The key feature of Bohr'[s spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton we will extend this to a general rotational motion to find quntized rotantized rotational energy of a diatomic molecule assuming it to be right . The rate to energy applied is Bohr's quantization condition
A diatomic molecute has moment of inertie `1`by Bohr's quantization condition its rotational energy in the `n^(th)` level `(n = 0 is not allowed ) `is

Options

A. `(1)/(n^(2)) ((h^(2))/(8 pi ^(2) 1))`
B. `(1)/(n) ((h^(2))/(8 pi ^(2) 1))`
C. `n ((h^(2))/(8 pi ^(2) 1))`
D. `n^(2) ((h^(2))/(8 pi ^(2) 1))`

Answer

D

Explanation

According to bohr's quantisation principle `L = (nh)/(2 pi) `
Rotational kinetic energy `= (1)/(2) 1 omwga^(2)= (1)/(2)I[(L)/(I)]^(3)`
` [:' L= 1omega]`
`= (1)/(2) (L^(2))/(I) = (1)/(2I) = (n^(2) h^(2))/(4 pi^(2)I) [2^(2)- 1^(2)] ` [Fram(i)]`
`hv = (3h^(2))/(8 pi^(2)I`
`rArr I = (3h)/(8 pi^(2)v) = (3 xx 2 pi^(2) xx (4)/(pi) xx 10^(11) = (3)/(16) xx 10^(-445)`
1.87 xx 10^(-46) kgm^(2) ``