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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
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 \({eV}\))
- A \(1.99\)
- B \(3.3\)
- C \(1.41\)
- D \(7.61\)
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…
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