In a Frank - Hertz experiment, an electron of energy 5.6 e V passes through mercury vapour and emerges with an energy 0.7 e V . The minimum wavelength of photons emitted by mercury atoms is close to:

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When electron pass through the mercury vapor, it losses some of its energy. The loss in $K E$ of electron $= (5 \cdot 6 - 0.7) e V$ $= 4.9 e V$

$∴$ energy of radiation emitted $= 4.9 e V$

$∴$ wavelength of radiation, $λ = \frac{1.24 \times 1 0 ^{4}}{4.9} A \approx 250 nm$

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