NCERT Solutions For Chapter 14: Semiconductor Electronics

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NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits are provided in the article below. A semiconductor is a material whose resistivity is between a conductor such as metallic copper and an insulator such as glass. Its resistivity falls as the temperature rises which is completely opposite to any metal.

Class 12 Physics Chapter 14 Semiconductor Electronics belong to Unit 9 - Electronic Devices which has a weightage of 7 marks in the CBSE Board examinations. Class 12 Physics Semiconductor Electronics NCERT Solutions covers the concepts of intrinsic and extrinsic semiconductors, P-n Junction, and Rectifiers

Download PDF: NCERT Solutions for Class 12 Physics Chapter 14

NCERT Solutions for Class 12 Physics Chapter 14

Class 12 Physics Chapter 14 – Topics Covered

Semiconductors have resistivity or conductivity in between that of metals and insulators.

ρ ~ 10^-5. 10⁶ Ωm, σ ~ 10⁺⁵ .10^-6 Sm^-1

Types of Semiconductors: There are two types of semiconductors – Elements Semiconductors and Compound Semiconductors.

(i) Elements Semiconductors are available in natural form, e.g. germanium and silicon.
(ii) Compound Semiconductors are made by compounding the metals, e.g. InP, CdS, polyaniline, GaAs, CdSe, anthracene, etc.

On the basis of purity, semiconductors are classified as intrinsic semiconductors and extrinsic semiconductors.
Intrinsic Semiconductors are pure semiconductor that does not have any significant dopant species present

n_e = n_h = n_i

where, n_e and n_h are the number densities of electrons and holes respectively and n_i is the intrinsic carrier concentration.

Extrinsic Semiconductors are pure semiconductors that are doped with an impurity.

Extrinsic semiconductors are classified into two types: p-type semiconductors and n-type semiconductors.

Formation of Depletion Region in p-n junction: During the formation of a p-n junction, due to the concentration gradient across the p and n sides, the holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side.

Formation of Depletion Region in p-n junction

Formation of Depletion Region in p-n junction

Also Read:

Chapter 14 Related Articles
Junction Transistor	Classification of Metals, Conductors, and Semiconductors	Zener Diode as Voltage Regulator
Full Wave Rectifier	Integrated Circuits	Half Wave Rectifier
Uses of Zener Diode	Difference between Diode and Rectifier	Digital Electronics and Logic Gates
Application of Junction Diode as a Rectifier	Semiconductor Diode	Zener Diode

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CBSE CLASS XII Related Questions

1.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:
- 5 V
- 3.5 V
- 2.5 V
- Zero V
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2.
A circular coil of 100 turns and radius $\left(\frac{10}{\sqrt{\pi}}\right)$ cm carrying current of $5.0\,\text{A}$ is suspended vertically in a uniform horizontal magnetic field of $2.0\,\text{T}$. The field makes an angle $30^\circ$ with the normal to the coil. Calculate:
(I) the magnetic dipole moment of the coil, and
(II) the magnitude of the counter torque that must be applied to prevent the coil from turning.
View Solution
3.
A current of $4.0\,\text{A}$ flows through a wire of length $1\,\text{m}$ and cross-sectional area $1.0\,\text{mm}^2$, when a potential difference of $2\,\text{V}$ is applied across its ends. Calculate the resistivity of the material of the wire.
View Solution
4.
A wire of length $L$ is bent round into (i) a square coil having $N$ turns and (ii) a circular coil having $N$ turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.
View Solution
5.
Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:
- $\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})$
- $\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})$
- $\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})$
- $\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})$
View Solution
6.
Consider a cylindrical conductor of length $l$ and area of cross-section $A$. Current $I$ is maintained in the conductor and electrons drift with velocity $\vec{v}_d$ ($|\vec{v}_d| = \frac{e|\vec{E}|}{m}\tau$), (where symbols have their usual meanings). Show that the conductivity $\sigma$ of the material of the conductor is given by
$$\sigma = \frac{ne^2}{m}\tau.$$
View Solution