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 :
- (A)1700 nm
- (B)2020 nm
- (C)220 nm
- (D)250 nm✓ Correct
Answer: (D)
Solution
The electron hands over the energy it loses to a mercury atom, so the excitation energy taken up is eV. When the atom falls back it releases this as a single photon, and the shortest wavelength corresponds to the whole eV coming out at once. Writing with eV\AA, the wavelength is \AA nm, which is close to nm.
JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option