Modern Physics Ques 263
- 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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Answer:
Correct Answer: 263.(a)
Solution:
Formula:
- 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.
BETA
