KCET · Physics · Semiconductors

The density of an electron -hole pair in a pure germanium is \( 3 \times 10^{16} \mathrm{~m}^{-3} \) at room temperature
. On doping with aluminium, the hole density increases to \( 4.5 \times 10^{(22)} \mathrm{m}^{-3} \). Now the electron
density \( \left(\right. \) in \( \left.m^{-3}\right) \) in doped germanium is

A \( 1 \times 10^{10} \)
B \( 2 \times 10^{10} \)
C \( 0.5 \times 10^{10} \)
D \( 4 \times 10^{10} \)

Verified Solution

Answer & Solution

Correct Answer

(B) \( 2 \times 10^{10} \)

Step-by-step Solution

Detailed explanation

Given, density of electron-hole pair in pure germanium \(=3 \times 10^{16} \mathrm{~m}^{-3}\)
On doping, hole density \(=4.5 \times 10^{22} \mathrm{~m}^{-3}\)
Electron density \(=?\)
We know \(n_{1}^{2}=n_{h} n_{e}\)
Where \(n_{i}\) is electron-hole density in pure semiconductor; \(n_{h}\) is hole density; \(n_{e}\) is electron density
\(\Rightarrow n_{e}=\frac{n_{i}^{2}}{n_{h}}=\frac{\left(3 \times 10^{16}\right)}{4.5 \times 10^{22}}=2 \times 10^{10} \mathrm{~m}^{-3}\)
Therefore, electron density in doped germanium will be \(2 \times 10^{10} \mathrm{~m}^{-3}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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