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

Answer

A

Explanation

`r_(1)=(m_(2)r)/(m_(1)+m_(2)),r_(2)=(m_(1)r)/(m_(1)+m_(2))`
`L=(I_(1)+I_(2))omega=(nh)/(2pi)" "......(i)`
`(I_(1)+I_(2))=((1)/(2)m_(1)r_(1)^(2)+(1)/(2)m_(2)r_(2)^(2))`
`=(m_(1)m_(2)r^(2))/(2(m_(1)+m_(2)))" ".......(ii)`
From (i), we have `omega=(nh)/(2pi(I_(1)+I_(2)))`
`:.K.E.=(1)/(2)(I_(1)+I_(2))omega^(2)`
`=(1)/(2)(I_(1)+I_(2))(n^(2)h^(2))/(4pi^(2)(I_(1)+I_(2))^(2))`
Using (i) we get `(n^(2)h^(2)(m_(1)+m_(2)))/(2m_(1)m_(2)r^(2))`