GATE 2022 EC Q.17: solution and answer

GATE 2022 · ECQ.17 · 1 mark · MCQ

In a non-degenerate bulk semiconductor with electron density $n = 1 0^{16} cm^{- 3}$ , the value of $E_{C} - E_{F n} = 200 meV$ , where $E_{C}$ and $E_{F n}$ denote the bottom of the conduction band energy and electron Fermi level energy, respectively. Assume thermal voltage as 26 meV and the intrinsic carrier concentration is $1 0^{10} cm^{- 3}$ . For $n = 0.5 \times 1 0^{16} cm^{- 3}$ , the closest approximation of the value of $(E_{C} - E_{F n})$ , among the given options, is ____.

(A)
226 meV
(B)
174 meV
(C)
218 meV
(D)
182 meV

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Answer and solution

ANSWER

(E_C - E_{Fn}) at n = 0.5×10^16 cm^-3 is approximately 218 meV — option (C).

Official answer: CSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

In non-degenerate (Boltzmann) statistics, the electron density is $n = N_{C} exp (- \frac{E _{C} - E _{F n}}{k T})$ , so $E_{C} - E_{F n} = k T ln (N_{C} / n)$ .
For $n_{1} = 1 0^{16} cm^{- 3}$ : $200 meV = 26 meV \times ln (N_{C} / n_{1})$ , giving $ln (N_{C} / n_{1}) = 200/26 = 7.6923$ .
For $n_{2} = 0.5 \times 1 0^{16} cm^{- 3} = n_{1} /2$ : $E_{C} - E_{F n 2} = 26 ln (N_{C} / n_{2}) = 26 [ln (N_{C} / n_{1}) + ln (n_{1} / n_{2})] = 26 [ln (N_{C} / n_{1}) + ln 2]$ .
Substituting: $E_{C} - E_{F n 2} = 200 meV + 26 meV \times ln 2$ .
$26 \times 0.6931 = 18.02 meV$ , so $E_{C} - E_{F n 2} \approx 218.0 meV$ .
This matches option (C), $218 meV$ . The intrinsic carrier concentration $n_{i}$ given in the stem is not needed for this shift calculation — it is a distractor here.

KEY IDEA

In non-degenerate (Boltzmann) statistics, $n = N_{C} exp (- \frac{E _{C} - E _{F n}}{k T})$ , so $E_{C} - E_{F n} = k T ln (N_{C} / n)$ . Since $N_{C}$ and $k T$ are unchanged between the two cases, going from $n_{1}$ to $n_{2} = n_{1} /2$ simply adds $k T ln 2$ to $E_{C} - E_{F n}$ — no need to solve for $N_{C}$ at all.

WATCH OUT FOR

Trying to use $n_{i}$ and the given thermal voltage of 26 meV to compute $E_{C} - E_{F n}$ from an intrinsic-level formula (mixing up $E_{i}$ and $E_{C}$ relations), or forgetting that halving $n$ only shifts the Fermi level by $k T ln 2$ relative to the first case — some solvers redo the full $N_{C}$ calculation and introduce rounding error instead of using the simple shift.

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GATE 2022 EC Q.17: solution and answer

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