A stationary helium ion emits a photon corresponding to the first line of Lyman series. That photon liberates a photoelectron form a stationary hydrogen atom in ground state. Find the velocity of photoelectron. Take mass of electron `=9.11xx10^(-31)kg` and ionisation energy of hydrogen atom=13.6ev.

Explanation

We neglect recoil effects. The energy of the first Lyman line photon emitted by `He^(+)` is
`4 ħR(1-(1)/(4))=3 ħR`
The velocity `v` of the photoelectron that this photon liberates is given by
`3 ħR=(1)/(2)mv^(2)+ ħR`
where ` ħ R` on the right is the binding energy of the `n=1` electron in `H` atom. Thus
`v=sqrt((4 ħR)/(m))=2sqrt(( ħR)/(m))=3.1xx10^(6)m//s`
Here `m` is the mass of the direction.