The electron and hole concentrations in an intrinsic semiconductor are $n_{i}$ per $c m^{3}$ at $300 K$ . Now, if acceptor impurities are introduced with a concentration of $N_{A}$ per $c m^{3}$ (where $N_{A} >> n_{i})$ , the electron concentration per $c m^{3}$ at $300 K$ will be

Open in App

Solution

Ans : (d)

By the law of electrical neutrality
$p = N_{D} = n + N_{A}$
as $N_{D} = 0$
$N_{A} >> n_{i} ≅ 0 p = N_{A}$
using mass action law $n p = n_{i}^{2}$
So, $n = \frac{n_{i}^{2}}{p} = \frac{n_{i}^{2}}{N_{A}}$

Suggest Corrections

Similar questions

N_{D} = 2.25 \times 10^{15} a t o m s / c m^{3}

T = 300 K

n_{i} = 1.5 \times 10^{10} c m^{- 3}

2.25 \times 10^{15} / c m^{3}

1.5 \times 10^{10} / c m^{3}

300 K

1.5 \times 10^{10} / c m^{3}

2.25 \times 10^{15} a t o m s / c m^{3}

An abrupt Si p-n junction has doping conncentration on p-side and n-side as N_{A} = 10^{18} / c m^{3} and N_{D} = 5 \times 10^{15} / c m^{3} respectively. Then the contact potential at the junction is (Assume intrinsic carrier concentration, n_{i} = 1.0 \times 10^{10} / c m^{3}, and k T = 0.026 V)

N

Δ n

Δ p

n_{i}

Join BYJU'S Learning Program