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  • An electron and proton are separated by a large distance. The electron starts approaching the proton with energy The proton captures the electrons and forms a hydrogen atom in second

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 \mathrm{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 \AA. What is the maximum kinetic energy of the emitted photoelectron?
 
Option: 1 7.61 \mathrm{eV}
 
Option: 2 1.41 \mathrm{eV}
Option: 3 3.3 \mathrm{eV}
Option: 4 \text { No photoelectron would be emitted }

Answers (1)

best_answer

Energy lost by electron=Energy of emitted photon

\begin{aligned} &\left(E_{i}-E_{f}\right)=\frac{h c}{\lambda} \\ &\left(3 \operatorname{ev}-\left(\frac{-13.6 \mathrm{eV}}{3^{2}}\right)\right]=\frac{h c}{\lambda} \\ \end{aligned}

\begin{aligned} &{\because \text { for second excitcd state }} \\ &\text { i.e., } For \ (n=3) \Rightarrow E_{n}=\frac{-13.6 \mathrm{eV}}{n^{2}}=\frac{-13.6 \mathrm{eV}}{3^{2}}=-1.51 eV \\ &\frac{h c}{\lambda}=4.51 \mathrm{eV} \end{aligned}

By Einstein's photoelectric equation
\begin{aligned} &\frac{h c}{\lambda}=\phi_{0}+K E_{\max } \\ &4.51 \mathrm{ev}=3.1 \mathrm{ev}+K E_{\text {max }} \\ &K E_{\max }=1.41 \mathrm{eV} \end{aligned}
The correct option is (2)

Posted by

vishal kumar

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