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.

Explanation

Energy of incident photon (E) is given by,
`E=(hc)/(lambda)`
`=((6.26xx10^(-34)Js)(3.0xx10^(8) ms^(-1)))/(150xx10^(-12) m)`
`=1.3252xx10^(15)J`
`~-13.252xx10^(-16)J`
Energy of the electron ejected (K.E)
`=(1)/(2) m_(e)v^(2)`
`(1)/(2)(9.10939xx10^(-31) kg)(1.5xx10^(7) ms^(-1))^(2)`
`= 10.2480 xx 10^(-17) J`
`= 1.025 xx10^(-16) J`
Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E – K.E
`= 13.252 xx 10^(-16) J -1.025 xx 10^(-16) J`
`= 12.227 xx 10^(-16) J`
`=( 12.227 xx 10^(-16))/(1.602xx10^(-19)eV`
`7.6xx10^(3)eV`
`(5 lambda_(0)-2000)/(4 lambda_(0)-2000)=((5.35)/(2.55))^(2)=(28.6225)/(6.5025)`
`5 lambda_(0)-2000)/(4 lambda_(0)-2000)=4.40177`
`17.6070 lambda_(0)=5 lambda_(0)=8803.537-2000`
`lambda_(0)=(6805.537)/(12.607)`
`lambda _(0)=539.8 nm`
`lambda_(0)=540 nm`