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

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CBSE CLASS XII Related Questions

1.
Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Both Assertion (A) and Reason (R) are false.
View Solution
2.
In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4I and zero respectively.
View Solution
3.
A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.
View Solution
4.
Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Both Assertion (A) and Reason (R) are false.
View Solution
5.
The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:
- P
- Q
- R
- S
View Solution
6.
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° in 0.2 s. The value of emf induced in the loop will be:
- 5 V
- 3.5 V
- 2.5 V
- Zero V
View Solution

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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.
Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

2.
In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4I and zero respectively.

3.
A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

4.
Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

5.
The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

6.
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° in 0.2 s. The value of emf induced in the loop will be:

Similar Physics Concepts

Comments

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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.Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

2.In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4I and zero respectively.

3. A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

4.Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

5.The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

6.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° in 0.2 s. The value of emf induced in the loop will be:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

2.
In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4I and zero respectively.

3.
A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

4.
Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

5.
The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

6.
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° in 0.2 s. The value of emf induced in the loop will be: