Question

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

• A. \(\frac{h}{2I\pi ^{2}}\)
• B. \(\frac{h^{2}}{2I\pi ^{2}}\)
• C. \(\frac{h^{2}}{2I^{2}\pi ^{2}}\)
• D. \(\frac{h}{2I^{2}\pi }\)

Hint:

Apply the Bohr condition L = n h / 2 pi and write kinetic energy as L squared over 2 I.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Bohr's quantisation says the angular momentum is \(L = \dfrac{nh}{2\pi}\). Rotational kinetic energy can be written in terms of \(L\) and the moment of inertia \(I\).

Step 2: Angular Momentum
For the second orbit, \(n = 2\), so \(L = \dfrac{2h}{2\pi} = \dfrac{h}{\pi}\).

Step 3: Kinetic Energy
\(K = \dfrac{1}{2}I\omega^2 = \dfrac{L^2}{2I}\)
\(K = \dfrac{1}{2I}\left(\dfrac{h}{\pi}\right)^2 = \dfrac{h^2}{2I\pi^2}\)

Step 4: Compare the options
This matches option (B).

Final Answer:
The rotational kinetic energy is h squared over 2 I pi squared. \[ \boxed{\text{(B) }\dfrac{h^{2}}{2I\pi^{2}}} \]