The recoil speed of a hydrogen atom after it emits a photon in going from $\mathrm{n}=5$ state to $\mathrm{n}=1$ state will be

As we know, mass of proton, $\lambda =\frac{25}{1.09\times {10}^7\times 24}$$\mathrm{m}=1.6\times {10}^{-27}\mathrm{kg}$ Using de-Broglie wavelength,

$\lambda =\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}\dots \text{.(i)}$ where, h is Planck’s constant $=6.63\times {10}^{-34}\mathrm{J}-\mathrm{s}$ and p is momentum of photon. By using Rydberg wavelength equation,

$\frac{1}{\lambda }=\mathrm{R}[\frac{1}{{\mathrm{n}}_2^2}-\frac{1}{{\mathrm{n}}_1^2}]$

$\frac{1}{\lambda }=1.09\times {10}^7[\frac{1}{1^2}-\frac{1}{25}]$

$\frac{1}{\lambda }=1.09\times {10}^7\left[\frac{24}{25}\right]$

Substituting values in Eq. (i), we get $\mathrm{v}=\frac{\mathrm{h}}{\mathrm{m}\lambda }=\frac{6.63\times {10}^{-34}\times 1.09\times {10}^7\times 24}{1.6\times {10}^{-27}\times 25}$

$=\frac{6.63\times 1.09\times 24}{1.6\times 2.5}=4.17{\mathrm{ms}}^{-1}$