Q 12-12-116JEE MainJEE Main 2019 (12 Jan, Shift 2)Easy
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: (A) $250$ nm
Energy given to a mercury atom: $5.6 - 0.7 = 4.9$ eV. The photon emitted on de-excitation has at most this energy:
$$\lambda_{\min} = \frac{1240}{4.9}\ \text{nm} \approx 253\ \text{nm} \approx 250\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics