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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 × 107 ms–1, calculate the energy with which it is bound to the nucleus.
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Solution 1
Energy of incident photon (E) is given by,
`E = ("hc")/lambda`
`= ((6.626xx10^(-34) " Js")(3.0 xx 10^8 " ms"^(-1)))/(150xx10^(-12) " m")`
`= 1.3252 xx 10^(-15)` J
`= 13.252 xx 20^(-16)` J
Energy of the electron ejected (K.E)
`= 1/2 "m"_"e""v"^2`
`=1/2(9.10939xx10^(-31) " kg")(1.5 xx 10^7 " ms"^(-1))^2`
= 10.2480 × 10–17 J
= 1.025 × 10–16 J
Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E – K.E
= 13.252 × 10–16 J – 1.025 × 10–16 J
= 12.227 × 10–16 J
`= (12.227xx10^(-16))/(1.602xx10^(-19))` eV
`= 7.6 xx 10^3` eV
`(5lambda_0 - 2000)/(4lambda_0 - 20000) = (5.35/2.55)^2 = 28.6225/6.5025`
`(5lambda_0 - 2000)/(4lambda_0- 2000) = 4.40177`
`17.6070lambda_0 - 5lambda_0 = 8803.537- 2000`
`lambda_0 = (6805.537)/12.607`
`lambda_0 = 539.8 "nm"`
`lamda_0 = 540 "nm"`
Solution 2
Energy of the incident photon= hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/(150×10-12m) = 13.25×10-16 J
Energy of the electron ejected = 1/2 mv2 = 1/2×(9.11×10-31kg)×(1.5×107ms-1)2 = 1.025×10-16 J
Energy with which the electron was bound to the nucleus = 13.25×10-16 J - 1.025×10-16 J
= 12.225×10-16 J = 12.225×10-16/1.602×10-19 eV = 7.63×103 eV
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