Question

In the Kronig penny Model, the energy of the lowest band at wave vector k = 0 is given by \( E = \frac{h^2 P}{4\pi^2 m a^2} \). This value of energy holds for which condition of Kronig penny potential (P):

• A. P >> 1
• B. P << -1
• C. P = 1
  (D) P << 1
• D.

Hint:

Remember the two important limits for the Kronig-Penney model:

\textbf{Weak Potential (P \(\ll\) 1):} Corresponds to the nearly-free electron model. The energy bands are wide, and the gaps are small.
\textbf{Strong Potential (P \(\gg\) 1):} Corresponds to the tight-binding model. The energy bands are narrow, and the gaps are large.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The Kronig-Penney model describes the behavior of an electron in a 1D periodic potential of square barriers. The parameter P, known as the barrier strength, is defined as \( P = \frac{mV_0ba}{\hbar^2} \), where \(V_0\) is the barrier height and \(b\) is the barrier width. The question asks for the condition on P for which a specific formula for the energy at k=0 is valid.
Step 2: Key Formula or Approach:
The general equation from the Kronig-Penney model relating energy (via \(\alpha\), where \(E = \hbar^2\alpha^2/2m\)) to the wave vector k is:
\[ \frac{P}{\alpha a} \sin(\alpha a) + \cos(\alpha a) = \cos(ka) \]
We need to solve this equation at k=0 under a certain approximation for P.
Step 3: Detailed Explanation:
The given energy formula is \( E = \frac{h^2 P}{4\pi^2 m a^2} \). Let's convert this using \(\hbar = h/2\pi\):
\[ E = \frac{(2\pi\hbar)^2 P}{(2\pi)^2 m a^2} = \frac{\hbar^2 P}{m a^2} \]
This formula is an approximation that holds in the nearly-free electron limit. This limit corresponds to a very weak periodic potential, which means the barrier strength P is very small (\(P \ll 1\)).
Let's verify this by solving the general equation for k=0 and \(P \ll 1\).
At k=0, \(\cos(ka) = 1\). The equation becomes:
\[ \frac{P}{\alpha a} \sin(\alpha a) + \cos(\alpha a) = 1 \]
For a weak potential, the energy E is small, meaning \(\alpha a\) is small. We can use the Taylor series expansions: \(\sin(x) \approx x - x^3/6\) and \(\cos(x) \approx 1 - x^2/2\).
\[ \frac{P}{\alpha a} \left(\alpha a - \frac{(\alpha a)^3}{6}\right) + \left(1 - \frac{(\alpha a)^2}{2}\right) \approx 1 \]
\[ P \left(1 - \frac{(\alpha a)^2}{6}\right) - \frac{(\alpha a)^2}{2} \approx 0 \]
\[ P \approx \frac{(\alpha a)^2}{6} P + \frac{(\alpha a)^2}{2} \]
Since \(P \ll 1\), the term with P on the right is negligible.
\[ P \approx \frac{(\alpha a)^2}{2} \implies (\alpha a)^2 \approx 2P \]
Now substitute this back into the energy formula:
\[ E = \frac{\hbar^2 \alpha^2}{2m} = \frac{\hbar^2 (\alpha a)^2}{2ma^2} \approx \frac{\hbar^2 (2P)}{2ma^2} = \frac{\hbar^2 P}{ma^2} \]
This matches the formula given in the question.
Step 4: Final Answer:
The derivation confirms that the given energy expression is valid in the weak potential limit, i.e., for \(P \ll 1\).