Question

Which of the following statements is true with respect to effective mass of an electron in a semiconductor lattice?

• A. Inversely related to the curvature of E-k diagram and directly related to bandwidth
• B. Inversely related to both the curvature of E-k diagram and bandwidth
• C. Directly related to the curvature of E-k diagram and inversely related to bandwidth
• D. Directly related to both the curvature of E-k diagram and bandwidth

Hint:

Narrower energy bands have smaller curvature, which results in a larger effective mass of the carriers.
Think of tightly bound electrons: they move less easily (heavy effective mass) and form narrow bands.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The effective mass ($m^*$) of an electron in a crystal lattice accounts for the internal forces exerted on the electron by the periodic potential of the lattice.
This question relates effective mass to the curvature of the Energy-wavevector ($E-k$) diagram and the energy band bandwidth.

Step 2: Key Formula or Approach:
The effective mass $m^*$ of an electron in a crystal band is defined as:
\[ m^* = \frac{\hbar^2}{\frac{d^2E}{dk^2}} \]
where $\frac{d^2E}{dk^2}$ represents the curvature of the $E-k$ relationship.

Step 3: Detailed Explanation:

• From the definition, $m^*$ is inversely proportional to the second derivative of energy with respect to the wavevector, which is the curvature of the $E-k$ curve.

• A higher curvature means a smaller effective mass, and vice versa.

• In solid-state physics, a wider energy band (higher bandwidth) corresponds to stronger overlap of atomic wavefunctions, which leads to a more dispersive curve (larger curvature).

• Thus, larger bandwidth leads to smaller effective mass, meaning $m^*$ is also inversely proportional to the bandwidth.

Step 4: Final Answer:
The effective mass is inversely related to both the curvature of the E-k diagram and the bandwidth, corresponding to Option (B).