From equation 42.7,the vibrational energy levels of a diatomic molecule are given by:\({E_n} = \left( {n + \frac{1}{2}} \right)\hbar \sqrt {\frac{{k'}}{{{m_r}}}} \)
Where\(k'\)is the force constant and\({m_r}\)is the reduced mass.
From equation 42.4,the reduced mass of a diatomic molecule of two atoms with masses\({m_1}\)and\({m_2}\)is given by:\({m_r} = \frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}\).
The vibrational energy difference between\(n = 0\)and\(n = 1\)levels is\(\Delta E = 0.463eV\), the mass if an oxygen atom is\({m_1} = 2.66 \times {10^{ - 26}}kg\)and the mass of a hydrogen atom is\({m_2} = 1.67 \times {10^{ - 27}}kg\).
From the first equation, the energy difference between\(n = 0\)and\(n = 1\)levels is given by:\(\Delta E = {E_1} - {E_0} = \hbar \omega \)
Solving for\(\omega \), we get:\(\omega = \frac{{\Delta E}}{\hbar }\)
Now, we plug our value for\(\Delta E\), so we get the angular frequency of vibrations for\(OH\)molecule:
\(\omega = \frac{{0.463 \times 1.6 \times {{10}^{ - 19}}}}{{1.055 \times {{10}^{ - 34}}}} = 7.02 \times {10^{14}}rad/s\)
We know that the relation between the angular frequency and the frequency is\(f = \frac{\omega }{{2\pi }}\); Thus, the frequency of vibration is:
\(f = \frac{{7.02 \times {{10}^{14}}}}{{2\pi }} = 1.12 \times {10^{14}}Hz\)
We plug our values for\({m_1}\)and\({m_2}\)into the second equation, so we get the reduced mass of\(OH\)molecule:
\({m_r} = \frac{{2.66 \times {{10}^{ - 26}} \times 1.67 \times {{10}^{ - 27}}}}{{2.66 \times {{10}^{ - 26}} + 1.67 \times {{10}^{ - 27}}}} = 1.57 \times {10^{ - 27}}kg\)
From the first equation, we can see that\(\omega = \sqrt {\frac{{k'}}{{{m_r}}}} \). Thus,\(k' = {m_r}{\omega ^2}\).
Finally, we plug our values for\({m_r}\)and\(\omega \), so we get the force constant:
\(\begin{aligned}k' = 1.57 \times {10^{ - 27}} \times {\left( {7.02 \times {{10}^{14}}} \right)^2}\\k' = 775N/m\end{aligned}\)
