SEMI CONDUCTOR DEVICES MCQs for NEET UG PHYSICS — Easy

1 Easy

Application +4 | -1

The ratio of RMS output voltage of full wave rectifier and half wave rectifier is

\sqrt{n}.

The value of

n

is? (The input voltage is sine wave).

View answer and explanation

Correct option: A

2

Option A (2) is correct because the ratio of RMS output voltage of a full-wave rectifier to a half-wave rectifier is

\sqrt{2}

.

Solution approach

Correct Answer + Core Reason: 2. The ratio of RMS output voltage of full wave rectifier and half wave rectifier is $\sqrt{2}$ , therefore n=2.

Step-by-Step Solution:

Step 1: Recall the RMS output voltage for a full-wave rectifier: $V_{RMS, Full} = \frac{V_m}{\sqrt{2}}$

Step 2: Recall the RMS output voltage for a half-wave rectifier: $V_{RMS, Half} = \frac{V_m}{2}$

Step 3: Calculate the ratio: $\frac{V_{RMS, Full}}{V_{RMS, Half}} = \frac{\frac{V_m}{\sqrt{2}}}{\frac{V_m}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Step 4: Given the ratio is $\sqrt{n}$ , then $\sqrt{n} = \sqrt{2}$ , therefore $n = 2$ .

Why other options are wrong:
- Option B (0): Incorrect ratio of RMS voltages.
- Option C (0): Incorrect ratio of RMS voltages.
- Option D (0): Incorrect ratio of RMS voltages.

Key Concepts:
- RMS voltage of full-wave rectifier
- RMS voltage of half-wave rectifier

Memory Aids:
- Remember the RMS voltage formulas for both rectifiers.

Quick Tips:
- Know the standard formulas for rectifier output voltages.

2 Easy

2 of 13Question permalink

Understanding +4 | -1

The depletion layer of a p-n junction:

View answer and explanation

Correct option: B

Acts like an insulating zone under reverse bias

Option B (Acts like an insulating zone under reverse bias) is correct because the depletion region is devoid of free charge carriers and thus acts as an insulator, especially under reverse bias which widens the region.

Solution approach

Correct Answer + Core Reason: Acts like an insulating zone under reverse bias. The depletion region is devoid of free carriers.

Step-by-Step Solution:

Step 1: Define the depletion layer. The depletion layer is a region in a p-n junction that is depleted of free charge carriers (electrons and holes).

Step 2: Explain its insulating properties. Because it lacks free charge carriers, the depletion layer acts as an insulator.

Step 3: Describe the effect of reverse bias. Reverse bias widens the depletion layer, increasing its resistance and insulating properties.

Why other options are wrong:
- Option A: Is of constant width irrespective of the bias: The width of the depletion layer changes with bias.
- Option C: Has a width that increases with an increase in forward bias: The width decreases with forward bias.
- Option D: Is depleted of ions: The depletion region contains immobile ions, not depleted of them.

Key Concepts:
- p-n junction
- Depletion layer
- Reverse bias
- Insulating properties

Memory Aids:
- Depletion = No carriers = Insulator

Quick Tips:
- Remember that the depletion region's width is bias-dependent.

3 Easy

3 of 13Question permalink

Understanding +4 | -1

When a

\mathbf{P}-\mathbf{N}

junction is reverse biased :

View answer and explanation

Correct option: A

Holes and electrons move away from the junction

Option A (Holes and electrons move away from the junction) is correct because when a p-n junction is reverse biased, the applied voltage increases the width of the depletion region, causing charge carriers to move away from the junction.

Solution approach

Correct Answer + Core Reason: Holes and electrons move away from the junction. Reverse bias widens the depletion region.

Step-by-Step Solution:

Step 1: Define reverse bias. Reverse bias is when the positive terminal of the voltage source is connected to the n-side and the negative terminal to the p-side of the p-n junction.

Step 2: Explain the effect on the depletion region. Reverse bias increases the width of the depletion region.

Step 3: Describe the movement of charge carriers. The increased electric field in the depletion region pulls holes away from the junction on the p-side and electrons away from the junction on the n-side.

