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

Step-by-step solution

When electron pass through the mercury vapor, it losses some of its energy. The loss in K E of electron = 5 ⋅ 6 - 0.7 e V = 4.9   e V ∴ energy of radiation emitted =   4.9   e V ∴ wavelength of radiation, λ = 1.24   ×   10 4 4.9 A ≈   250   n m