Modern Physics 7 Question 69

17. 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

(Main 2019, 12 Jan II)

(a) 250nm

(b) 2020nm

(c) 1700nm

(d) 220nm

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Solution:

  1. Minimum wavelength occurs when mercury atom deexcites from highest energy level.

Maximum possible energy absorbed by mercury atom

=ΔE=5.60.7=4.9eV

Wavelength of photon emitted in deexcitation is

λ=hcE1240eVnm4.9eV250nm

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 ( 4.9eV ) of it’s kinetic energy. This experiment shows existence of quantum energy levels.



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