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)

• A. \(\frac{9h^{2}}{8\pi ^{2}I}\)
• B. \(\frac{3\text{ }\mathrm{h}^{2}}{8\pi ^{2}\mathrm{I}}\)
• C. \(\frac{6h^{2}}{8\pi I}\)
• D. \(\frac{12h^{2}}{7\pi ^{2}I}\)

Hint:

Use the quantized angular momentum L = nh/2 pi and the formula for rotational kinetic energy L squared over 2I.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Bohr's quantization condition for a rotating molecule: \(L=I\omega=\frac{nh}{2\pi}\).

Step 2: Rotational kinetic energy
\(K=\frac{1}{2}I\omega^2=\frac{L^2}{2I}=\frac{n^2h^2}{8\pi^2I}\).

Step 3: Third orbit
For \(n=3\), \(K=\frac{9h^2}{8\pi^2I}\).

Step 4: Compare the options
This is option (A).

Final Answer:
The rotational kinetic energy is \(\frac{9h^2}{8\pi^2I}\). \[ \boxed{\text{(A) }\dfrac{9h^{2}}{8\pi^{2}I}} \]