The expression for the effective mass is derived as

Question

Examoogle · Accepted Answer

Option A: $m^* = \frac{\hbar^2}{\frac{d^2E}{dk^2}}$

Explanation & Concepts:
The effective mass ($m^*$) of a charge carrier in a crystal lattice is defined as the inverse of the curvature of the energy-momentum ($E-k$) dispersion relation. It represents the inertial properties of electrons or holes moving through a periodic potential, accounting for the lattice forces.

Working Principle: In a crystal, an electron is subjected to internal periodic forces. According to the semi-classical model, the force on an electron due to an external field is $F = m^*a$. By differentiating the group velocity $v_g = \frac{1}{\hbar} \frac{dE}{dk}$ with respect to time and relating it to acceleration $\frac{dv_g}{dt}$, we derive that the acceleration is proportional to the second derivative of energy with respect to the wave vector $k$. Thus, $m^* = \hbar^2 \left( \frac{d^2E}{dk^2} \right)^{-1}$.

Key Formulas: $m^* = \hbar^2 \left( \frac{d^2E}{dk^2} \right)^{-1}$ — Definition of effective mass; $v_g = \frac{1}{\hbar} \frac{dE}{dk}$ — Group velocity of the electron

Key Points: The effective mass is inversely proportional to the curvature of the $E-k$ diagram.; A sharper curvature (larger second derivative) implies a smaller effective mass.; The effective mass can be negative near the top of the energy band, corresponding to holes.; It is a tensor quantity in non-spherical energy bands, but simplifies to a scalar in parabolic bands.

Advantages: Simplifies complex many-body lattice interactions into a single particle model.; Allows usage of classical drift-diffusion equations for semiconductor transport.

Disadvantages: Only strictly valid near band extrema.; Fails to capture full quantum mechanical tunneling or high-energy non-parabolicity without corrections.

Applications: Determining mobility of carriers in semiconductors.; Calculating density of states (DOS) effective mass.; Analyzing transistor switching speeds.

Additional Info: The parameter $\hbar$ is the reduced Planck constant ($h/2\pi$).. Option B is dimensionally incorrect as it lacks the squared $\hbar$ term.. Option C suggests a negative mass which only applies to specific concave regions of the Brillouin zone, not the general definition.

Answer: A is correct — The effective mass is defined as the inverse of the second derivative of the energy $E$ with respect to the wave vector $k$, scaled by $\hbar^2$.

Exam Tip: Remember that the effective mass is inversely related to the curvature: a 'flat' band means a heavy mass (low mobility), while a 'curved' band means a light mass (high mobility).

Examoogle · Answer

Option B: $m^* = \frac{\hbar}{\frac{d^2E}{dk^2}}$

Examoogle · Answer

Option C: $m^* = -\frac{\hbar^2}{\frac{d^2E}{dk^2}}$

Examoogle · Answer

Option D: none

Correct Answer

Discussion (0)

Correct Answer

Related Questions

Discussion (0)