From equation\(42.7\),the vibrational energy levels of a diatomic molecule are given by: \({E_l} = \left( {n + \frac{1}{2}} \right)\hbar \omega = \left( {n + \frac{1}{2}} \right)\hbar \sqrt {\frac{{k'}}{{{m_r}}}} \)
Where\(n\)is the vibrational quantum number.
From equation \(42.4\), the reduced mass of diatomic molecule with two atoms of masses\({m_1}\)and\({m_2}\)is given by:\({m_r} = \frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}\)
From equation\(38.2\),the energy of a photon of wavelength\(\lambda \)is given by:\(E = \frac{{hc}}{\lambda }\)
Given that the mass of deuterium is\({m_1} = 3.34 \times {10^{ - 27}}kg\). From the Bridging Problem, the mass of a fluorine atom is\({m_2} = 3.15 \times {10^{ - 26}}kg\), the mass of the hydrogen\({m_3} = 1.67 \times {10^{ - 27}}kg\)and the vibration frequency for\(HF\)is\({f_H} = 1.24 \times {10^{14}}Hz\).
We plug our values for\({m_1}\)and\({m_2}\)into the second equation, so we get the reduced mass of\(HF\)molecule with the deuterium:
\(\begin{aligned}{m_{r,D}} = \frac{{3.34 \times {{10}^{ - 27}} \times 3.15 \times {{10}^{ - 26}}}}{{3.34 \times {{10}^{ - 27}} + 3.15 \times {{10}^{ - 26}}}}\\{m_{r,D}} = 3.02 \times {10^{ - 27}}kg\end{aligned}\)
Then, we plug our values for\({m_2}\)and\({m_3}\)into the second equation, so we get the reduced mass of\(HF\)molecule with the normal hydrogen:
\(\begin{aligned}{m_{r,H}} = \frac{{1.67 \times {{10}^{ - 27}} \times 3.15 \times {{10}^{ - 26}}}}{{1.67 \times {{10}^{ - 27}} + 3.15 \times {{10}^{ - 26}}}}\\{m_{r,D}} = 1.59 \times {10^{ - 27}}kg\end{aligned}\)
From the first equation, the angular frequency in terms of the reduced mass is given by:\(\omega = \sqrt {\frac{{k'}}{{{m_r}}}} \)
We know that the relation between the vibration frequency and the angular frequency is\(f = \frac{\omega }{{2\pi }}\), thus:\(f = \frac{1}{{2\pi }}\sqrt {\frac{{k'}}{{{m_r}}}} \)
So, the vibration frequency is inversely proportional to the square root of the reduced mass,\(f \propto \frac{1}{{\sqrt {{m_r}} }}\).
So, for two molecules with the same force constant, we have:\(\frac{{{f_2}}}{{{f_1}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} \).
Applying this relation to the two\(HF\)molecules, the one with normal hydrogen and the other with deuterium, we get:\({f_2} = {f_1}\sqrt {\frac{{{m_{r,H}}}}{{{m_{r,D}}}}} \).
Now, we plug our values for\({f_1},{m_{r,H}}\)and\({m_{r,D}}\), so we get the vibration frequency of the molecule with the deuterium:
\(\begin{aligned}{f_2} = \left( {1.24 \times {{10}^{14}}} \right)\sqrt {\frac{{1.59 \times {{10}^{ - 27}}}}{{3.02 \times {{10}^{ - 27}}}}} = 89.86 \times {10^{12}}Hz\\{f_2} \approx 90 \times {10^{12}}Hz\end{aligned}\)