Why other options are wrong:
- Option B: Holes and electrons move towards the junction: This happens in forward bias.
- Option C: Movement of holes and electrons stops: There is still a small reverse saturation current due to minority carriers.
- Option D: Width of depletion region decreases: This happens in forward bias.

Key Concepts:
- p-n junction
- Reverse bias
- Depletion region
- Charge carriers (holes and electrons)

Memory Aids:
- Reverse = Away

Quick Tips:
- Remember the effect of forward and reverse bias on the depletion region width.

4 Easy

4 of 13Question permalink

Application +4 | -1

The acceptor level of a p-type semiconductor is

6 \mathrm{~eV}

. The maximum wavelength of light which can create a hole would be : Given

\mathrm{hc}=1242 \mathrm{~eV} \mathrm{~nm}

.

View answer and explanation

Correct option: C

207 nm

Option C (207 nm) is correct because the maximum wavelength corresponds to the minimum energy required to create a hole, which is equal to the acceptor level energy. Using the formula

E = \frac{hc}{\lambda}

, we can solve for

\lambda

.

Solution approach

Correct Answer + Core Reason: 207 nm. This is calculated using the energy-wavelength relationship.

Step-by-Step Solution:

Step 1: Identify the given information: Acceptor level energy $E = 6 \text{{ eV}}$ , $hc = 1242 \text{{ eV nm}}$ .

Step 2: Recall the relationship between energy and wavelength: $E = \frac{hc}{\lambda}$ .

Step 3: Rearrange the formula to solve for wavelength: $\lambda = \frac{hc}{E}$ .

Step 4: Substitute the given values: $\lambda = \frac{1242 \text{{ eV nm}}}{6 \text{{ eV}}}$ .

Step 5: Calculate the wavelength: $\lambda = 207 \text{{ nm}}$ .

Why other options are wrong:
- Option A: 407 nm: This is incorrect; it might be a calculation error.
- Option B: 414 nm: This is incorrect; it might be a calculation error.
- Option D: 103.5 nm: This is incorrect; it might be a calculation error, perhaps dividing by 12 instead of 6.

Key Concepts:
- Energy levels in semiconductors
- Energy-wavelength relationship: $E = \frac{hc}{\lambda}$

Memory Aids:
- Remember the energy-wavelength formula.

Quick Tips:
- Always use consistent units when applying the energy-wavelength formula.

5 Easy

5 of 13Question permalink

Understanding +4 | -1

With the fall in temperature, the electric conductivity of intrinsic semiconductors

View answer and explanation

Correct option: B

Decreases

Option B (Decreases) is correct because as temperature falls, the conductivity of intrinsic semiconductors decreases due to reduced thermal excitation of electrons.

Solution approach

Correct Answer + Core Reason: Decreases. Conductivity of intrinsic semiconductors decreases with temperature.

Step-by-Step Solution:

Step 1: Define intrinsic semiconductors. Intrinsic semiconductors are pure semiconductors with no impurities.

Step 2: Understand the effect of temperature on conductivity. In intrinsic semiconductors, conductivity is proportional to the concentration of charge carriers (electrons and holes).

Step 3: Explain the relationship between temperature and charge carrier concentration. As temperature decreases, fewer electrons have enough energy to jump from the valence band to the conduction band, reducing the concentration of charge carriers.

Step 4: Conclusion: With a fall in temperature, the electric conductivity of intrinsic semiconductors decreases.

Why other options are wrong:
- Option A: Conductivity increases with temperature in semiconductors, not decreases.
- Option B: Conductivity decreases with temperature in intrinsic semiconductors.
- Option C: Conductivity does not first decrease and then increase with a fall in temperature.
- Option D: Conductivity changes with temperature in semiconductors.

Key Concepts:
- Intrinsic Semiconductors
- Electrical Conductivity
- Temperature Dependence of Conductivity

Memory Aids:
- Think of it like freezing: as things get colder, it's harder for electrons to move.

Quick Tips:
- Remember that the conductivity of semiconductors is temperature-dependent.

6 Easy

6 of 13Question permalink

Remembering +4 | -1

Integrated chip (IC) is an electronic device that contains

View answer and explanation

Correct option: D

Transistors, resistors and capacitors

Option D (Transistors, resistors and capacitors) is correct because an integrated circuit (IC) typically contains transistors, resistors, and capacitors fabricated on a single semiconductor chip.

