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In a Frank -Hertz experiment , an electron of energy 5.6eV passes through mercury Vapour and emerges with an electron 0.7eV . The minimum wavelength of photon emitted by mercury atom close to?
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In a Frank -Hertz experiment , an electron of energy 5.6eV passes thro...
Frank-Hertz Experiment:
The Frank-Hertz experiment is an experimental verification of the quantization of energy levels in the atoms. It is based on the collision of electrons with atoms in a vapor.
Given:
Energy of electron before collision (E1) = 5.6 eV
Energy of electron after collision (E2) = 0.7 eV
Explanation:
In the Frank-Hertz experiment, the electrons are accelerated towards the mercury atoms by applying a potential difference. During the collision, the electrons lose some of their energy, which is absorbed by the mercury atoms. This results in the excitation of the mercury atoms to higher energy levels.
The energy of the photon emitted during the de-excitation of the mercury atom is given by the energy difference between the two energy levels. The minimum wavelength of the emitted photon can be calculated using the following formula:
λmin = hc/(E1 - E2)
where h is the Planck constant and c is the speed of light.
Substituting the given values, we get:
λmin = (6.626 x 10^-34 Js x 3 x 10^8 m/s)/(5.6 eV - 0.7 eV)
λmin = 184.3 nm
Therefore, the minimum wavelength of the photon emitted by the mercury atom is 184.3 nm.
Conclusion:
The Frank-Hertz experiment provides a practical demonstration of the quantization of energy levels in atoms. The experiment can be used to determine the energy levels of atoms and the wavelengths of the photons emitted during their de-excitation.
Community Answer
In a Frank -Hertz experiment , an electron of energy 5.6eV passes thro...
Energy retained by mercury vapour = 5.6ev−0.7ev
= 4.9ev
therefore:-
12400 ÷ 4.9 = 2500A (ans)
= 4.9ev
therefore:-
12400 ÷ 4.9 = 2500A (ans)
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In a Frank -Hertz experiment , an electron of energy 5.6eV passes through mercury Vapour and emerges with an electron 0.7eV . The minimum wavelength of photon emitted by mercury atom close to?
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