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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 $4000\, \mathring {{A}}$. What is the maximum kinetic energy of the emitted photoelectron ? (In ${eV}$)
$1.99$
$3.3$
$1.41$
$7.61$
Solution
Initially, energy of electron $=+3 {eV}$
Finally, in $2^{\text {nd }}$ excited state,
$E=-\frac{(13.6\, {eV})}{3^{2}}$
$=-1.51 \,{eV}$
Loss in energy is emitted as photon,
So, photon energy $\frac{{hc}}{\lambda}=4.51\, {eV}$
No, photoelectric effect equation
${KE}_{\max }=\frac{{hc}}{\lambda}-\phi=4.512-\left(\frac{{hc}}{\lambda_{{m}}}\right)$
$=4.51\, {eV}-\frac{12400\, {eV} \stackrel{\circ}{{A}}}{4000\, \stackrel{\circ}{{A}}}$
$=1.41\, {eV}$
