Question

When in Kronig-Penney Potential, P tends to infinity, the energy spectrum is given by

• A. $a^2 = \frac{2mE}{\hbar^2}$
• B. $a^2 = \frac{4mE}{\hbar^2}$
• C. $a^2 = \frac{6mE}{\hbar^2}$
• D. $a^2 = \frac{8mE}{\hbar^2}$

Hint:

In Kronig-Penney model, the energy spectrum simplifies to a^2 =2mE/ħ^2 under high potential.

Answer 1

The Correct Option is A

In the Kronig-Penney model, for $P \rightarrow \infty$:
\[ E \propto \frac{\hbar^2 k^2}{2m} \]
The relation simplifies to:
\[ a^2 = \frac{2mE}{\hbar^2} \]