If the photon of the wavelength 150 pm strikes an atom, one of its inner bound electrons is ejected out with a velocity of 1.5 x 107 m s-1. Calculate the energy with which it is bound to the nucleus.
If the photon of the wavelength 150 pm strikes an atom, one of its inner bound electrons is ejected out with a velocity of 1.5 x 107 m s-1. Calculate the energy with which it is bound to the nucleus.
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1 Answer
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λ = 150 pm, v = 1.5 x 107 m s-1
Kinetic energy, K.E. = ½ mv2 = ½ x 9.1 x 10-31 kg x (1.5 x 107 ms-1)2
= [ (9.1 x 1.5 x 1.5) / 2] x 10-31 +14
= 10.2375 x 10-17 J = 1.02375 x 10-16 J
K.E = hc/ λ = (6.626 x 10-34 kg m2 s-1) / (3 x 108 ms-1) / (1.5 x 10-10 m)
= [ (6.626 x 3) x 10-34+8+10] / 1.5
= 13.252 x 10-16 J
We know, E = W0 + K.E.
W0 = E – K.E. = (13.252 – 1.024) x 10-16 J
= 12.228 x 10-16 J
= 12.228 10-16 / 1.602 x 10-19
= 7.63 x 103 eV
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