Question

In Kronig-Penney model, the energy of lower band at \(k=0\) for \(P<<1\) is given by:

• A. \(0\)
• B. \(\dfrac{h^2P}{ma}\)
• C. \(\dfrac{hP}{ma}\)
• D. \(\dfrac{h^2P}{ma^2}\)

Hint:

In Kronig-Penney model, weak barrier means \(P<<1\), and the lower band energy at \(k=0\) is proportional to \(\frac{h^2P}{ma^2}\).

Answer 1

The Correct Option is D

Concept:
In the Kronig-Penney model, allowed energy bands are obtained from the periodic potential condition.

Step 1: Understand the condition.
The question asks for the lower band energy at: \[ k=0 \] and for weak barrier: \[ P<<1 \]

Step 2: Use the weak barrier result.
For weak potential barrier, the lower band energy at \(k=0\) is proportional to: \[ \frac{h^2P}{ma^2} \]

Step 3: Select correct option. \[ E=\frac{h^2P}{ma^2} \] \[ \therefore \text{Correct Answer is (D)} \]