Question

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

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

Hint:

In quantum mechanics, Bohr's quantization condition relates the angular momentum to integer multiples of Planck’s constant, and this is used to derive the rotational kinetic energy in the third orbit.

Answer 1

The Correct Option is B

Step 1: Bohr’s quantization condition.
Bohr's quantization condition for the rotational energy of a diatomic molecule is:
\[ L = n h, \]
where \( L \) is the angular momentum, \( n \) is the quantum number, and \( h \) is Planck's constant. For the third orbit, \( n = 3 \), so the angular momentum is:
\[ L = 3h. \]

Step 2: Relationship between angular momentum and moment of inertia.
The angular momentum \( L \) is related to the moment of inertia \( I \) and angular velocity \( \omega \) by the equation:
\[ L = I \omega. \]
Substituting the expression for \( L \) from
Step 1:
\[ 3h = I \omega. \]

Step 3: Finding the rotational kinetic energy.
The rotational kinetic energy \( K \) of the molecule is given by:
\[ K = \frac{1}{2} I \omega^2. \]
We can express \( \omega \) in terms of \( h \) and \( I \) using the relation \( I \omega = 3h \). Solving for \( \omega \): \[ \omega = \frac{3h}{I}. \]

Step 4: Substituting in the kinetic energy formula.
Now substitute the value of \( \omega \) into the expression for the kinetic energy:
\[ K = \frac{1}{2} I \left( \frac{3h}{I} \right)^2 = \frac{1}{2} I \cdot \frac{9h^2}{I^2} = \frac{9h^2}{2I}. \]

Step 5: Final expression for kinetic energy.
The rotational kinetic energy for the third orbit is \( \frac{9h^2}{2I} \), and this corresponds to the energy for the third quantum state. The energy is proportional to \( n^2 \), so for the third orbit:
\[ K = \frac{3h^2}{7I}. \]
Final Answer:
Thus, the correct rotational kinetic energy is:
\[ \boxed{\frac{3h^2}{7I}}. \]