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:

Question

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:, 250 n m, 1700 n m, 220 n m, 2020 n m

MARKS App · Accepted Answer

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

Search any question & find its solution

Looking for more such questions to practice?