Q23. The electrical conductivity of a semiconductor increases when electromagnetic radiation of wave

Q: What is the correct answer for CUET UG Physics Q23?

The correct answer is a. For a semiconductor, the electrical conductivity increases when incident radiation provides enough energy to excite electrons from the valence band to the conduction band. The minimum energy required for this is the band gap (E_g). This occurs when the photon energy (E_photon) is greater than or equal to E_g. The longest wavelength (λ_max) of radiation that can cause this effect corresponds to the

Q: Q23. The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap for the semiconductor is: (a) 0.5…

The correct answer is a. For a semiconductor, the electrical conductivity increases when incident radiation provides enough energy to excite electrons from the valence band to the conduction band. The minimum energy required for this is the band gap (E_g). This occurs when the photon energy (E_photon) is greater than or equal to E_g. The longest wavelength (λ_max) of radiation that can cause this effect corresponds to the

Q: What are the options for this CUET UG Physics question?

A. 0.5 eV · B. 0.7 eV · C. 0.9 eV · D. 1.1 eV

Question

Q23. The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap for the semiconductor is:
(a) 0.5 eV
(b) 0.7 eV
(c) 0.9 eV
(d) 1.1 eV

FirstInTest · Accepted Answer

For a semiconductor, the electrical conductivity increases when incident radiation provides enough energy to excite electrons from the valence band to the conduction band. The minimum energy required for this is the band gap (E_g). This occurs when the photon energy (E_photon) is greater than or equal to E_g. The longest wavelength (λ_max) of radiation that can cause this effect corresponds to the minimum photon energy, i.e., E_g = hc/λ_max.Given λ_max = 2480 nm = 2480 × 10⁻⁹ m.h (Planck's constant) = 6.626 × 10⁻³⁴ J·sc (speed of light) = 3 × 10⁸ m/sE_g = (6.626 × 10⁻³⁴ J·s * 3 × 10⁸ m/s) / (2480 × 10⁻⁹ m)E_g = (1.9878 × 10⁻²⁵ J·m) / (2480 × 10⁻⁹ m)E_g ≈ 8.015 × 10⁻²⁰ J.To convert this energy to electron volts (eV), divide by the elementary charge (e = 1.602 × 10⁻¹⁹ J/eV).E_g_eV = (8.015 × 10⁻²⁰ J) / (1.602 × 10⁻¹⁹ J/eV) ≈ 0.500 eV.This is approximately 0.5 eV.

Answer

0.5 eV

Answer

0.7 eV

Answer

0.9 eV

Answer

1.1 eV

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