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Photoelectrons are emitted when 4000 Å radiation is incident on a surface of work function 1 . 9 eV . These photoelectrons pass through a region having α -particles to form H e + ion, emitting a single photon. In this process, H e + ions thus formed are in their fourth excited state. The energy released during the combination of H e + ions is

Options

  1. A5 . 38 eV
  2. B3 . 38 eV
  3. C2 . 38 eV
  4. D1 . 38 eV

Correct answer

B. 3 . 38 eV

Step-by-step solution

The total energy of electron initially = h c λ − w = 1.18 Final total energy = − 2.2 The difference of energy when photoelectron recombines with α -particle = 1.18 - - 2.2 = 3.38 e V

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