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Question

(Marking scheme for all questions will be: correct answer +3 and wrong answer -1)

The number of photons emitted when electrons in a H-atom make transition from a higher energy state to lower energy state, whose difference in angular momentum is h/π , are made to fall incidently on sodium metal (work function, W=2.3 eV). The maximum possible kinetic energy of emitted photoelectrons is:

A
7.9 eV
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B
0.25 eV
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C
10.45 eV
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D
9.79 eV
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Solution

The correct option is D 9.79 eV
Difference in angular momentum=hπ
(n2n1)h2π=hπ
n2n1=2 (Difference in shell no.)
For photoelectric effect to be observed,
Energy of photon > Work function (2.3 eV)
Two photons are possible in H-atom where difference in shell number is 2 and energy >2.3 eV
Ephoton=12.09 eV (From 3 1 transition) and 2.55 eV (From 4 2 transition)
Maximum KE of photoelectron will correspond to maximum energy of the incident photon.
(KE)max=12.092.3=9.79 eV

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