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JEE Physics

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

434 Q 9 Topics Take Test
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Difficulty: All Easy Medium Hard 51–60 of 434
Topics in JEE Physics
Q.51 Medium Semiconductors
In a light-emitting diode (LED), the color of emitted light depends primarily on:
A Doping concentration
B Bandgap energy of the material
C Forward bias voltage
D Junction area
Correct Answer:  B. Bandgap energy of the material
EXPLANATION

The wavelength of emitted light λ = hc/E_g, where E_g is the bandgap. Different materials (GaAs, GaN, AlGaAs) emit different colors based on their bandgap energy.

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Q.52 Medium Semiconductors
Which semiconductor material has the highest electron mobility at room temperature?
A Silicon
B Germanium
C Gallium Arsenide
D Indium Phosphide
Correct Answer:  C. Gallium Arsenide
EXPLANATION

GaAs has electron mobility ~8500 cm²/Vs compared to Si (~1350 cm²/Vs), making it superior for high-frequency and high-speed applications.

Test
Q.53 Medium Semiconductors
The transconductance g_m of a MOSFET in saturation is:
A ∂I_D/∂V_GS
B ∂I_D/∂V_DS
C I_D/V_GS
D V_GS/I_D
Correct Answer:  A. ∂I_D/∂V_GS
EXPLANATION

Transconductance g_m = ∂I_D/∂V_GS measures how much the drain current changes with gate voltage, a key figure of merit in MOSFET amplifier design.

Test
Q.54 Medium Semiconductors
In an n-channel MOSFET operating in saturation region, the drain current is approximately:
A I_D = (W/L) × C_ox × (V_GS - V_T)^2 / 2
B I_D = (W/L) × C_ox × (V_GS - V_T)
C I_D = (W/L) × C_ox × V_DS
D I_D is independent of V_GS
Correct Answer:  A. I_D = (W/L) × C_ox × (V_GS - V_T)^2 / 2
EXPLANATION

In saturation, the MOSFET acts as a voltage-controlled current source with I_D proportional to (V_GS - V_T)^2, independent of V_DS beyond saturation voltage.

Test
Q.55 Medium Semiconductors
In a MOSFET, the threshold voltage V_T increases with:
A Decrease in oxide thickness
B Increase in doping concentration of substrate
C Decrease in channel length
D Increase in temperature
Correct Answer:  B. Increase in doping concentration of substrate
EXPLANATION

Higher substrate doping increases the surface potential needed to invert the channel, raising V_T. This is described by V_T = V_T0 + γ(√(2φ_f + V_SB) - √(2φ_f)).

Test
Q.56 Medium Semiconductors
In a BJT operating in active mode, which of the following is correct?
A Base-emitter junction is reverse biased
B Base-emitter junction is forward biased and base-collector junction is reverse biased
C Both junctions are forward biased
D Both junctions are reverse biased
Correct Answer:  B. Base-emitter junction is forward biased and base-collector junction is reverse biased
EXPLANATION

In active mode, the BE junction is forward biased to inject carriers, while the BC junction is reverse biased to collect these carriers, ensuring transistor amplification.

Test
Q.57 Medium Semiconductors
The diode equation for current through a p-n junction is I = I_s(e^(qV/kT) - 1). Here, I_s is:
A Saturation current
B Source current
C Static current
D Signal current
Correct Answer:  A. Saturation current
EXPLANATION

I_s is the reverse saturation current, representing the leakage current when the junction is reverse biased. It depends on temperature and doping concentrations.

Test
Q.58 Medium Semiconductors
In a p-i-n photodiode, the intrinsic region serves to:
A Increase the dark current
B Extend the depletion region and improve photon collection
C Reduce the operating voltage
D Increase the forward bias conductance
Correct Answer:  B. Extend the depletion region and improve photon collection
EXPLANATION

The intrinsic (lightly doped) region in a p-i-n diode extends the depletion width, increasing quantum efficiency for photon absorption and collection. This improves responsivity and frequency response.

Test
Q.59 Medium Semiconductors
Compared to a p-n junction diode, a Schottky diode has:
A Higher forward voltage drop and faster switching
B Lower forward voltage drop and faster switching
C Higher forward voltage drop and slower switching
D Lower reverse saturation current
Correct Answer:  B. Lower forward voltage drop and faster switching
EXPLANATION

Schottky diodes have lower barrier height (typically 0.3-0.5V vs 0.7V for Si), resulting in lower forward voltage. Faster switching due to majority carrier conduction (no minority carrier storage).

Test
Q.60 Medium Semiconductors
In a metal-semiconductor (Schottky) junction, the barrier height is primarily determined by:
A Metal work function minus semiconductor electron affinity
B Semiconductor bandgap energy
C Applied reverse bias voltage
D Temperature of operation
Correct Answer:  A. Metal work function minus semiconductor electron affinity
EXPLANATION

Schottky barrier height Φ_B ≈ Φ_M - χ, where Φ_M is metal work function and χ is semiconductor electron affinity. This makes Schottky junctions useful for various applications with tunable barriers.

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