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

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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

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CBSE CLASS XII Related Questions

1.
Define electric flux through a surface. Give the significance of a Gaussian surface. A charge outside a Gaussian surface does not contribute to total electric flux through the surface. Why?
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Show that the energy required to build up the current \( I \) in a coil of inductance \( L \) is \( \frac{1}{2} L I^2 \).
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5.
A charge \( -6 \, \mu C \) is placed at the center B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge \( +5 \, \mu C \) is moved from point C to point A along the circumference. Calculate the work done on the charge.
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NCERT Solutions For Chapter 14: Semiconductor Electronics

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Class 12 Physics Chapter 14 – Topics Covered

CBSE CLASS XII Related Questions

1.
Define electric flux through a surface. Give the significance of a Gaussian surface. A charge outside a Gaussian surface does not contribute to total electric flux through the surface. Why?

2.
Show that the energy required to build up the current \( I \) in a coil of inductance \( L \) is \( \frac{1}{2} L I^2 \).

3.
Define ‘Mass defect’ and ‘Binding energy’ of a nucleus. Describe ‘Fission process’ on the basis of binding energy per nucleon.

4.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

6.
A 1 cm segment of a wire lying along the x-axis carries a current of 0.5 A along the \( +x \)-direction. A magnetic field \( \vec{B} = (0.4 \, \text{mT} \hat{j}) + (0.6 \, \text{mT} \hat{k}) \) is switched on. The force acting on the segment is:

Similar Physics Concepts

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NCERT Solutions For Chapter 14: Semiconductor Electronics

NCERT Solutions for Class 12 Physics Chapter 14

Class 12 Physics Chapter 14 – Topics Covered

CBSE CLASS XII Related Questions

1.Define electric flux through a surface. Give the significance of a Gaussian surface. A charge outside a Gaussian surface does not contribute to total electric flux through the surface. Why?

2.Show that the energy required to build up the current \( I \) in a coil of inductance \( L \) is \( \frac{1}{2} L I^2 \).

3.Define ‘Mass defect’ and ‘Binding energy’ of a nucleus. Describe ‘Fission process’ on the basis of binding energy per nucleon.

4.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

6.A 1 cm segment of a wire lying along the x-axis carries a current of 0.5 A along the \( +x \)-direction. A magnetic field \( \vec{B} = (0.4 \, \text{mT} \hat{j}) + (0.6 \, \text{mT} \hat{k}) \) is switched on. The force acting on the segment is:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Define electric flux through a surface. Give the significance of a Gaussian surface. A charge outside a Gaussian surface does not contribute to total electric flux through the surface. Why?

2.
Show that the energy required to build up the current \( I \) in a coil of inductance \( L \) is \( \frac{1}{2} L I^2 \).

3.
Define ‘Mass defect’ and ‘Binding energy’ of a nucleus. Describe ‘Fission process’ on the basis of binding energy per nucleon.

4.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

6.
A 1 cm segment of a wire lying along the x-axis carries a current of 0.5 A along the \( +x \)-direction. A magnetic field \( \vec{B} = (0.4 \, \text{mT} \hat{j}) + (0.6 \, \text{mT} \hat{k}) \) is switched on. The force acting on the segment is: