A diatomic molecules is made of two masses `m_(2) and m_(2)` which are separated by a distance r. We calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by (n is an integer)

Options

A. `((m_(1)+m_(2))^(2)n^(2)h^(2))/(2m_(1)^(2) m_(2)^(2) r^(2))`
B. `(n^(2)h^(2))/(2(m_(1)+m_(2))r^(2))`
C. `(2n^(2)h^(2))/((m_(1)+m_(2))r^(2))`
D. `((m_(1)+m_(2))n^(2)h^(2))/(2m_(1)m_(2)r^(2))`

Answer

D

Explanation

`m_(1)r_(1)=m_(2)r_(2)`
`r_(1)+r_(2)=r`
`r_(1)=(m_(2)r)/(m_(1)+m_(2)), r_(2)=(m_(1)r)/(m_(1)+m_(2))`
`epsi=1/2 I omega^(2)`
`=1/2 (m_(1)r_(1)^(2)+m_(2)r_(2)^(2))omega^(2) ........(i)`
`mvr=(nh)/(2pi)=I omega`
`rArr omega=(nh)/(2pi I)`
`epsi=1/2 I. (n^(2)h^(2))/(4pi^(2)I^(2))`
`=(n^(2)h^(2))/(8pi^(2)) (1)/((m_(1)r_(1)^(2)+m_(2)r_(2)^(2)))`
`=(n^(2)h^(2))/(8pi^(2)) (1)/(m_(1)(m_(2)^(2)r^(2))/((m_(!)+m_(2))^(2))+m_(2) (m_(1)^(2)r^(2))/((m_(1)+m_(2))^(@)))`
`=(n^(2)h^(2))/(8pi^(2)r^(2)) ((m_(1)+m_(2))^(2))/(m_(1)m_(2)(m_(1)+m_(2)))=((m_(1)+m_(2))n^(2)h^(2))/(8pir^(2)m_(1)m_(2))`
`=((m_(1)+m_(2))n^(2)h^(2))/(2m_(1)m_(2)r^(2))`