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If a semiconductor has an intrinsic carrier concentration of 1.41 × 10¹⁶/m³. When doped with 10²¹/m³ phosphorous atoms, then the concentration of holes / m³ at room temperature will be.

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1.41 × 10¹⁶

2 × 10¹¹

2 × 10²¹

1.41 × 10¹⁰

answer is B.

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Detailed Solution

$For any extrinsic semiconductor$

${(n_{i})}^{2} = n_{e} n_{h}$

$n_{i} = intrinsic charge carrier concentration$

$n_{e} = electron concentration$

$n_{h} = hole concentration$

$n_{h} = \frac{{(1.41 \times 10^{16})}^{2}}{10^{21}} = 2 \times 10^{11}$

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