Physics · 2019
JEE Main · 12 January 2019, Shift 2 · Q25
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…
In a Frank-Hertz experiment, an electron of energy $\displaystyle 5.6$ eV passes through mercury vapour and emerges with an energy $\displaystyle 0.7$ eV . The minimum wavelength of photons emitted by mercury atoms is close to :
Official answer
(4)
$\displaystyle 250$ nm
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.