A diatomic molecule is madde of two masses `m_(1) and m_(2)` which are separated by a distance `r` . If we calculate its rotational energy by appliying Bohr's rule of angular momemtum 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

The energy of the system of two atoms of diatomic
molecule `E = (1)/(2) 1 omega ^(2)`
where `1=` moment of inertia
`omega = Angular velocity= (L)/(1)`
`L = Angularmomentum `
`1 = (1)/(2) (m_(1) r_(1)^(2) + m^(2)r_(2)^(2)`
`Then E = (1)/(2) (m_(1) r_(1)^(2) + m^(2)r_(2)^(2))omega^(2)` ....(i)
` E = (1)/(2) (m_(1) r_(1)^(2) + (m_(1) r_(1)^(2) )(m^(2)r_(2)^(2)) (L^(2))/(1^(2))`
`L = n (nh)/(2n)` (According Bohr's Hydrogen )
` E = (1)/(2) (m_(1) r_(1)^(2) + (m_(1) r_(1)^(2))(L^(2))/( (m_(1) r_(1)^(2) + (m_(1) r_(2)^(2))^(2))`
E = (1)/(2) (L^(2))/(2(m_(1) r_(1)^(2) + (m_(1) r_(2)^(2))= (n^(2)h^(2))/(8 pi ^(2)(m_(1) r_(1)^(2) + (m_(1) r_(2)^(2))`
`E = ((m_(1) + m_(2))n^(2)h^(2))/(8 pi ^(2) r^(2) m_(1) m_(2)) [:' eta = (m_(2)r)/( m_(1) + m_(2)) r_(2) = ((m_(2)r)/( m_(1) + m_(2)) r_(2))]`