Einstein relation: mobility–diffusion conversion at 300 K A semiconductor has carrier mobility μ = 0.40 m^2/(V·s) at 300 K. Using the Einstein relation, compute the diffusion constant D (in m^2/s), taking kT/q ≈ 0.026 V at 300 K.

Introduction / Context:
The Einstein relation links random (diffusive) carrier motion to drift under an electric field. Converting between mobility μ and diffusion coefficient D is routine in device modeling (diodes, BJTs, MOSFETs) and semiconductor transport problems.

Given Data / Assumptions:

• Mobility μ = 0.40 m^2/(V·s).
• Temperature T = 300 K.
• Thermal voltage: kT/q ≈ 0.026 V at 300 K.
• Non-degenerate semiconductor and low-field conditions.

Concept / Approach:
The Einstein relation for carriers is:

D = μ * (kT / q).

It follows from detailed balance between drift and diffusion in thermal equilibrium and assumes Maxwell–Boltzmann statistics.

Step-by-Step Solution:
Write the formula: D = μ * (kT/q).Substitute numbers: D = 0.40 * 0.026.Compute: 0.40 * 0.026 = 0.0104 m^2/s.Round to two significant figures consistent with data: D ≈ 0.01 m^2/s.

Verification / Alternative check:
Dimensional check: μ in m^2/(V·s) times V gives m^2/s—correct. Typical electron D at 300 K ranges from 0.01 to 0.04 m^2/s depending on μ, so the value is reasonable.

Why Other Options Are Wrong:
0.40 and 0.16 ignore the small thermal voltage factor. 0.04 corresponds to μ ≈ 1.54 m^2/(V·s). 0.0026 would result from μ = 0.10 m^2/(V·s), not 0.40.

Common Pitfalls:
Using kT (in joules) without dividing by q, or confusing electron and hole mobilities. Always convert to the thermal voltage kT/q in volts.

Final Answer:
0.01