AP EAMCET · PHYSICS · Semiconductors
A pure semiconductor crystal has \(8 \times 10^{28} \frac{\text { atoms }}{\mathrm{m}^3}\). It is doped by \(2 \mathrm{ppm}\) concentration of pentavalent atoms. The number of holes formed in the semiconductor crystal is (Intrinsic carrier concentration, \(\left.n_i=1 \times 10^{16} \mathrm{~m}^{-3}\right)\).
- A \(4.3 \times 10^9 \mathrm{~m}^{-3}\)
- B \(6.25 \times 10^8 \mathrm{~m}^{-3}\)
- C \(2.5 \times 10^9 \mathrm{~m}^{-3}\)
- D \(125 \times 10^8 \mathrm{~m}^{-3}\)
Answer & Solution
Correct Answer
(B) \(6.25 \times 10^8 \mathrm{~m}^{-3}\)
Step-by-step Solution
Detailed explanation
Number of atoms per \(\mathrm{m}^3\) in semiconductor crystal \(=8 \times 10^{28}\) atoms \(/ \mathrm{m}^3\) Number of doped atoms \(=2 \mathrm{ppm}=2\) atoms out of \(10^6\) atoms. \(\therefore\) Numebr of doped atoms in \(8 \times 10^{28}\) atoms \(/ \mathrm{m}^3\)…
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