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An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 Å. What is the maximum kinetic energy of the emitted photoelectron?
Correct Option: 2,
Initially, energy of electron = +3eV
finally, in $2^{\text {nd }}$ excited state,
energy of electron $=-\frac{(13.6 \mathrm{eV})}{3^{2}}$
$=-1.51 \mathrm{lV}$
Loss in energy is emitted as photon,
So, photon energy $\frac{\mathrm{hc}}{\lambda}=4.51 \mathrm{eV}$
Now, photoelectric effect equation
$\mathrm{KE}_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi=4.51-\left(\frac{\mathrm{hc}}{\lambda_{\text {th }}}\right)$
$=4.51 \mathrm{eV}-\frac{12400 \mathrm{eV A}}{4000 A}$
$=1.41 \mathrm{eV}$
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