Question 311: Atoms & Nuclei

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)

Accepted Answer

The energy of the system of two atoms of diatomic
molecule $$E = {1 \over 2}I\omega $$
where $$I=$$ moment of inertia
$$\omega  = $$ Angular velocity $$ = {L \over I}.$$
$$L=$$ Angular momentum
$$I = {1 \over 2}\left( {{m_1}{r_1}^2 + {m^2}{r_2}^2} \right)$$
Thus, $$E = {1 \over 2}\left( {{m_1}{r_1}^2 + m{}_2{r_2}^2} \right){\omega ^2}\,\,\,\,\,\,\,\,\,...\left( i \right)$$
$$E = {1 \over 2}\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right){{{L^2}} \over {{I^2}}}$$
$$L = n{{nh} \over {2n}}$$ (According Bohr's Hypothesis)
$$E = {1 \over 2}\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right){{{L^2}} \over {{{\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right)}^2}}}$$
$$E = {1 \over 2}{{{L^2}} \over {\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right)}}$$
$$ = {{{n^2}{h^2}} \over {8{\pi ^2}\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right)}}$$
$$E = {{\left( {{m_1} + {m_2}} \right){n^2}{h^2}} \over {8{\pi ^2}{r^2}{m_1}{m_2}}}$$
$$\left[ {\,\,} \right.$$ as $$\left. {\,\,\,{r_1} = {{{m_2}r} \over {{m_1} + {m_2}}};\,{r_2} = {{{m_2}r} \over {{m_1} + {m_2}}}\,\,} \right]$$

Answer

$${{{{\left( {{m_1} + {m_2}} \right)}^2}{n^2}{h^2}} \over {2m_1^2m_2^2{r^2}}}$$

Answer

$${{{n^2}{h^2}} \over {2\left( {{m_1} + {m_2}} \right){r^2}}}$$

Answer

$${{2{n^2}{h^2}} \over {\left( {{m_1} + {m_2}} \right){r^2}}}$$

Answer

$${{\left( {{m_1} + {m_2}} \right){n^2}{h^2}} \over {2{m_1}{m_2}{r^2}}}$$

Atoms & Nuclei

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)

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