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In a Frank-Hertz experiment,an electron ...

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

A

2020 nm

B

1700 nm

C

220 nm

D

250 nm

Text Solution

Verified by Experts

The correct Answer is:
D
`lambda` (in nm) `= (1240)/((5.6 - 0.7)eV) = 250 nm`
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