An electron in the third energy level of an excited `He^(o+)` ion return back to the ground sate .The photon emitted in the process is absorbed by a stationary hydrogen atom in the process is absermine by a stationary hydrogen atom in the ground state .Determine the velocity of the photoelectron ejected from the hydrogen atom in metre per second

Explanation

Transition occurs from `n = 2` to `n = 1`
energy released during transition
`Delta E = 13.6(Z)^(2)[(1)/(1^(2)) - (1)/(3^(2))]" "He^(o+)(Z = 2)`
`= 13.6 xx 2^(2)((8)/(9)) = 13.6 xx 4 xx (8)/(9) = 48.35 J`
The energy is abserbed by a stationary hydrogen atom
`Delta E = H underset ((n = 1))"atom"]` = e("electron ejected")
`:. (Delta E - phi) = KE = (1)/(2) mv^(2)`
or `(48.35 - 13.6) = (1)/(2) xx 9.1 xx 10^(-31) xx v^(3)`
`or v = ((34.73)/(4.05 xx 10^(-31)))^(1//2) = 3.49 xx 10^(6)ms^(-1)`