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 it energy will be ( 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

Rotational kinetic energy of the two body system rotating about their centre of mass is RKE=`(1)/(2)mu omega^(2)r^(2)`,
where `mu=(m_(1)m_(2))/(m_(1)+m_(2))=`reduces mass and angular momentum, `L=mu omegar^(2)=(nh)/(2pi)`
`therefore RKE=(1)/(2)muomega^(2)r^(2)=(1)/(2)mu*((nh)/(2pimur^(2)))^(2)r^(2)`
`=(n^(2)h^(2))/(8pi^(2)mur^(2))=(n^(2)h^(2))/(2mur^(2))`
`=((m_(1)+m_(2))n^(2)h^(2))/(2m_(1)m_(2)r^(2))" "["Here, h^(2)=(lamda)/(4pi)]`