CBSE Class 12 Physics Chapter 14 Semiconductor Electronics NCERT Solutions

The correct statement is (c).

An n-type semiconductor is created by doping pure silicon with pentavalent impurity atoms, such as Phosphorus. These pentavalent atoms have five valence electrons. When they replace silicon atoms in the crystal lattice, four of their valence electrons form covalent bonds with the neighboring silicon atoms, and the fifth electron becomes loosely bound and is free to move. Therefore, in an n-type semiconductor, electrons are the majority charge carriers, and holes are the minority charge carriers. Statement (c) accurately reflects this: holes are minority carriers, and pentavalent atoms are the dopants.

The correct statement is (d).

A p-type semiconductor is formed by doping pure silicon with trivalent impurity atoms, such as Aluminium. These trivalent atoms have three valence electrons. When they replace silicon atoms, they form covalent bonds with three neighboring silicon atoms, but there is one bond that lacks an electron. This absence of an electron is called a hole, which can accept an electron from a neighboring atom, effectively moving the hole. Thus, in a p-type semiconductor, holes are the majority charge carriers, and electrons are the minority charge carriers. Statement (d) correctly states that holes are majority carriers and trivalent atoms are the dopants.

The correct statement is (c).

Carbon, silicon, and germanium are all Group 14 elements and have four valence electrons. However, the energy required to excite an electron from the valence band to the conduction band, known as the energy band gap ($E\_g$), varies significantly among them. Carbon (in its diamond form) has the largest energy band gap, making it an excellent insulator. Silicon has a smaller band gap than carbon, classifying it as a semiconductor. Germanium has the smallest band gap among the three, making it more conductive than silicon at room temperature. The relationship between their energy band gaps is:

$$(E\_g)\_C\; >\; (E\_g)\_\{Si\}\; >\; (E\_g)\_\{Ge\}$$

Therefore, statement (c) is true.

The correct statement is (c).

In a semiconductor, charge carriers tend to move from a region where their concentration is high to a region where their concentration is low. This process is called diffusion. In the p-region of a p-n junction, the concentration of holes (which are the majority carriers) is significantly higher than in the n-region. Conversely, the concentration of electrons (majority carriers in the n-region) is higher in the n-region than in the p-region. Therefore, in an unbiased p-n junction, holes naturally diffuse from the p-region (where they are abundant) across the junction into the n-region (where they are scarce). Similarly, electrons diffuse from the n-region to the p-region. Option (c) correctly identifies the higher concentration of holes in the p-region as the primary reason for their diffusion into the n-region.

The correct statement is (c).

A potential barrier is naturally formed at the junction of a p-type and an n-type semiconductor due to the diffusion of charge carriers, which creates a depletion region. When a forward bias is applied, the positive terminal of the external voltage source is connected to the p-region and the negative terminal to the n-region. This applied voltage opposes the internally developed potential barrier. As the forward bias voltage increases, it effectively reduces the height of the potential barrier. Once the applied voltage exceeds the barrier potential, the barrier is significantly lowered, allowing a large number of majority carriers to cross the junction, resulting in a substantial forward current. Therefore, applying a forward bias lowers the potential barrier.