Chapter 12 of 14

Updated 6 August 2026

NCERT Solutions

Semiconductor Electronics: Materials, Devices and Simple Circuits — NCERT Solutions

Madhya Pradesh Board · Class 12 · Physics

NCERT Solutions for Semiconductor Electronics: Materials, Devices and Simple Circuits, Madhya Pradesh Board Class 12 Physics: 6 textbook questions solved.

105 questions60 flashcards9 formulas & key relations5 concepts

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Exercises

14.1In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.Show solution

Correct Option: (c) Holes are minority carriers and pentavalent atoms are the dopants.

Justification:
In an n-type semiconductor, silicon (tetravalent) is doped with pentavalent atoms (donors) such as As, Sb, or P. Each dopant atom donates one extra electron to the conduction band. Therefore:

Electrons are the majority carriers.
Holes are the minority carriers.
The dopants are pentavalent atoms.

Option (a) is wrong because it states trivalent atoms are dopants (that would give p-type). Option (b) is wrong because electrons are majority carriers, not minority. Option (d) is wrong because holes are not majority carriers in n-type.

Hence, option (c) is correct.

14.2Which of the statements given in Exercise 14.1 is true for p-type semiconductors?Show solution

Correct Option: (d) Holes are majority carriers and trivalent atoms are the dopants.

Justification:
In a p-type semiconductor, silicon (tetravalent) is doped with trivalent atoms (acceptors) such as B, Al, or In. Each dopant atom creates a hole (electron vacancy) in the valence band. Therefore:

Holes are the majority carriers.
Electrons are the minority carriers.
The dopants are trivalent atoms.

This matches statement (d) from Exercise 14.1, which is the correct answer for p-type semiconductors.

14.3Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to

(E_g)_C

(E_g)_{Si}

and

(E_g)_{Ge}

. Which of the following statements is true?
(a)

(E_g)_{Si} < (E_g)_{Ge} < (E_g)_C

(b)

(E_g)_C < (E_g)_{Ge} > (E_g)_{Si}

(c)

(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}

(d)

(E_g)_C = (E_g)_{Si} = (E_g)_{Ge}

Show solution

Correct Option: (c) $(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$

Given/Known values:

$(E_g)_C \approx 5.4\,\text{eV}$ (Carbon/Diamond — insulator)
$(E_g)_{Si} \approx 1.1\,\text{eV}$ (Silicon — semiconductor)
$(E_g)_{Ge} \approx 0.7\,\text{eV}$ (Germanium — semiconductor)

Concept: As we go down Group 14 of the periodic table (C → Si → Ge), the atomic size increases, the outermost electrons are farther from the nucleus, and the interatomic bonding becomes weaker. This results in a decrease in the energy band gap.

Conclusion:
$( E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$

Hence, option (c) is correct.

14.4In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.

Correct Option: (c) hole concentration in p-region is more as compared to n-region.

Justification:
Diffusion is the movement of charge carriers from a region of higher concentration to a region of lower concentration. In a p-n junction:

The p-region has a very high concentration of holes.
The n-region has a very low concentration of holes.

Due to this concentration gradient, holes diffuse from the p-region to the n-region. This is purely a diffusion process driven by the concentration difference, not by electric attraction or an externally applied potential difference.

Option (a) is incorrect because free electrons do not attract holes across the junction in an unbiased state. Option (b) is incorrect because in an unbiased junction there is no externally applied potential difference.

Hence, option (c) is correct.

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14.5When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.

Correct Option: (c) lowers the potential barrier.

Justification:
In forward bias, the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side. The applied voltage opposes the built-in potential barrier at the junction.

The external voltage pushes majority carriers (holes from p-side and electrons from n-side) towards the junction.
This reduces (lowers) the depletion layer width and consequently lowers the potential barrier.
As a result, a significant majority carrier current flows across the junction.

Option (a) is wrong — raising the barrier occurs in reverse bias. Option (b) is wrong — majority carrier current is not reduced to zero; in fact it increases substantially.

Hence, option (c) is correct.

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14.6In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.

Given:

Input frequency, $f_{in} = 50\,\text{Hz}$

Case 1: Half-Wave Rectifier

Concept: In a half-wave rectifier, only one half-cycle (either positive or negative) of the AC input is allowed to pass through the diode. The other half-cycle is blocked. Therefore, for every one complete cycle of input, there is one pulse of output.

$f_{output} = f_{input} = 50\,\text{Hz}$

Output frequency of half-wave rectifier = 50 Hz

Case 2: Full-Wave Rectifier

Concept: In a full-wave rectifier, both half-cycles of the AC input are utilised (both are converted to the same polarity). Therefore, for every one complete cycle of input, there are two pulses of output.

$f_{output} = 2 \times f_{input} = 2 \times 50 = 100\,\text{Hz}$

Output frequency of full-wave rectifier = 100 Hz

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Frequently Asked Questions

What are the important topics in Semiconductor Electronics: Materials, Devices and Simple Circuits for Madhya Pradesh Board Class 12 Physics?

Key topics in Semiconductor Electronics: Materials, Devices and Simple Circuits include Classification of Solids and Semiconductor Types, Intrinsic Semiconductor and Charge Carriers, Extrinsic Semiconductor and Doping, p-n Junction Formation and Depletion Region. Study these first, then practise questions on each for the Madhya Pradesh Board Class 12 board exam.

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How should I revise Semiconductor Electronics: Materials, Devices and Simple Circuits for the Madhya Pradesh Board Class 12 board exam?

Learn the core ideas first, then work through the 105 practice questions on Semiconductor Electronics: Materials, Devices and Simple Circuits. Revise definitions regularly and use flashcards for quick recall before the exam.

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Semiconductor Electronics: Materials, Devices and Simple Circuits — NCERT Solutions

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Exercises

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