JEE Main 2019 Physics question
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:
Answer: (B)
Solution
ΔE = 4.9 eV ⇒ λ = 1240/4.9 ≈ 253 nm.