Q81 / 90 · JEE Main 2019 (12 Jan Shift 2) · Physics

In a Frank - Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to:

  • Correct answer
    A
    250 nm
  • B
    1700 nm
  • C
    220 nm
  • D
    2020 nm

Solution

When electron pass through the mercury vapor, it losses some of its energy. The loss in KE of electron =5⋅6-0.7eV =4.9 eV

∴ energy of radiation emitted = 4.9 eV

∴ wavelength of radiation, λ=1.24 × 1044.9A≈ 250 nm

Hence, the correct answer is option: 'A'.

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