Question 486: Centre of Mass and Rotational Motion

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 $\mathrm{h}=$ Planck's constant)

Accepted Answer

Step 1: Recall the quantisation condition
Bohr's quantisation condition for angular momentum:
$ L = n \frac{h}{2\pi}, \quad n=1,2,3,\dots $
Step 2: Express the rotational energy
Rotational kinetic energy is:
$ E = \frac{L^2}{2I} $
Substitute $ L = n \frac{h}{2\pi} $:
$ E_n = \frac{1}{2I}\left(n \frac{h}{2\pi}ight)^2 $
$ E_n = \frac{n^2 h^2}{8 \pi^2 I} $
Step 3: For the second orbit (n=2)
$ E_2 = \frac{(2)^2 h^2}{8 \pi^2 I} = \frac{4 h^2}{8 \pi^2 I} = \frac{h^2}{2 I \pi^2} $
Ã¢ÂÂ Final Answer:
$ \boxed{	ext{Option B: } \; \frac{h^2}{2 I \pi^2}} $

Answer

$\frac{\mathrm{h}}{2 \mathrm{I} \pi^2}$

Answer

$\frac{h^2}{2 I \pi^2}$

Answer

$\frac{\mathrm{h}^2}{2 \mathrm{I}^2 \pi^2}$

Answer

$\frac{\mathrm{h}}{2 \mathrm{I}^2 \pi}$

Centre of Mass and Rotational Motion

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 \mathrm{h}= Planck's constant)

Choose the correct answer: