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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2021
- Medium
- MCQ
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 \({eV}\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(1.41\)
Step-by-step Solution
Detailed explanation
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
\({KE}_{\max }=\frac{{hc}}{\lambda}-\phi=4.512-\left(\frac{{hc}}{\lambda_{{m}}}\right)\)
\(=4.51\, {eV}-\frac{12400\, {eV} \stackrel{\circ}{{A}}}{4000\, \stackrel{\circ}{{A}}}\)
\(=1.41\, {eV}\)
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