If the photon of the wavlength 150 pm strikes an atom and one of tis inner bound electron is ejectred out with a velocity of `1.5 xx 10^(7) ms^(-7)` calculate the energy with which it is bound to the nucleus .

Explanation

Calculation of kinetic energy
mass of electron (m ) = 9.1 `xx 10^(31) kg`
velocity of ejected electron ( v) = 1.5 `xx 10^(7) nm^(1)`
`KE = (1)/(92) mv^(2)`
`=(1)/(2) (9.1 xx 10^(31) kg (1.5 xx 10^(7) ms^(-1) )`
`=10 .2375 xx 10^(17)`
`1.02375 xx 10^(16) kg m^(2) s^(2)`
Applying energy
`=1.3252 xx 10^(15) J = 13.252 xx 10^(16) J`
Energy bond to the nucleus `(E_(0))` = minimum energy required for ejectred electron (e )
`E_(0) = (E- KE)`
`=(13.252 xx 10^(16) J) (1.02375xx 10^(16)J)`
`=(12.228 xx 10^(16)) + (1.602 xx 10^(19))`
`=7.5483 xx 10^(3) eV`