Solution approach

Correct Answer + Core Reason: Transistors, resistors and capacitors. Integrated circuits contain all these components.

Step-by-Step Solution:

Step 1: Understand Integrated Circuits: Integrated circuits are miniaturized electronic circuits manufactured on a semiconductor material.

Step 2: Identify Components: ICs can contain transistors, resistors, capacitors, and diodes.

Step 3: Conclusion: An IC contains transistors, resistors, and capacitors.

Why other options are wrong:
- Option A: ICs contain more than just transistors.
- Option B: ICs contain more than just resistors.
- Option C: ICs contain more than just transistors and resistors.

Key Concepts:
- Integrated Circuits
- Electronic Components

Memory Aids:
- ICs are complex circuits.

Quick Tips:
- Remember the basic components of an IC.

7 Easy

7 of 13Question permalink

Understanding +4 | -1

If a LED forward biased, then

View answer and explanation

Correct option: A

electrons from the

n

- type side cross the

p-n

junction and recombine with holes in the

p

- type side

Option A (electrons from the

n

- type side cross the

p-n

junction and recombine with holes in the

p

- type side) is correct because this is the fundamental process that produces light in an LED.

Solution approach

Correct Answer + Core Reason: electrons from the $n$ - type side cross the $p-n$ junction and recombine with holes in the $p$ - type side. This recombination emits light.

Step-by-Step Solution:

Step 1: Define LED operation. An LED emits light when electrons and holes recombine at the p-n junction.

Step 2: Explain forward biasing. In forward bias, electrons from the n-type side and holes from the p-type side are pushed towards the junction.

Step 3: Describe recombination. When electrons and holes meet at the junction, they recombine, releasing energy in the form of photons (light).

Step 4: Conclusion: Electrons from the n-type side cross the p-n junction and recombine with holes in the p-type side, emitting light.

Why other options are wrong:
- Option A: This is the correct explanation of LED operation.
- Option B: Electrons and holes do not simply neutralize each other in the depletion region; they recombine and emit light.
- Option C: This is an incomplete description of the process.
- Option D: The correct answer is described in option A.

Key Concepts:
- LED Operation
- P-N Junction
- Recombination
- Forward Bias

Memory Aids:
- LED = Light Emitting Diode: Light comes from electron-hole recombination.

Quick Tips:
- Remember the basic principle of LED operation: electron-hole recombination at the p-n junction.

8 Easy

8 of 13Question permalink

Application +4 | -1

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than

6200 \, \mathop {\text{A}}\limits^ \circ,

is incident on it. The band gap in (eV) for the semiconductor is

View answer and explanation

Correct option: B

2

2 is correct because the band gap energy is calculated as

E = \frac{{hc}}{\lambda}

, and substituting the given wavelength yields approximately 2 eV.

Solution approach

Correct Answer + Core Reason: 2. The band gap is calculated using the formula $E = \frac{{hc}}{\lambda}$ .

Step-by-Step Solution:

Step 1: Identify the given values: $\lambda = 6200 \, \mathop{\text{A}}\limits^\circ = 6200 \times 10^{-10} \text{{ m}}$ , $hc = 12400 \text{{ eV}} \, \mathop{\text{A}}\limits^\circ$ .

Step 2: Convert the wavelength to Angstroms: $\lambda = 6200 \, \mathop{\text{A}}\limits^\circ$ .

Step 3: Use the formula: $E = \frac{{hc}}{\lambda}$

Step 4: Substitute the values: $E = \frac{{12400 \text{{ eV}} \, \mathop{\text{A}}\limits^\circ}}{{6200 \, \mathop{\text{A}}\limits^\circ}}$

Step 5: Calculate the band gap: $E = 2 \text{{ eV}}$

Why other options are wrong:
- Option A: Incorrect calculation.
- Option C: Incorrect calculation.
- Option D: Incorrect calculation.

Key Concepts:
- Energy band gap
- Wavelength of radiation
- Relationship between energy and wavelength: $E = \frac{{hc}}{\lambda}$

Memory Aids:
- Remember the formula: $E = \frac{{hc}}{\lambda}$

Quick Tips:
- Pay attention to units and use the correct value of hc.

