Modern Physics 7 Question 87
17. 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
(Main 2019, 12 Jan II)
(a) $250 nm$
(b) $2020 nm$
(c) $1700 nm$
(d) $220 nm$
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Solution:
- Minimum wavelength occurs when mercury atom deexcites from highest energy level.
$\therefore$ Maximum possible energy absorbed by mercury atom
$$ =\Delta E=5.6-0.7=4.9 eV $$
Wavelength of photon emitted in deexcitation is
$$ \lambda=\frac{h c}{E} \approx \frac{1240 eVnm}{4.9 eV} \approx 250 nm $$
NOTE
Frank-Hertz experiment was the first electrical measurement to show quantum nature of atoms. In a vacuum tube energatic electrons are passed through thin mercury vapour film. It was discovered that when an electron collided with a mercury atom, it loses only a specific quantity ( $4.9 eV$ ) of it’s kinetic energy. This experiment shows existence of quantum energy levels.
