Semiconductor Physics - NEET Physics Questions
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Semiconductor Physics

Question 51: easy

In the \( I-V \) characteristics of a silicon \( p-n \) junction diode, the current varies from 10 mA to 20 mA when applied voltage varies from 1 V to 1.2 V in the linear portion of forward biasing. The dynamic resistance of the diode will be:

1. 10 ohm
2. 20 ohm
3. 30 ohm
4. 40 ohm
View Answer

Dynamic resistance is defined as \( r_d = \frac{\Delta V}{\Delta I} \). Here, \( \Delta V = 1.2 \text{ V} - 1 \text{ V} = 0.2 \text{ V} \) and \( \Delta I = 20 \text{ mA} - 10 \text{ mA} = 10 \times 10^{-3} \text{ A} \). Thus, \( r_d = \frac{0.2}{10^{-2}} = 20 \ \Omega \).

Question 52: 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 53: 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 54: 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 55: easy

Assertion (A): Photo cell is also called electric eye.


Reason (R): Photo cell can see the things placed in front of it.


 

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

A photocell is indeed commonly referred to as an 'electric eye' due to its light-sensing ability. So, (A) is true. However, a photocell merely detects the presence or intensity of light; it does not 'see' or form images of objects in front of it in the way a biological eye does. Therefore, (R) is false.

Question 56: 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 57: 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 58: easy

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


Reason (R): In forward bias external battery reduced the internal electric field in depletion layer.


 

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 forward bias, the external voltage opposes the built-in potential barrier, effectively reducing the internal electric field across the depletion region. This reduction in the electric field causes the depletion region to narrow. Hence, both Assertion (A) and Reason (R) are true, and (R) correctly explains (A).

Question 59: easy

Assertion (A): Bridge full wave rectifier is more used than centre tap full wave rectifier.


Reason (R): In bridge full wave rectifier four diodes are used.


 

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

Bridge full-wave rectifiers are more popular because they do not require a costly center-tapped transformer and offer higher output voltage. So Assertion (A) is true. Reason (R) is also true as a bridge rectifier uses four diodes. However, the number of diodes is not the primary reason for its preference over a center-tap rectifier.

Question 60: easy

Assertion (A): The semiconductor used for fabrication of visible LED must at least have a band gap of 1.8 eV.


Reason (R): The spectral range of visible light is from 0.4 eV to 1.8 eV.

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

For visible light emission, the photon energy must be within the visible spectrum, which corresponds to energies from approximately \(1.8\) to \(3.1 \text{ eV}\). Thus, Assertion (A) is true. Reason (R) states the range from \(0.4\) to \(1.8 \text{ eV}\), which is incorrect for visible light.