devarshi-dt-logo

Question:

In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapor and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to:

2020nm

220nm

1700nm

250nm

Solution:

The correct option is C 250nm
Energy retained by mercury vapor = 5.6 eV - 0.7 eV = 4.9 eV
λ = hc/E = (12400 eV Å) / 4.9 eV = 2530 Å ≈ 250 nm