Find the velocity of photoelectrons liberated by electromagnetic radiation of wavelength `lambda=18.0nm` from stationary `He^(+)` ions in the ground state.

Explanation

By conservation of enery
`(1)/(2)mv^(2)=(2pi ħ c)/(lambda)-E_(b)`
where `E_(b)=4 ħ R` is the binding energy of the electron in the ground state of `He^(+)`. (Recoil of `He^(++)` nucleus is neglected). Then
`v=sqrt((2)/(m)((2pi ħc)/(lambda)-E_(b)))`
Substitution gives
`v= 2.25xx10^(6)m//s`