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Question

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107ms1, calculate the energy with which it is bound to the nucleus.

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Solution

E=hcλ=(6.626×1034Js)(3.0×108ms1)150×1012m=1.3252×1015J=13.252×1016J
Energy of the electron ejected (K.E)
=12mev2=12(9.10939×1031kg)(1.5×107ms1)2=10.2480×1017J=1.025×1016J
Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E – K.E
= 13.252 × 1016 J – 1.025 × 1016 J
= 12.227 × 1016 J
12.227×10161.602×1019eV=7.6×103eV


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