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The electrical conductivity of a semiconductor increases when electromagnetic radiation of a wavelength shorter than 2480   nm is incident on it. The bandgap in (eV) for the semiconductor is [take hc = 12400   eV   A ∘ ]

Options

  1. A0.5 eV
  2. B0.7 eV
  3. C1.1 eV
  4. D2.5 eV

Correct answer

A. 0.5 eV

Step-by-step solution

λ max = 2 4 8 0  nm = 2 4 8 0 0  Å Energy (in eV ) = 12400   e V   Å λ E = 1 2400 2 4 8 0 0  eV E= 0 . 5  eV

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