9 Easy

9 of 13Question permalink

Application +4 | -1

A light emitting diode (LED) is fabricated using Ga. As semiconducting material whose band gap is

1.42 \mathrm{~eV}

. The wavelength of light emitted from the LED is:

View answer and explanation

Correct option: C

875 nm

Option C (875 nm) is correct because the wavelength of light emitted from an LED is related to its band gap energy by the formula

\lambda = \frac{hc}{E_g}

, where

E_g

is the band gap energy.

Solution approach

Correct Answer + Core Reason: 875 nm. This is calculated using the band gap energy and the formula relating energy and wavelength.

Step-by-Step Solution:

Step 1: Identify the given information: Band gap energy $E_g = 1.42 \text{{ eV}}$ , $hc = 1242 \text{{ eV nm}}$ .

Step 2: Recall the relationship between energy and wavelength: $E_g = \frac{hc}{\lambda}$ .

Step 3: Rearrange the formula to solve for wavelength: $\lambda = \frac{hc}{E_g}$ .

Step 4: Substitute the given values: $\lambda = \frac{1242 \text{{ eV nm}}}{1.42 \text{{ eV}}}$ .

Step 5: Calculate the wavelength: $\lambda \approx 875 \text{{ nm}}$ .

Why other options are wrong:
- Option A: 650 nm: This is incorrect; it might be a calculation error.
- Option B: 1243 nm: This is incorrect; it might be a misunderstanding of the formula.
- Option D: 1400 nm: This is incorrect; it might be a calculation error.

Key Concepts:
- Band gap energy
- Energy-wavelength relationship: $E = \frac{hc}{\lambda}$
- LED operation

Memory Aids:
- Remember the energy-wavelength formula.

Quick Tips:
- Always use consistent units when applying the energy-wavelength formula.

10 Easy

10 of 13Question permalink

Understanding +4 | -1

In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be:

View answer and explanation

Correct option: D

60 Hz

Option D (60 Hz) is correct because in half-wave rectification, the output frequency is the same as the input frequency.

Solution approach

Correct Answer + Core Reason: 60 Hz. The output frequency in half-wave rectification is the same as the input frequency.

Step-by-Step Solution:

Step 1: Define half-wave rectification. Half-wave rectification allows only one half-cycle of the AC voltage to pass, blocking the other half-cycle.

Step 2: Understand the frequency relationship. The frequency of the output waveform is the same as the input waveform because the time period of the repeating pattern remains unchanged.

Step 3: Conclusion: If the input frequency is 60 Hz, the output frequency is also 60 Hz.

Why other options are wrong:
- Option A: 120 Hz is the output frequency for full-wave rectification, not half-wave.
- Option B: The output frequency is not zero.
- Option C: 30 Hz is not the output frequency.
- Option D: 60 Hz is the correct output frequency.

Key Concepts:
- Half-Wave Rectification
- Frequency

Memory Aids:
- Half-wave = Half the wave is used, but the frequency stays the same.

Quick Tips:
- Remember that half-wave rectification does not change the frequency of the signal.

11 Easy

11 of 13Question permalink

Understanding +4 | -1

If we add impurity to a metal those atoms also deflect electrons. Therefore

View answer and explanation

Correct option: B

The electrical and thermal conductivities both decrease

The electrical and thermal conductivities both decrease because impurities scatter electrons, hindering their movement.

Solution approach

Correct Answer + Core Reason: The electrical and thermal conductivities both decrease. Impurities in a metal scatter electrons, reducing both electrical and thermal conductivity.

Step-by-Step Solution:

Step 1: Recall the effect of impurities on electron flow in metals.

Step 2: Impurities act as scattering centers for electrons.

Step 3: Scattering reduces the mean free path of electrons.

Step 4: Reduced mean free path decreases both electrical and thermal conductivity.

Step 5: Conclusion: Both conductivities decrease.

Why other options are wrong:
- Option A: Impurities hinder, not enhance, electron flow.
- Option C: Thermal and electrical conductivities are related; if one decreases, the other typically does as well.
- Option D: Thermal conductivity also decreases due to increased electron scattering.

