In a Franck-Hertz experiment,an electron of e | vedclass

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(D) The energy lost by the electron during the collision with the mercury atom is given by the difference in its initial and final kinetic energies.En

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$250$

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(D) The energy lost by the electron during the collision with the mercury atom is given by the difference in its initial and final kinetic energies.Energy lost,$\Delta E = 5.6 \ eV - 0.7 \ eV = 4.9 \ eV$.This energy is absorbed by the mercury atom,which then emits a photon when it de-excites. The energy of the emitted photon is $E = 4.9 \ eV$.The relationship between the energy of a photon and its wavelength $\lambda$ is given by $\lambda = \frac{hc}{E}$.Using the approximation $hc \approx 1240 \ eV \cdot nm$,we have:$\lambda = \frac{1240 \ eV \cdot nm}{4.9 \ eV} \approx 253 \ nm$.Rounding to the nearest given option,the value is $250 \ nm$.

In a Franck-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 ............. $nm$.

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