Question

In a Frank-Hertz experiments, 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. $2020nm$
• B. $220nm$
• C. $250nm$
• D. $1700nm$

Answer 1

Answer: (C)

$250nm$

Energy retained by mercury vapour $=5.6ev-0.7ev=4.9ev$

$\frac{12400}{4.9}=2500A$