Question:
In a Frank-Hertz experiment, an electron of energy $5.6 \mathrm{eV}$ passes through mercury vapour and emerges with an energy $0.7 \mathrm{eV}$. The minimum wavelength of photons emitted by mercury atoms is close to :
Correct Option: , 4
Solution:
(4) Using, wavelength, $\lambda=\frac{12375}{\Delta \mathrm{E}}$
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