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NEET PHYSICSSEMICONDUCTOR ELECTRONICSEasy

Question

A p-n photodiode is made of a material with a band gap of 2.0 eV2.0\text{ eV}. The minimum frequency of the radiation that can be absorbed by the material is nearly:

A

10×1014 Hz10 \times 10^{14}\text{ Hz}

B

5×1014 Hz5 \times 10^{14}\text{ Hz}

C

1×1014 Hz1 \times 10^{14}\text{ Hz}

D

20×1014 Hz20 \times 10^{14}\text{ Hz}

Step-by-Step Solution

  1. Condition for Absorption: For a photodiode to absorb radiation, the energy of the incident photon (EE) must be greater than or equal to the band gap energy (EgE_g) of the material. Therefore, the minimum energy required is E=EgE = E_g.
  2. Identify Given Data: Band gap energy, Eg=2.0 eVE_g = 2.0\text{ eV} Convert eV to Joules: Eg=2.0×1.6×1019 J=3.2×1019 JE_g = 2.0 \times 1.6 \times 10^{-19}\text{ J} = 3.2 \times 10^{-19}\text{ J}
  • Planck's constant, h6.63×1034 J sh \approx 6.63 \times 10^{-34}\text{ J s}
  1. Calculate Minimum Frequency (ν\nu): Since E=hνE = h\nu, the minimum frequency is ν=Egh\nu = \frac{E_g}{h} ν=3.2×1019 J6.63×1034 J s0.482×1015 Hz=4.82×1014 Hz\nu = \frac{3.2 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 0.482 \times 10^{15}\text{ Hz} = 4.82 \times 10^{14}\text{ Hz}
  2. Conclusion: The calculated frequency 4.82×1014 Hz4.82 \times 10^{14}\text{ Hz} is closest to 5×1014 Hz5 \times 10^{14}\text{ Hz}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from SEMICONDUCTOR ELECTRONICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSSEMICONDUCTOR ELECTRONICSphotodiodematerialminimumfrequencyradiation

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