Read the following passage and answer the questions.
A thin wafer of p-type silicon is made by adding boron, a trivalent element, to pure silicon. About one boron atom is added for every million silicon atoms. Pure silicon has about \(5\times10^{28}\) atoms per cubic metre. Its intrinsic carrier concentration \(n_i\) at room temperature is about \(1.5\times10^{16}\ \mathrm{m^{-3}}\). Next, a larger and carefully measured amount of phosphorus, a pentavalent element, is added to one part of the wafer. This part becomes n-type. The wafer now has a p-region and an n-region with a junction between them. As soon as the junction forms, charge carriers start to cross it. After a short time, the net flow of carriers stops and a potential difference appears across the junction. Finally, metal contacts are fixed at the two ends, and the device is joined to a dc supply.
Far from the junction, the hole concentration in the p-region is about \(N_A\), the number of boron atoms per unit volume. In thermal equilibrium, the electron concentration \(n_e\) in this part of the p-region is about
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(n_i^{2}/N_A\)
In thermal equilibrium, the electron and hole concentrations in a semiconductor are linked. Their product is always equal to \( n_i^{2} \). Far from the junction, each boron atom has given one hole, so \( n_h \approx N_A \).
\( \begin{aligned} n_en_h &= n_i^{2} \\ n_e &= \frac{n_i^{2}}{n_h} \\ &= \frac{n_i^{2}}{N_A} \end{aligned} \)
From the above, \( n_e=n_i^{2}/N_A \). Since \( N_A \) is much larger than \( n_i \), this value is far smaller than \( n_i \). Hence, electrons are the minority carriers in the p-region.
NCERT (2026-27):
Extrinsic Semiconductor – “… The electron and hole concentration in a semiconductor in thermal equilibrium is given by”
\( n_en_h=n_i^{2} \)\( \qquad (14.5) \)