GATE 2025 EC Q.46: solution and answer

GATE 2025 · ECQ.46 · 2 marks · MCQ

The intrinsic carrier concentration of a semiconductor is $2.5 \times 1 0^{16} m^{- 3}$ at 300 K. If the electron and hole mobilities are $0.15 m^{2} / Vs$ and $0.05 m^{2} / Vs$ , respectively, then the intrinsic resistivity of the semiconductor (in k $Ω \cdot$ m) at 300 K is ________. (Charge of an electron $e = 1.6 \times 1 0^{- 19} C$ .)

(A)
1.65
(B)
1.25
(C)
0.85
(D)
1.95

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Answer and solution

ANSWER

The intrinsic resistivity of the semiconductor at 300 K is 1.25 kΩ·m, option (B).

Official answer: BSource: Official answer key, GATE 2025 EC (IIT Roorkee)

For an intrinsic semiconductor, the electron concentration equals the hole concentration, both equal to $n_{i}$ . Both carrier types contribute to conduction, so the intrinsic conductivity is $σ_{i} = e n_{i} (μ_{n} + μ_{p})$
Substitute the given values: $e = 1.6 \times 1 0^{- 19} C$ , $n_{i} = 2.5 \times 1 0^{16} m^{- 3}$ , and $μ_{n} + μ_{p} = 0.15 + 0.05 = 0.20 m^{2} / Vs$ .
$σ_{i} = (1.6 \times 1 0^{- 19}) (2.5 \times 1 0^{16}) (0.20) = 8 \times 1 0^{- 4} S/m$
The resistivity is the reciprocal of conductivity: $ρ_{i} = \frac{1}{σ _{i}} = \frac{1}{8 \times 1 0 ^{- 4}} = 1250 Ω \cdot m$
Converting to k $Ω \cdot m$ : $ρ_{i} = 1.25 k Ω \cdot m$ which matches option (B).

KEY IDEA

In an intrinsic semiconductor, electrons and holes exist in equal concentration $n_{i}$ , so both carrier types contribute to conductivity: $σ_{i} = e n_{i} (μ_{n} + μ_{p})$ , and resistivity is simply the reciprocal, $ρ_{i} = 1/ σ_{i}$ .

WATCH OUT FOR

Using only the electron mobility (or only the hole mobility) instead of the sum $μ_{n} + μ_{p}$ , or forgetting to convert the final answer from $Ω \cdot m$ to $k Ω \cdot m$ (giving 1250 instead of 1.25).

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 9 Carrier transport questions, with notes

GATE 2025 EC Q.46: solution and answer

Answer and solution

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