JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 13)
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
توضیح
The minimum wavelength of emitted photons is
$$\lambda $$ = $${{1240} \over {5.6 - 0.7}}nm$$ = 250 nm
$$\lambda $$ = $${{1240} \over {5.6 - 0.7}}nm$$ = 250 nm