We know that momentum of a photon, $p=\frac{h}{\lambda}$
$\Rightarrow m v=\frac{h}{\lambda} \Rightarrow v=\frac{h}{m \lambda} $..(i)
$[\because p=m v]$
Given, hydrogen atom emits a photon during a transition from $n=4$ to $n=2$, According to Bohr's formula for $H$ -atom, the wavelength of the emitted photon is given by
$\therefore \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
where, $R=$ Rydberg's constant $=1.097 \times 10^{7} m ^{-1}$
$\Rightarrow \frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$
$\Rightarrow \lambda=4.86 \times 10^{-7} m$
$\because$ Mass of a proton $=1.6 \times 10^{-27} kg$
Now, from Eq. (i), we get
Hence, recoil speed of photon,
$v = \frac{6.62 \times 10^{-34}}{1.6 \times 10^{-27} \times 4.86 \times 10^{-7}}$
$ = 0.8513\,m/s$
or $v \approx 0.814\,ms^{-1}$