P-N Junction Diode and its applications - NEET Physics Chapterwise MCQs & PYQs

NEET P-N Junction Diode and its applications MCQs & PYQs

Question 11:

easy

Consider the following statements (A) and (B) and identify the correct answer.


(A) A zener diode is connected in reverse bias, when used as a voltage regulator.


(B) The potential barrier of \(p\)-\(n\) junction lies between 0.1 V to 0.3 V.

A Zener diode regulates voltage when reverse-biased in the breakdown region (A is correct). The potential barrier for a Silicon p-n junction is around 0.7 V, which is outside the 0.1 V to 0.3 V range (B is incorrect).

Question 12:

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:

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 13:

easy

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


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


 

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 14:

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.

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.

Question 15:

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.


 

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 16:

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.


 

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 17:

easy

Assertion (A): Working principle of photodiode and photocell is same.


Reason (R): Biasing circuit for photodiode and photocell is same.


 

Assertion (A) is false; photodiodes rely on \(p-n\) junction semiconductor physics, while photocells (e.g., photoemissive cells) often rely on the photoelectric effect. Reason (R) is also false as their biasing circuits differ significantly (e.g., photodiode often reverse biased, photocell can be forward biased or unbiased).

Question 18:

easy

Assertion (A): GaAs is preferred for making solar panels.


Reason (R): \(\Delta E_g\) for GaAs is \(1.5\text{ eV}\), and sun’s radiation has highest intensity around this energy level.


 

Assertion (A) is true in terms of performance; GaAs solar cells offer very high efficiency. Reason (R) is also true; the bandgap of GaAs (approx. \(1.42\text{ eV}\,) is near \(1.5\text{ eV}\), which matches well with the peak intensity of the solar spectrum. (R) correctly explains (A) as this optimal bandgap is key to its high efficiency.

Question 19:

easy

Assertion (A): In LED \(e^-\)-hole pair recombination gives us photon.


Reason (R): In LED \(e^-\)-hole pair recombination occurs in depletion region.


 

Assertion (A) is true; LEDs emit photons when electrons and holes recombine. Reason (R) is false; while carriers cross the depletion region, significant recombination leading to light emission primarily occurs in the quasi-neutral regions (or active layer) under forward bias, not mainly within the depletion region itself.

Question 20:

easy

Assertion (A): In a N-type semiconductor, the number of holes get reduced.


Reason (R): Rate of recombination of holes would increase due to the increase in the number of electrons


 

Assertion (A) is true; in an N-type semiconductor, donor doping increases electron concentration, which, by mass action law (\(np=n_i^2\)), reduces the equilibrium hole concentration. Reason (R) is true; the increased number of electrons in an N-type semiconductor leads to a higher rate of recombination with the minority holes. (R) correctly explains (A) as this increased recombination helps establish and maintain the lower equilibrium hole concentration.