When electron pass through the mercury vapor, it losses some of its energy. The loss in KE of electron =(5⋅6−0.7)eV =4.9eV
∴ energy of radiation emitted =4.9eV
∴ wavelength of radiation, λ=4.91.24×104A≈250nm
JEE Main 2019 — Physics Modern Physics
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:
Held on 12 Jan 2019 · Verified 6 Jul 2026.
250nm
1700nm
220nm
2020nm
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