91Ó°ÊÓ

From equation\(42.9\),the energy levels of a diatomic molecule including the vibration­-rotation energies are given by:\({E_{n,l}} = l\left( {l + 1} \right)\frac{{{\hbar ^2}}}{{2I}} + \left( {n + \frac{1}{2}} \right)\hbar \omega \)

Where\(l\)is the rotational quantum number and\(n\)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\(42.6\),the moment of inertia of a diatomic molecule about an axis through its centre of mass is given by:\(I = {m_r}r_0^2\)

Where\({r_0}\)is the distance between centres of molecule’s two atoms.

From equation\(38.2\),the energy of a photon of wavelength\(\lambda \)is given by:\(E = \frac{{hc}}{\lambda }\)

Given that the equilibrium separation for\(HI\)molecule is\({r_0} = 0.16nm\), its vibration frequency is\(f = 6.93 \times {10^{13}}Hz\), the mass of the hydrogen atom is\({m_1} = 1.67 \times {10^{ - 27}}kg\)and the mass of iodine atom is\({m_2} = 2.11 \times {10^{ - 25}}kg\).

First, we plug our values for\({m_1}\)and\({m_2}\)into the second equation, so we get the reduced mass of the molecule:

\(\begin{aligned}{}{m_r} = \frac{{1.67 \times {{10}^{ - 27}} \times 2.11 \times {{10}^{ - 25}}}}{{1.67 \times {{10}^{ - 27}} + 2.11 \times {{10}^{ - 25}}}}\\{m_r} = 1.66 \times {10^{ - 27}}kg\end{aligned}\)

Now, we plug our values for\({m_r}\)and\({r_0}\)into the third equation, so we get the moment of inertia of\(HI\)about a perpendicular axis through its centre of mass:

\(\begin{aligned}{}I = 1.66 \times {10^{ - 27}} \times {\left( {0.16 \times {{10}^{ - 9}}} \right)^2}\\I = 4.24 \times {10^{ - 47}}kg.{m^2}\end{aligned}\)

The relation between the frequency and the angular frequency is\(\omega = 2\pi f\), so the angular frequency of \(HI\) molecule is:

\(\begin{aligned}{}\omega = 2\pi \left( {6.93 \times {{10}^{13}}} \right)\\\omega = 4.35 \times {10^{14}}rad/s\end{aligned}\)

Now, we plug our values for\(\omega \)and\(I\)into the first equation, so we get the energy levels in terms of the quantum numbers only:

\(\begin{aligned}{E_{n,l}} = l\left( {l + 1} \right)\frac{{{{\left( {1.055 \times {{10}^{ - 34}}} \right)}^2}}}{{2\left( {4.24 \times {{10}^{ - 47}}} \right)}} + \left( {n + \frac{1}{2}} \right)\left( {1.055 \times {{10}^{ - 34}}} \right)\left( {4.35 \times {{10}^{14}}} \right)\\{E_{n,l}} = l\left( {l + 1} \right)\left( {1.31 \times {{10}^{ - 22}}} \right) + \left( {n + \frac{1}{2}} \right)\left( {4.59 \times {{10}^{ - 20}}} \right)\end{aligned}\)

The energy of the emitted photon from a transition equals the energy difference between the two levels between which the transition occurs,\({E_\gamma } = \Delta E\).

And we can use the fifth equation to calculate the difference in energy between the levels of the transition.

Thus,

For the\(n = 1,l = 1 \to n = 0,l = 0\), the energy of the emitted photon is:

\(\begin{aligned}{E_{\gamma ,i}} = {E_{1,1}} - {E_{0,0}}\\{E_{\gamma ,i}} = \left[ {\left( 1 \right)\left( 2 \right) - 0} \right]\left( {0.819meV} \right) + \left[ {1 + \frac{1}{2} - 0 - \frac{1}{2}} \right]\left( {286.989eV} \right)\\{E_{\gamma ,i}} = 288.627meV\end{aligned}\)

Now, we plug this value into the fourth equation and evaluate for\(\lambda \), so we get the wavelength of the emitted photon from the\(n = 1,l = 1 \to n = 0,l = 0\)transition:

\(\begin{aligned}{\lambda _i} = \frac{{hc}}{{{E_{\gamma ,i}}}} = \frac{{\left( {4.136 \times {{10}^{ - 15}}} \right)\left( {3 \times {{10}^8}} \right)}}{{288.627meV}}\\{\lambda _i} = 4.299\mu m\end{aligned}\)

For the\(n = 1,l = 2 \to n = 0,l = 1\), the energy of the emitted photon is:

\(\begin{aligned}{E_{\gamma ,ii}} = {E_{1,2}} - {E_{0,1}}\\{E_{\gamma ,ii}} = \left[ {\left( 2 \right)\left( 3 \right) - \left( 1 \right)\left( 2 \right)} \right]\left( {0.819meV} \right) + \left[ {1 + \frac{1}{2} - 0 - \frac{1}{2}} \right]\left( {286.989eV} \right)\\{E_{\gamma ,ii}} = 290.266meV\end{aligned}\)

Now, we plug this value into the fourth equation and evaluate for\(\lambda \), so we get the wavelength of the emitted photon from the\(n = 1,l = 2 \to n = 0,l = 1\)transition:

\(\begin{aligned}{}{\lambda _{ii}} = \frac{{hc}}{{{E_{\gamma ,ii}}}} = \frac{{\left( {4.136 \times {{10}^{ - 15}}} \right)\left( {3 \times {{10}^8}} \right)}}{{290.266meV}}\\{\lambda _i} = 4.275\mu m\end{aligned}\)

For the\(n = 2,l = 2 \to n = 1,l = 3\), the energy of the emitted photon is:

\(\begin{aligned}{E_{\gamma ,iii}} = {E_{2,2}} - {E_{1,3}}\\{E_{\gamma ,iii}} = \left[ {\left( 2 \right)\left( 3 \right) - \left( 3 \right)\left( 4 \right)} \right]\left( {0.819meV} \right) + \left[ {2 + \frac{1}{2} - 1 - \frac{1}{2}} \right]\left( {286.989eV} \right)\\{E_{\gamma ,iii}} = 282.0726meV\end{aligned}\)

Now, we plug this value into the fourth equation and evaluate for\(\lambda \), so we get the wavelength of the emitted photon from the\(n = 2,l = 2 \to n = 1,l = 3\)transition:

\(\begin{aligned}{\lambda _{iii}} = \frac{{hc}}{{{E_{\gamma ,iii}}}} = \frac{{\left( {4.136 \times {{10}^{ - 15}}} \right)\left( {3 \times {{10}^8}} \right)}}{{282.0726meV}}\\{\lambda _i} = 4.399\mu m\end{aligned}\)