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If the photon of the wavelength `150 p m` strikes an atom and one of its inner bound electrons is ejected out with a velocity of `1.5xx10^(7) ms^(-1)`, calculate the energy with which it is bound to the nucleus. |
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Answer» Energy of incident photon`=(hc)/lambda` `=((6.626xx10^(-34)J s)(3.0xx10^(8) m s^(-1)))/((150xx10^(-12)m))` `=13.25xx10^(-19)J` Energy of the electron ejected `=1/2 mv^(2)` `=1/2(9.11xx10^(-31)kg)(1.5xx10^(7)ms^(-1))^(2)` `=1.025xx10^(-16)J` Energy with which the electron was bound to the nucleus `=13.25xx10^(-16)J-1.025xx10^(-16)J` `=12.225xx10^(-16)J` `=(12.225xx10^(-16))/(1.602xx10^(-19))eV=7.63xx10^(3) eV` |
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