NCERT Solutions For Chapter 14: Semiconductor Electronics

Jasmine Grover Content Strategy Manager

Content Strategy Manager

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.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 $\Omega$), (2 V, 4 $\Omega$) and (6 V, 2 $\Omega$) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.
View Solution
2.
A current carrying circular loop of area A produces a magnetic field $ B $ at its centre. Show that the magnetic moment of the loop is $ \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} $.
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3.
Figure shows variation of Coulomb force (F) acting between two point charges with $ \frac{1}{r^2} $, $ r $ being the separation between the two charges $ (q_1, q_2) $ and $ (q_2, q_3) $. If $ q_2 $ is positive and least in magnitude, then the magnitudes of $ q_1, q_2 $, and $ q_3 $ are such that:
- $ q_2<q_1<q_3 $
- $ q_3<q_1<q_2 $
- $ q_1<q_2<q_3 $
- $ q_2<q_3<q_1 $
View Solution
4.
A parallel plate capacitor has plate area $ A $ and plate separation $ d $. Half of the space between the plates is filled with a material of dielectric constant $ K $ in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.
View Solution
5.
A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle $i$ such that it is refracted out grazing the face AD. Find the value of angle $i$.
$A rectangular glass slab ABCD (refractive index 1.5)$
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6.
The electric field at a point in a region is given by $ \vec{E} = \alpha \frac{\hat{r}}{r^3} $, where $ \alpha $ is a constant and $ r $ is the distance of the point from the origin. The magnitude of potential of the point is:
- $ \frac{\alpha}{r} $
- $ \frac{\alpha r^2}{2} $
- $ \frac{\alpha}{2r^2} $
- $ -\frac{\alpha}{r} $
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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.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

4.
A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

6.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point 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.Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

4.A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

5.A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle \(i\) such that it is refracted out grazing the face AD. Find the value of angle \(i\).

6.The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

4.
A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

6.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is: