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An electron in an excited hydrogen atom makes a transition from n=2 to n=1. If the photon emitted in the process strikes the surface of a photo sensitive material, the maximum K.E. 4.8 eV

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4.8 eV

6 eV

3.42 eV

5 eV

answer is B.

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Detailed Solution

hv = Energy emitted photon
$= 13.6 (1- \frac{1}{2^{2}}) e V = 10.2 e V$
$∴ {(KE)}_{\max} = hv - ϕ = (10.2 - 4.2) e V = 6 e V$

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