Question 426: Atoms & Nuclei

Question

Using Bohr's quantization condition, the rotational kinetic energy in the third orbit for a diatomic molecule is ( $h=$ Planck's constant, $\mathrm{I}=$ moment of inertia of diatomic molecule)

Accepted Answer

We are tasked to find the
rotational kinetic energy in the third orbit
for a
diatomic molecule
using Bohr's quantization condition.
Step 1: BohrÃ¢ÂÂs quantization condition for angular momentum
$ L = n \frac{h}{2\pi}, \quad n = 1,2,3,\dots $
Step 2: Rotational kinetic energy in terms of angular momentum
For a rotating rigid body (diatomic molecule modeled as a rigid rotor), the rotational kinetic energy is
$ E = \frac{L^2}{2I} $
Step 3: Substitute quantized angular momentum
$ E_n = \frac{1}{2I} \left( n \frac{h}{2\pi} \right)^2 $
$ E_n = \frac{n^2 h^2}{8 \pi^2 I} $
Step 4: Energy in the
third orbit
$(n = 3)$
$ E_3 = \frac{3^2 h^2}{8 \pi^2 I} = \frac{9h^2}{8 \pi^2 I} $
Ã¢ÂÂ Final Answer:
Option
A
:
$ \boxed{\frac{9 h^2}{8 \pi^2 I}} $

Answer

$\frac{9 h^2}{8 \pi^2 I}$

Answer

$\frac{3 \mathrm{~h}^2}{8 \pi^2 \mathrm{I}}$

Answer

$\frac{6 h^2}{8 \pi I}$

Answer

$\frac{12 h^2}{7 \pi^2 I}$

Atoms & Nuclei

Using Bohr's quantization condition, the rotational kinetic energy in the third orbit for a diatomic molecule is ( h= Planck's constant, \mathrm{I}= moment of inertia of diatomic molecule)

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