Atoms & Nuclei

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75 of 239

Question 75 of 239

In a Frank - Hertz experiment, an electron of energy 5.6eV5.6 eV passes through mercury vapour and emerges with an energy 0.7eV.0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to:

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Answer: (A) 250nm250nm

When electron pass through the mercury vapor, it losses some of its energy. The loss in KEKE of electron =(5⋅6−0.7)eV=\left(5\cdot 6-0.7\right)eV =4.9eV=4.9eV

∴∴ energy of radiation emitted =4.9eV=4.9eV

∴∴ wavelength of radiation, λ=1.24×1044.9A≈250nm\lambda =\dfrac{1.24 \times {10}^{4}}{4.9}A\approx 250 nm

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