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

Hint: 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 }} $ .