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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a
4.8 eV
b
6 eV
c
3.42 eV
d
5 eV
answer is B.
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Detailed Solution
hv = Energy emitted photon= 13.6 (1-122) eV=10.2 eV∴(KE)max=hv–ϕ=(10.2 -4.2) eV = 6 eV
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