Question

How do I calculate the force constant, zero-point energy, and the energy level spacings for $^{12}{C^{16}}O$ if ${\tilde \omega _e} = 2170c{m^{ - 1}}$ ?

Answer 1

In here, we are dealing with anharmonic oscillator. The zero-point energy is the lowest energy of the molecule in the ground state denoted by ${E_0}$ i.e., the energy ${E_v}$ as the function of the vibrational quantum number $v$ at $v = 0$ . The energy of the harmonic oscillator can be given as ${E_v} = h{v_0}\left( {v + \dfrac{1}{2}} \right)$ .

Complete answer:
The information given to us is ${\tilde \omega _e} = {\tilde v_e} = 2170c{m^{ - 1}}$
First we will need to convert ${\tilde v_0} \to {v_0}$ . The prior is in $c{m^{ - 1}}$ whereas the latter is in ${s^{ - 1}}$ . The conversion can be done by using the formula: ${\tilde v_0} = {\tilde v_e} - 2{\tilde v_e}{\chi _e}$
We need to find the value of ${\tilde v_e}{\chi _e} = 13.28831$ . substituting in the above equation we get; ${\tilde v_0} = 2170c{m^{ - 1}} - 2(13.28831c{m^{ - 1}}) = 2143.4c{m^{ - 1}}$
On converting this into ${s^{ - 1}}$ for finding the value of energy ${E_v}$ . The conversion can be given as: ${v_0} = {\tilde v_0}c$ where c is the speed/velocity of light in cm/s which is equal to $2.998 \times {10^{10}}cm/s$ . Substituting the vale in the above formula for conversion of ${\tilde v_0} \to {v_0}$
${v_0} = (2143.4c{m^{ - 1}})(2.998 \times {10^{10}}cm/s)$
${v_0} = 6.426 \times {10^{13}}{s^{ - 1}}$
The zero point energy is the energy at v=0. Hence the zero point energy will be equal to: ${E_0} = h{v_0}\left( {0 + \dfrac{1}{2}} \right) = \dfrac{1}{2}h{v_0}$
${E_0} = \dfrac{1}{2}(6.626 \times {10^{ - 34}}J.s)(6.426 \times {10^{13}}{s^{ - 1}})$
${E_0} = 2.129 \times {10^{ - 20}}J$
This is the Zero Point energy. The energy level spacing can be given as the difference of energy between two levels. It can be given as $\Delta E = {E_1} - {E_0} = {E_{v + 1}} - {E_v}$
$\Delta E = h{v_0}\left( {v + 1 + \dfrac{1}{2}} \right) - h{v_0}\left( {v + \dfrac{1}{2}} \right)$
$\Delta E = h{v_0}\left[ {\left( {v + \dfrac{3}{2}} \right) - \left( {v + \dfrac{1}{2}} \right)} \right]$
$\Delta E = h{v_0} = (6.626 \times {10^{ - 34}}J.s)(6.426 \times {10^{13}}{s^{ - 1}})$
$\Delta E = 4.258 \times {10^{ - 20}}J$
Next, we need to find the force constant ‘k’. For this we’ll use the formula ${v_0} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{\mu }}$ , where $\mu$ is the reduced mass. The reduced mass is given as: $\mu = \dfrac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$ . The molar masses should be in kg/mol. The compound given to us $^{12}{C^{16}}O$ .
The reduced mass will be $= \dfrac{{12 \times 16}}{{12 + 16}} = 6.857g/mol = 0.00685kg/mol$
The force constant k from the above equation can be given as: $k = \mu {(2\pi {v_0})^2} = \dfrac{{0.00685}}{{6.022 \times {{10}^{23}}}} \times {(2\pi \times 6.426 \times {10^{13}}{s^{ - 1}})^2}$
$k = 1856.92kg/{s^2} = 1856.92N/m$ .

Note:
The formula ${v_0} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{\mu }}$ is derived from the formula $\omega = \sqrt {\dfrac{k}{m}}$ . $\omega$ is the angular frequency. We will use the reduced mass here and treat it as one effective mass. We know that $\dfrac{\omega }{{2\pi }} = {v_0} = \dfrac{1}{T}$ where T is the time period in seconds. Combining the equation with ${v_0} = {\tilde v_0}c$ we get, ${\tilde v_0} = \dfrac{1}{{2\pi c}}\sqrt {\dfrac{k}{\mu }} \to {v_0} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{\mu }}$ .