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

According to Bohr's rule of quantization of angular momentum,
`mvr=(nh)/(2pi)=nh`
But `v=r omega, :. m(r omega)r=nh`
`omega=(nh)/(mr^(2))`
Rotational energy of molecule
`=1/2Iomega^(2)=1/2(mr^(2)) ((nh)/(mr^(2)))^(2)`
`E=(n^(2)h^(2))/(2mr^(2))`
As `m=(m_(1)m_(2))/(m_(1)+m_(2))`
`:. E= (n^(2)h^(2)(m_(1)+m_(2)))/(2m_(1)m_(2)r^(2))`