ISRO EC 2025 Q.68: solution and answer

ISRO 2025 · ECQ.68 · 1 mark · MCQ

In an N-type semiconductor, the Fermi level E_F equals :

(A)
E_C – kT ln (N_C / N_D)
(B)
E_C – kT ln (N_D / N_C)
(C)
E_V + kT ln (N_C / N_D)
(D)
E_V – kT ln (N_C / N_D)

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Answer and solution

ANSWER

The Fermi level is $E_{F} = E_{C} - k T ln \frac{N _{C}}{N _{D}}$ , which is option (a).

Official answer: ASource: ISRO's final answer key of 15 May 2026

At room temperature essentially all donor atoms in an N-type semiconductor are ionised, so the electron concentration is $n \approx N_{D}$ .
For a non-degenerate semiconductor the electron concentration is $n = N_{C} exp (- \frac{E _{C} - E _{F}}{k T})$ .
Set $N_{D} = N_{C} exp (- \frac{E _{C} - E _{F}}{k T})$ and take the natural log: $ln \frac{N _{D}}{N _{C}} = - \frac{E _{C} - E _{F}}{k T}$ .
Rearranging gives $E_{F} = E_{C} + k T ln \frac{N _{D}}{N _{C}} = E_{C} - k T ln \frac{N _{C}}{N _{D}}$ .
Since $N_{D} < N_{C}$ , the log term $ln (N_{C} / N_{D})$ is positive, so $E_{F}$ lies a little below $E_{C}$ , as expected for N-type material.
Option (a) matches. Option (b) inverts the ratio and would put $E_{F}$ above $E_{C}$ . Options (c) and (d) are referenced to $E_{V}$ , the valence band edge, and do not give a level near $E_{C}$ .

KEY IDEA

In an N-type semiconductor $n \approx N_{D} = N_{C} e^{- (E_{C} - E_{F}) / k T}$ , so $E_{F}$ sits below $E_{C}$ by $k T ln (N_{C} / N_{D})$ .

WATCH OUT FOR

Inverting the ratio inside the logarithm, which places $E_{F}$ above $E_{C}$ , or using the $E_{V}$ -referenced expression that belongs to P-type material.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 3 ISRO Semiconductors questions, with notes

ISRO EC 2025 Q.68: solution and answer

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