Key Concepts:
- Electrical conductivity
- Thermal conductivity
- Electron scattering
- Mean free path

Memory Aids:
- Impurities = Impediments to electron flow.

Quick Tips:
- Remember that impurities generally reduce conductivity in metals.

12 Easy

12 of 13Question permalink

Application +4 | -1

To write the decimal number 37 in binary, how many binary digits are required

View answer and explanation

Correct option: B

6

6 is correct because 37 in decimal is 100101 in binary, which requires 6 digits.

Solution approach

Correct Answer + Core Reason: 6. Converting 37 to binary requires 6 digits.

Step-by-Step Solution:

Step 1: Convert the decimal number 37 to binary.

Step 2: Divide 37 by 2 repeatedly and record the remainders:
- 37 / 2 = 18 remainder 1
- 18 / 2 = 9 remainder 0
- 9 / 2 = 4 remainder 1
- 4 / 2 = 2 remainder 0
- 2 / 2 = 1 remainder 0
- 1 / 2 = 0 remainder 1

Step 3: Read the remainders in reverse order: 100101

Step 4: Count the number of binary digits: 6

Why other options are wrong:
- Option A: 5 digits are not sufficient to represent 37 in binary.
- Option C: 7 digits are not required to represent 37 in binary.
- Option D: 4 digits are not sufficient to represent 37 in binary.

Key Concepts:
- Decimal to binary conversion
- Binary number system

Memory Aids:
- Remember the division method for decimal to binary conversion.

Quick Tips:
- Practice converting decimal numbers to binary and vice versa.

13 Easy

13 of 13Question permalink

Application +4 | -1

Conductivity of a photodiode starts changing only if the wavelength of incident light is less than

660 \mathrm{~nm}

. The band gap of photodiode is found to be

\left(\frac{X}{8}\right) \mathrm{eV}

. The value of

X

is (Given,

h=6.6 \times 10^{-34} \mathrm{~Js}, e=1.6 \times 10^{-19} \mathrm{C}

)

View answer and explanation

Correct option: D

15

Option D (15) is correct because the band gap energy can be calculated from the wavelength using

E = \frac{{hc}}{{\lambda}}

, and then solving for X.

Solution approach

Correct Answer + Core Reason: 15. The band gap energy is calculated from the given wavelength and equated to $\frac{X}{8}$ to find X.

Step-by-Step Solution:

Step 1: Identify the given information: Wavelength $\lambda = 660 \text{{ nm}} = 660 \times 10^{-9} \text{{ m}}$ , Planck's constant $h = 6.6 \times 10^{-34} \text{{ Js}}$ , electron charge $e = 1.6 \times 10^{-19} \text{{ C}}$ . Band gap $E_g = \frac{X}{8} \text{{ eV}}$ .

Step 2: Use the formula for energy of a photon: $E = \frac{{hc}}{{\lambda}}$

Step 3: Substitute the values: $E = \frac{{(6.6 \times 10^{-34} \text{{ Js}})(3 \times 10^8 \text{{ m/s}})}}{{660 \times 10^{-9} \text{{ m}}}}$

Step 4: Calculate the energy in Joules: $E = \frac{{19.8 \times 10^{-26}}}{{660 \times 10^{-9}}} \text{{ J}} = 3 \times 10^{-19} \text{{ J}}$

Step 5: Convert the energy to electron volts: $E(\text{{eV}}) = \frac{{3 \times 10^{-19} \text{{ J}}}}{{1.6 \times 10^{-19} \text{{ C}}}} = 1.875 \text{{ eV}}$

Step 6: Equate the energy to the given band gap: $1.875 = \frac{X}{8}$

Step 7: Solve for X: $X = 1.875 \times 8 = 15$

Why other options are wrong:
- Option A (13): Incorrect calculation or wrong formula used.
- Option B (11): Incorrect calculation or wrong formula used.
- Option C (21): Incorrect calculation or wrong formula used.

Key Concepts:
- Energy of a photon: $E = \frac{{hc}}{{\lambda}}$
- Conversion between Joules and electron volts

Memory Aids:
- Remember the formula $E = \frac{{hc}}{{\lambda}}$

Quick Tips:
- Always convert units to SI units before substituting into formulas.

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