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JEE Physics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

165 Q 9 Topics Take Test
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Difficulty: All Easy Medium Hard 31–40 of 165
Topics in JEE Physics
Q.31 Hard Semiconductors
The noise figure of a semiconductor amplifier is lowest when operating at:
A Maximum frequency
B Optimum source impedance
C Minimum supply voltage
D Maximum collector current
Correct Answer:  B. Optimum source impedance
EXPLANATION

Minimum noise figure occurs at optimum source impedance that matches the device's noise characteristics, typically provided in device specifications.

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Q.32 Hard Semiconductors
The current gain (β) of a BJT at constant IC depends on:
A Only collector current
B Temperature and base-emitter voltage drop
C Collector voltage only
D Load resistance
Correct Answer:  B. Temperature and base-emitter voltage drop
EXPLANATION

β varies with temperature and VBE changes, following the Ebers-Moll model. Early voltage causes slight VCE dependence.

Test
Q.33 Hard Semiconductors
In a tunnel diode, negative differential resistance occurs because:
A Tunneling current increases with voltage
B Drift current dominates at higher voltages
C Tunneling probability decreases while diffusion current increases
D Majority carrier concentration increases
Correct Answer:  C. Tunneling probability decreases while diffusion current increases
EXPLANATION

In tunnel diode's NDR region, as voltage increases, tunneling current decreases (fewer states to tunnel into) while diffusion current increases, causing net decrease in total current.

Test
Q.34 Hard Semiconductors
Which combination of semiconductor parameters would result in the highest intrinsic carrier concentration at 300K?
A Large Eg and high effective mass
B Small Eg and high density of states
C Large Eg and small density of states
D Negative bandgap energy
Correct Answer:  B. Small Eg and high density of states
EXPLANATION

ni = √(Nc·Nv)·exp(-Eg/2kT) where Nc, Nv depend on effective masses. Small Eg and high density of states both exponentially increase ni. GaAs has higher ni than Si due to smaller Eg.

Test
Q.35 Hard Semiconductors
The specific contact resistance of a metal-semiconductor junction is proportional to:
A exp(φB/kT) where φB is Schottky barrier height
B exp(-φB/kT)
C Directly proportional to doping concentration
D Inversely proportional to bandgap
Correct Answer:  B. exp(-φB/kT)
EXPLANATION

Contact resistance ρc ∝ exp(φB/kT)/Nc where φB is barrier height. Higher barrier leads to exponentially higher resistance following Thermionic emission theory.

Test
Q.36 Hard Semiconductors
In an avalanche photodiode (APD), the internal gain (multiplication factor M) increases significantly due to:
A Increased forward bias voltage
B Impact ionization creating secondary carriers in high electric field region
C Increased bandgap energy
D Decrease in depletion width
Correct Answer:  B. Impact ionization creating secondary carriers in high electric field region
EXPLANATION

APD operates in high reverse bias where impact ionization (collision ionization) produces secondary electron-hole pairs, creating avalanche multiplication and signal amplification internally.

Test
Q.37 Hard Semiconductors
The fill factor of a solar cell is defined as:
A Ratio of maximum power to product of Voc and Isc
B Ratio of efficiency to bandgap energy
C Ratio of short circuit current to open circuit voltage
D Product of voltage and current at maximum power point
Correct Answer:  A. Ratio of maximum power to product of Voc and Isc
EXPLANATION

Fill Factor (FF) = Pmax/(Voc × Isc) = (Vm × Im)/(Voc × Isc). It indicates how close the I-V curve is to a rectangle, typically 0.7-0.85 for practical solar cells.

Test
Q.38 Hard Semiconductors
In a forward-biased p-n junction diode at room temperature, if the applied voltage is increased from 0.3V to 0.6V, by approximately what factor does the forward current increase? (assume kT/e ≈ 26mV)
A 2 times
B 10 times
C 100 times
D 1000 times
Correct Answer:  D. 1000 times
EXPLANATION

Forward current I = I₀exp(eV/kT). Ratio = exp(e×0.3/kT) = exp(0.3/0.026) ≈ exp(11.5) ≈ 10^5. For ΔV = 0.3V more, increase is exp(11.5) ≈ 1000 times.

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Q.39 Hard Semiconductors
In a heterojunction like AlGaAs/GaAs, the main advantage is:
A Reduced lattice mismatch strain
B Carrier confinement due to larger bandgap material creating potential wells
C Increased doping concentration
D Lower operating temperature requirement
Correct Answer:  B. Carrier confinement due to larger bandgap material creating potential wells
EXPLANATION

Heterojunctions use materials with different bandgaps to create band offsets that confine carriers (electrons and holes) to specific regions, improving device efficiency. AlGaAs has larger Eg than GaAs, confining carriers in GaAs.

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Q.40 Hard Semiconductors
The temperature coefficient of bandgap for silicon is approximately:
A Positive: +2 meV/K
B Negative: -2 meV/K
C Zero at 300 K
D Depends only on doping concentration
Correct Answer:  B. Negative: -2 meV/K
EXPLANATION

Bandgap of semiconductors decreases with increasing temperature. For Si: dEg/dT ≈ -2.3 meV/K near 300 K. This is derived from Varshni equation and is crucial for device design.

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