$(\Delta E)$ Releases when photon going from $n =5$ to
$n =\Delta E =(13.6-0.54) eV =13.06 eV$
$P _{ i }= P _{ f }$ (By linear momentum conservation)
$0=\frac{ h }{\lambda}- Mv = V _{ \text{Recoil} }=\frac{ h }{\lambda M } \dots$(i)
$\& \,\Delta E =\frac{ hc }{\lambda}=\frac{ hc }{\lambda M } \times M \Rightarrow McV _{\text { Recoil }}$
$V _{ \text{Recoil} }=\frac{\Delta E }{ Mc }=\frac{13.06 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27} \times 3 \times 10^{8}}$
$=4.17 m / sec$
