Question

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

Options

• `"h"/(2"l"pi^2)`

• `"h"^2/(2"l"pi^2)`

• `"h"^2/(2"l"^2pi^2)`

• `"h"/(2"l"^2pi)`

Answer 1

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

Explanation:

The diatomic molecule's angular momentum is

`"L"="l"omega="nh"/(2pi)`

For second orbit, `"L"=(2"h")/(2pi)="h"/pi`

Rotational energy `=1/2"l"omega^2=("l"omega)^2/(2"l")="L"^2/(2"l")=("h"/pi)^2xx1/(2"l")`

`="h"^2/(2"l"pi^2)`