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

Answer

`sqrt((2)/(m_(e ))[(10^(9)hc)/(18)-54.4 e])=2.2xx10^(6)m//s`

Explanation

`E_(n)=(-136z^(2))/(n^(2))`
For `He^(+) z=2`
`:. E_(n)= -(54.4)/(n^(2))eV`
Ground state energy of `He^(+)`
`E_(1)= -54.4 eV " "|E_(1)|=54.4 eV`
`:. phi=54.4 eV` work function for `He^(+)`
incident energy `=(hc)/(lambda)=(6.6xx10^(-34)xx3xx10^(8))/(18xx10^(-9)xx1.6xx10^(-19))=68.75 eV`
`(KE)_(max)= E-phi`
`:. KE_(max)=678.75-54.4=14.36 eV`
`:. ((1)/(2)mv^(2))_(max)=14.35xx1.6xx10^(-19)`
or `V_(max)=sqrt((2xx14.35xx1.6xx10^(-19))/(9.1xx10^(-3)))`
`rArr V_(max)=sqrt(5.04xx10^(12))" "=2.2xx10^(6)m//s`