Question
A diatomic molecule is made of two masses $${m_1}$$ and $${m_2}$$ which are separated by a distance $$r.$$ If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by : ($$n$$ is an integer)
A.
$$\frac{{{{\left( {{m_1} + {m_2}} \right)}^2}{n^2}{h^2}}}{{2m_1^2m_2^2{r^2}}}$$
B.
$$\frac{{{n^2}{h^2}}}{{2\left( {{m_1} + {m_2}} \right){r^2}}}$$
C.
$$\frac{{2{n^2}{h^2}}}{{\left( {{m_1} + {m_2}} \right){r^2}}}$$
D.
$$\frac{{\left( {{m_1} + {m_2}} \right){n^2}{h^2}}}{{2{m_1}{m_2}{r^2}}}$$
Answer :
$$\frac{{\left( {{m_1} + {m_2}} \right){n^2}{h^2}}}{{2{m_1}{m_2}{r^2}}}$$
Solution :
The energy of the system of two atoms of diatomic molecule $$E = \frac{1}{2}I{\omega ^2}$$ where $$I = $$ moment of inertia
$$\omega = {\text{Angular velocity}} = \frac{L}{I},$$
$$L = $$ Angular momentum
$$I = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)$$
$$\eqalign{
& {\text{Thus,}}\,E = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right){\omega ^2}\,......\left( {\text{i}} \right) \cr
& E = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)\frac{{{L^2}}}{{{I^2}}} \cr
& L = n\hbar \,\,\left( {{\text{According Bohr's Hypothesis}}} \right) \cr
& E = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)\frac{{{L^2}}}{{{{\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)}^2}}} \cr
& E = \frac{1}{2}\frac{{{L^2}}}{{\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)}} = \frac{{{n^2}{h^2}}}{{8{\pi ^2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)}} \cr
& E = \frac{{\left( {{m_1} + {m_2}} \right){n^2}{h^2}}}{{8{\pi ^2}{r^2}{m_1}{m_2}}} \cr} $$