Question

Using Bohr's quantization condition, what is the rotational energy in the second orbit for a diatomic molecule. (I = moment of inertia of diatomic molecule, h = Planck's constant)

Options

• `"h"/(2"I"^2 pi)`

• `"h"^2/(2"I" pi)`

• `"h"/(2"I" pi^2)`

• `"h"^2/(2"I"^2 pi^2)`

Answer 1

`"h"^2/(2"I" pi)`

Explanation:

According to Bohr's quantization condition

`"L" = "nh"/(2pi) = (2"h")/(2pi) = "h"/pi`     [For 2^nd orbit]

`"Rotational KE" = 1/2 "L"^2/"I" = 1/2 "h"^2/("I"pi^2)`