

Substitute the value of Planck’s constant, $v_{\text{e}}$ and $x_{\text{e}}{v}_{\text{e}}$ in equation (10).

$\begin{array}{rll}\Delta E\left(v=0\to v=3\right)& =& \left(\begin{array}{l}3\times 6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 8962.94\times {10}^{10}{\text{s}}^{-1}\\ -12\times 6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 156.04\times {10}^{10}{\text{s}}^{-1}\end{array}\right)\\ & =& 178165.32\times {10}^{-24}\text{J}-12407.05\times {10}^{-24}\text{J}\\ & =& 1.657\times {10}^{-19}\text{J}\end{array}$

The frequency of a transition in ${\text{cm}}^{-1}$ is calculated by the formula shown below.

$v=\frac{\Delta E}{hc}$ …(11)

Substitute the value of $\Delta E$, Planck’s constant and speed of light in equation (11).

$\begin{array}{rll}v& =& \frac{1.657\times {10}^{-19}\text{J}}{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^{10}\text{cm}{\text{s}}^{-1}}\\ & =& 8341.69{\text{cm}}^{-1}\end{array}$

Substitute $v=4$ in equation (1).

$\begin{array}{rll}E& =& h{v}_{\text{e}}\left(4+\frac{1}{2}\right)-h{v}_{\text{e}}{x}_{\text{e}}{\left(4+\frac{1}{2}\right)}^2\\ & =& \frac{9}{2}h{v}_{\text{e}}-\frac{81}{4}h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(12)

Subtract equation (2) from equation (12).

$\begin{array}{rll}\Delta E\left(v=0\to v=4\right)& =& \left(\frac{9}{2}h{v}_{\text{e}}-\frac{81}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)-\left(\frac{1}{2}h{v}_{\text{e}}-\frac{1}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)\\ & =& 4h{v}_{\text{e}}-20h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(13)

Substitute the value of Planck’s constant, $v_{\text{e}}$ and $x_{\text{e}}{v}_{\text{e}}$ in equation (13).

$\begin{array}{rll}\Delta E\left(v=0\to v=4\right)& =& \left(\begin{array}{l}4\times 6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 8962.94\times {10}^{10}{\text{s}}^{-1}\\ -20\times 6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 156.04\times {10}^{10}{\text{s}}^{-1}\end{array}\right)\\ & =& 237553.76\times {10}^{-24}\text{J}-20678.42\times {10}^{-24}\text{J}\\ & =& 2.169\times {10}^{-19}\text{J}\end{array}$

The frequency of a transition in ${\text{cm}}^{-1}$ is calculated by the formula shown below.

$v=\frac{\Delta E}{hc}$ …(14)

Substitute the value of $\Delta E$, Planck’s constant and speed of light in equation (14).

$\begin{array}{rll}v& =& \frac{2.169\times {10}^{-19}\text{J}}{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^{10}\text{cm}{\text{s}}^{-1}}\\ & =& 10919.20{\text{cm}}^{-1}\end{array}$

Substitute $v=5$ in equation (1).

$\begin{array}{rll}E& =& h{v}_{\text{e}}\left(5+\frac{1}{2}\right)-h{v}_{\text{e}}{x}_{\text{e}}{\left(5+\frac{1}{2}\right)}^2\\ & =& \frac{11}{2}h{v}_{\text{e}}-\frac{121}{4}h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(15)

Subtract equation (2) from equation (15).

$\begin{array}{rll}\Delta E\left(v=0\to v=5\right)& =& \left(\frac{11}{2}h{v}_{\text{e}}-\frac{121}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)-\left(\frac{1}{2}h{v}_{\text{e}}-\frac{1}{4}h{v}_{\text{e}}{x}_{\text{e}}\right)\\ & =& 5h{v}_{\text{e}}-30h{v}_{\text{e}}{x}_{\text{e}}\end{array}$ …(16)

Substitute the value of Planck’s constant, $v_{\text{e}}$ and $x_{\text{e}}{v}_{\text{e}}$ in equation (16).

$\begin{array}{rll}\Delta E\left(v=0\to v=5\right)& =& \left(\begin{array}{l}5\times 6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 8962.94\times {10}^{10}{\text{s}}^{-1}\\ -30\times 6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 156.04\times {10}^{10}{\text{s}}^{-1}\end{array}\right)\\ & =& 296942.20\times {10}^{-24}\text{J}-31017.63\times {10}^{-24}\text{J}\\ & =& 2.659\times {10}^{-19}\text{J}\end{array}$

The frequency of a transition in ${\text{cm}}^{-1}$ is calculated by the formula shown below.

$v=\frac{\Delta E}{hc}$ …(17)

Substitute the value of $\Delta E$, Planck’s constant and speed of light in equation (17).

$\begin{array}{rll}v& =& \frac{2.659\times {10}^{-19}\text{J}}{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.9979\times {10}^{10}\text{cm}{\text{s}}^{-1}}\\ & =& 13385.96{\text{cm}}^{-1}\end{array}$

Conclusion

The values of the five transitions for $\text{HCl}$ using equation for the energy of a Morse oscillator are $2885.61{\text{cm}}^{-1}$, $5668.52{\text{cm}}^{-1}$, $8341.69{\text{cm}}^{-1}$, $10919.20{\text{cm}}^{-1}$ and $13385.96{\text{cm}}^{-1}$. The predicted values are very close to the experimental values.