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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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Solution
The correct option is B 253 nm
E1 = 5.6 eV
E2 = 0.7 eV
ΔE = 5.6- 0.7 = 4.9 eV
λ = hcΔE= 12424.9≃253 nm
E1 = 5.6 eV
E2 = 0.7 eV
ΔE = 5.6- 0.7 = 4.9 eV
λ = hcΔE= 12424.9≃253 nm