Question:
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 :
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 :
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The energy lost by the electron is absorbed by the mercury atom and later emitted as a photon; use \(\lambda = hc/E\).
Updated On: Nov 5, 2026
- \(2020\ \mathrm{nm}\)
- \(250\ \mathrm{nm}\)
- \(1700\ \mathrm{nm}\)
- \(220\ \mathrm{nm}\)
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The Correct Option is B
Solution and Explanation
Energy lost by the electron = 5.6 - 0.7 = 4.9 eV, which is the energy given to the mercury atom.
The atom de-excites by emitting a photon; the maximum photon energy is 4.9 eV, so the minimum wavelength is \(\lambda_{min} = \dfrac{hc}{E} = \dfrac{1240\ \mathrm{eV\,nm}}{4.9\ \mathrm{eV}} \approx 253\ \mathrm{nm}\).
This is closest to 250 nm.
Answer: (2) \(250\ \mathrm{nm}\)
The atom de-excites by emitting a photon; the maximum photon energy is 4.9 eV, so the minimum wavelength is \(\lambda_{min} = \dfrac{hc}{E} = \dfrac{1240\ \mathrm{eV\,nm}}{4.9\ \mathrm{eV}} \approx 253\ \mathrm{nm}\).
This is closest to 250 nm.
Answer: (2) \(250\ \mathrm{nm}\)
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