The recoil speed of a hydrogen atom after it emits a photon in...

$\left(\DeltaE\right)$ Releases when photon going from $n = 5 to n = \DeltaE$

$= \left(13 . 6 - 0 . 54\right)eV = 13 . 06eV$

${P}_{i} = {P}_{f}$ (By linear momentum conservation)

$0 = \dfrac{h}{\lambda} - Mv = {V}_{Recoil} = \dfrac{h}{\lambdaM} . . . \left(1\right)$

$\text{\text{ and } } \DeltaE = \dfrac{hc}{\lambda} = \dfrac{hc}{\lambdaM} \times M ⇒ Mc{V}_{Recoil}$

${V}_{Recoil} = \dfrac{\DeltaE}{Mc} = \dfrac{13 . 06 \times 1 . 6 \times {10}^{- 19}}{1 . 67 \times {10}^{- 27} \times 3 \times {10}^{8}} = 4 . 17m{s}^{- 1}$