Properties of Semiconductors - NEET Physics Questions
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Properties of Semiconductors

Question 11: easy

A silicon (Si) specimen is doped with aluminium (Al). The concentration of acceptor atoms is \(10^{18}\text{ m}^{-3}\). Given that the intrinsic carrier concentration is \(10^{16}\text{ m}^{-3}\), the concentration of electrons in the specimen is

1. 10^{14} m^{-3}
2. 10^{16} m^{-3}
3. 10^{18} m^{-3}
4. 10^{12} m^{-3}
View Answer

Formula: \(n_e n_h = n_i^2\). Since the semiconductor is heavily p-doped, \(n_h approx N_a = 10^{18}\text{ m}^{-3}\). Therefore, \(n_e = frac{n_i^2}{n_h} = frac{(10^{16})^2}{10^{18}} = 10^{14}\text{ m}^{-3}\).

Question 12: easy

Consider the following statements:


(a) At 0 K, semiconductor behaves as perfect conductor.


(b) Semiconductors have negative temperature coefficient of resistance.


Choose the correct statement.

1. Only (a)
2. Only (b)
3. Both (a) & (b)
4. Neither (a) nor (b)
View Answer

At 0 K, semiconductors behave as insulators since no electrons are in the conduction band. The resistance of a semiconductor decreases with increasing temperature, meaning it has a negative temperature coefficient of resistance. Thus, only statement (b) is correct.

Question 13: easy

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than \(6200 A^0\), is incident on it. The band gap in (eV) for the semiconductor is

1. 1
2. 2
3. 0.7
4. 1.1
View Answer

The band gap is related to the threshold wavelength by the formula \(E_g = \frac{12400}{\lambdaΒ  in A^0}\text{ eV}\). Substituting \(\lambda = 6200Β  A^0\), we get \(E_g = \frac{12400}{6200} = 2\text{ eV}\).

Question 14: easy

Consider the following statements


(i) An intrinsic semiconductor will behave as insulator at \(T = 0text{ K}\).


(ii) Doping pure silicon with trivalent impurities gives p-type semiconductors.


(iii) The majority carriers in n-type semiconductors are electrons.


(iv) Solar cell works when it is in forward bias.


The correct statements(s) is/are

1. Only (i) and (ii)
2. Only (i) and (iii)
3. Only (i), (ii) and (iii)
4. Only (i), (ii) and (iv)
View Answer

At \(T = 0text{ K}\), all electrons are in the valence band, so an intrinsic semiconductor acts as an insulator. Trivalent doping creates p-type, and n-type has majority electron carriers. Solar cells do not require external bias to operate. Thus, statements (i), (ii), and (iii) are correct.

Question 15: easy

Assertion (A): The conductivity of an intrinsic semiconductor depends on its temperature.


Reason (R): No important electronic device can be developed using intrinsic semi conductor.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

In an intrinsic semiconductor, conductivity increases with temperature due to increased generation of electron-hole pairs. So Assertion (A) is true. Intrinsic semiconductors have limited practical use due to low conductivity, making Reason (R) true in terms of 'important' devices. However, (R) does not explain (A).

Question 16: easy

Assertion (A): The number of electrons in n-type semiconductor is higher than the number of electrons in a pure silicon semiconductor.


Reason (R): The law of mass action is applicable only to n-type semiconductors.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

In n-type semiconductors, donor impurities increase the number of free electrons, making Assertion (A) true. The law of mass action (\(n_e n_h = n_i^2\)) is a fundamental principle applicable to all types of semiconductors (intrinsic, n-type, p-type), so Reason (R) is false.

Question 17: easy

Assertion (A): Electron hole recombination takes place in P-region and N-region of PN Junction diode except in depletion region.


Reason (R): Electric field in depletion region oppose the diffusion.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is generally true in simplified models, where most recombination occurs in the quasi-neutral P and N regions. While some recombination does occur in the depletion region, its contribution to the overall current is often considered secondary for typical forward-biased diodes.


Reason (R) is true; the electric field in the depletion region acts as a barrier, opposing the diffusion of majority carriers across the junction. (R) does not explain the spatial distribution of recombination described in (A).

Question 18: easy

Assertion (A): Width of depletion region is reduced in forward bias.


Reason (R): In n-type semiconductor majority charge carriers are free electrons while in p-type they are holes.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

When a p-n junction is forward biased, the applied voltage counteracts the internal electric field, causing majority carriers to move towards the junction and reducing the depletion region's width. Thus, Assertion (A) is true.


Reason (R) correctly states the majority carriers in n-type (electrons) and p-type (holes) semiconductors, which is also true. However, Reason (R) does not explain why the depletion region width reduces under forward bias. Hence, both are true, but R is not the correct explanation for A.

Question 19: easy

Assertion (A): The drift current in a p-n junction is from n-side to p-side.


Reason (R): The diffusion current in a p-n junction is from p-side to n-side.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

In a p-n junction, the built-in electric field points from the n-side to the p-side, causing drift current to flow from n-side to p-side. Thus, Assertion (A) is true. Due to concentration gradients, diffusion current arises from the movement of majority carriers, resulting in a net diffusion current from p-side to n-side. Thus, Reason (R) is also true.


These are two distinct current mechanisms, and R does not explain A.

Question 20: easy

Assertion (A): P-type semiconductor has high density of holes in valence band while N-type semiconductor has high density of electrons in conduction band.


Reason (R): In N-type semiconductor, as the density of donor atoms (N_D) is increased, the fermi energy level shifts towards the valence band.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

P-type semiconductors have a high concentration of holes in the valence band, and N-type semiconductors have a high concentration of electrons in the conduction band. Thus, Assertion (A) is true.


In an N-type semiconductor, increasing the donor atom density (N_D) increases the electron concentration, causing the Fermi energy level to shift closer to the conduction band, not the valence band. Hence, Reason (R) is false.