GATE 2025 EC Q.48: solution and answer

GATE 2025 · ECQ.48 · 2 marks · MCQ

The electron mobility $μ_{n}$ in a non-degenerate germanium semiconductor at 300 K is 0.38 m²/Vs.

The electron diffusivity $D_{n}$ at 300 K (in cm²/s, rounded off to the nearest integer) is ______.

(Consider the Boltzmann constant $k_{B} = 1.38 \times 1 0^{- 23}$ J/K and the charge of an electron $e = 1.6 \times 1 0^{- 19}$ C.)

See the question in the paper

Answer and solution

ANSWER

The correct answer is (B) 98 cm²/s.

Apply the Einstein relation: $\frac{D _{n}}{μ _{n}} = V_{T} = \frac{k _{B} T}{q}$
Compute the thermal voltage using the given constants: $V_{T} = \frac{( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{1.6 \times 1 0 ^{- 19}} = 0.02588 V$
Compute diffusivity in SI units: $D_{n} = μ_{n} V_{T} = 0.38 \times 0.02588 = 9.83 \times 1 0^{- 3} m^{2} / s$
Convert to $cm^{2} / s$ (1 m $^{2}$ = $1 0^{4}$ cm $^{2}$ ): $D_{n} = 9.83 \times 1 0^{- 3} \times 1 0^{4} = 98.3 cm^{2} / s$
Round off to the nearest integer: $D_{n} \approx 98 cm^{2} / s$ , matching option (B)

KEY IDEA

The Einstein relation links diffusivity and mobility: $D_{n} / μ_{n} = V_{T} = k_{B} T / q$ , the thermal voltage.

WATCH OUT FOR

Forgetting to convert from m²/s to cm²/s (multiply by 10⁴), which would give an answer near 0.0098 and no matching option, or misreading μn's units.