An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000$\mathop A\limits^o$. What is the maximum kinetic energy of the emitted photoelectron?
Solution
initially, energy of electron = + 3eV<br><br>finally, in 2<sup>nd</sup> excited state,<br><br>energy of electron = $- {{(13.6eV)} \over {{3^2}}}$<br><br>$= - 1.51eV$<br><br>Loss in energy is emitted as photon, <br><br>So, photon energy ${{hc} \over \lambda } = 4.51eV$<br><br>Now, photoelectric effect equation <br><br>$$K{E_{\max }} = {{hc} \over \lambda } - \phi = 4.51 - \left( {{{hc} \over {{\lambda _{th}}}}} \right)$$<br><br>$= 4.51eV - {{12400eV\mathop A\limits^o } \over {4000\mathop A\limits^o }}$<br><br>$= 1.41eV$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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