A stationary $H{e}^{+}$  ion emitted a photon corresponding to the first line of the Lyman series. The photon liberates electron from a stationary hydrogen atom in the ground state. The velocity of the liberated electron is $3.1\times {10}^n$ m/s. Find n (You can make necessary approximations)

given by 
 $\Delta E=13.6(\frac{1}{n_1^2}-\frac{1}{n_2^2}.)eV\mathrm{...}\left(1\right)$
This transition energy is shared between recoiling helium ion and photon 
From conservation of energy.  $\Delta E=jf+K.E.\mathrm{....}\left(2\right)$
Where K.E. is kinetic energy of recoiling atom and hf is energy of photon
From conservation of momentum,
 ${\overrightarrow{P}}_{Photon}=-{\overrightarrow{P}}_{He}\mathrm{...}\left(3\right)$
 $=\frac{2\Delta E}{\sqrt{1+\frac{2\Delta E}{m{c}^2}}+1}\mathrm{...}\left(4\right)$
$\Delta E=13.6(\frac{1}{1^2}-\frac{1}{2^2})eV=40.8eV\mathrm{...}\left(5\right)$
$\frac{1}{2}m{v}^2=(40.8-13.6)eV=37.2eV$
$\therefore v=3.1\times {10}^{6}m/s$
 $\therefore x=6$