GATE 2024 EC Q.63: solution and answer

GATE 2024 · ECQ.63 · 2 marks · NAT

A non-degenerate n-type semiconductor has 5% neutral dopant atoms. Its Fermi level is located at 0.25 eV below the conduction band ( $E_{C}$ ) and the donor energy level ( $E_{D}$ ) has a degeneracy of 2. Assuming the thermal voltage to be 20 mV, the difference between $E_{C}$ and $E_{D}$ (in eV, rounded off to two decimal places) is ____.

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Answer and solution

ANSWER

E_C − E_D ≈ 0.18 eV.

Official answer: 0.17 to 0.19Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

The neutral-donor occupation probability, including the donor level's degeneracy $g$ , is $f_{D} = \frac{1}{1 + \frac{1}{g} exp ( \frac{E _{D} - E _{F}}{k T} )}$ , with $g = 2$ here.
5% of donor atoms are neutral (still hold their electron), so $f_{D} = 0.05$ .
Substituting: $0.05 = \frac{1}{1 + \frac{1}{2} exp ( \frac{E _{D} - E _{F}}{k T} )}$ .
Rearranging: $1 + \frac{1}{2} exp (\frac{E _{D} - E _{F}}{k T}) = \frac{1}{0.05} = 20$ .
So $exp (\frac{E _{D} - E _{F}}{k T}) = 2 (20 - 1) = 38$ .
Taking the natural log: $\frac{E _{D} - E _{F}}{k T} = ln (38) = 3.638$ . With $k T = V_{T} = 20 mV$ , $E_{D} - E_{F} = 3.638 \times 0.02 eV = 0.0728 eV$ .
Given $E_{C} - E_{F} = 0.25 eV$ , so $E_{F} = E_{C} - 0.25 eV$ .
Then $E_{D} = E_{F} + 0.0728 = E_{C} - 0.25 + 0.0728 = E_{C} - 0.1772 eV$ .
Therefore $E_{C} - E_{D} = 0.1772 eV \approx 0.18 eV$ (rounded to two decimal places).

KEY IDEA

Use the degeneracy-corrected donor occupation (neutral-donor) probability, $f_{D} = \frac{1}{1 + \frac{1}{g} exp ( \frac{E _{D} - E _{F}}{k T} )}$ , set it to the given neutral fraction, solve for $E_{D} - E_{F}$ , then combine with $E_{C} - E_{F}$ to get $E_{C} - E_{D}$ .

WATCH OUT FOR

Using the plain Fermi–Dirac form without the donor degeneracy factor $g = 2$ (dropping the $1/ g$ term), or confusing whether "5% neutral" refers to $f_{D}$ (occupied/neutral) versus $1 - f_{D}$ (ionized), which flips the sign of the final answer.

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GATE 2024 EC Q.63: solution and answer

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