CBSE Class 12 Physics NCERT Solutions: Semiconductor Electronics

The correct statement is (c).

In an n-type semiconductor, the doping is done using pentavalent atoms (like Phosphorus or Arsenic). These pentavalent atoms have five valence electrons. When doped into silicon (which has four valence electrons), four electrons form covalent bonds with neighboring silicon atoms, and the fifth electron becomes free, contributing to electrical conductivity. Therefore, in n-type silicon, electrons are the majority charge carriers, and holes are the minority charge carriers. Statement (c) correctly identifies holes as minority carriers and pentavalent atoms as dopants, although it omits that electrons are the majority carriers.

The correct statement is (d).

A p-type semiconductor is created by doping pure silicon with trivalent atoms (like Boron or Aluminum). These trivalent atoms have three valence electrons. When they replace silicon atoms, they form covalent bonds with three neighboring silicon atoms, but there is a deficiency of one electron to complete the fourth bond. This deficiency is called a 'hole'. These holes act as positive charge carriers. Therefore, in a p-type semiconductor, holes are the majority carriers, and electrons are the minority carriers. Statement (d) correctly identifies holes as majority carriers and trivalent atoms as the dopants.

The correct statement is (c).

Carbon, silicon, and germanium are all elements from Group 14 of the periodic table, possessing four valence electrons. However, they differ significantly in their electrical properties due to their distinct energy band gaps ($E\_g$). The energy band gap represents the minimum energy required to excite an electron from the valence band to the conduction band. Carbon, in its diamond form, has a very large energy band gap, classifying it as an insulator. Silicon has a smaller, moderate band gap, making it a semiconductor. Germanium has an even smaller band gap, also classifying it as a semiconductor but with higher conductivity than silicon at room temperature. The approximate values are: $(E\_g)\_C\; \backslash approx\; 5.5\; \backslash text\{\; eV\}$, $(E\_g)\_\{Si\}\; \backslash approx\; 1.1\; \backslash text\{\; eV\}$, and $(E\_g)\_\{Ge\}\; \backslash approx\; 0.7\; \backslash text\{\; eV\}$. Therefore, the relationship between their energy band gaps is $(E\_g)\_C\; >\; (E\_g)\_\{Si\}\; >\; (E\_g)\_\{Ge\}$.

The correct statement is (c).

In a semiconductor diode, a p-n junction is formed by joining a p-type semiconductor and an n-type semiconductor. In the p-type material, holes are the majority charge carriers, while in the n-type material, electrons are the majority charge carriers. Due to the difference in concentration across the junction, charge carriers tend to move from the region of higher concentration to the region of lower concentration. Specifically, the concentration of holes is much higher in the p-region than in the n-region. This concentration gradient drives the diffusion of holes from the p-region into the n-region. Similarly, electrons diffuse from the n-region to the p-region. While free electrons in the n-region do attract holes, and a potential difference is established, the primary driving force for diffusion in an unbiased junction is the concentration difference.

The correct statement is (c).

When a p-n junction is forward biased, the positive terminal of the external voltage source is connected to the p-region, and the negative terminal is connected to the n-region. This applied external voltage opposes the built-in potential barrier that exists across the junction. As a result, the effective potential barrier height is lowered. This reduction in the potential barrier allows a larger number of majority charge carriers (holes from the p-region and electrons from the n-region) to diffuse across the junction, leading to a significant forward current. Therefore, forward bias lowers the potential barrier, it does not raise it, nor does it reduce the majority carrier current to zero.