Question

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 ______.

पर्याय

• 1700 nm

• 2020 nm

• 220 nm

• 250 nm

Answer 1

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 250 nm.

Explanation:

Usmg, wavelength, λ = `12375/(Delta"E")`

or, λ = `12375/4.9` 

= 250 nm