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NEET PHYSICSEasy

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=2.0×1.6×10−19 J=3.2×10−19 JE_g = 2.0 \times 1.6 \times 10^{-19}\text{ J} = 3.2 \times 10^{-19}\text{ J}
  • Planck's constant, h≈6.63×10−34 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=3.2×10−19 J6.63×10−34 J s≈0.482×1015 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}.
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