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Question

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 ËšA. What is the maximum kinetic energy of the emitted photoelectron?

A
7.61 eV
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B
1.41 eV
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C
3.3 eV
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D
No photoelectron would be emitted
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Solution

The correct option is B 1.41 eV
Initially, energy of electron =+3 eV
Finally, in 2nd excited state, energy of electron is given as
13.6 eVn2=(13.6 eV)32=1.51 eV ..(1) [As n=3 for 2nd excited state]
So, the energy of resulting photon will be
hcλ=(3(1.51)) eV=4.51 eV ..(2)

Now, from photoelectric effect equation we have
K.Emax=hcλϕ=4.51(hcλthresold)
=4.51124004000=1.41 eV

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