Exams › JEE Main › General
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 :
- 2020 nm
- 250 nm
- 1700 nm
- 220 nm
Correct answer: 250 nm
Related JEE Main General questions
⚔️ Practice JEE Main General free + battle 1v1 →