An isolated hydrogen atom emits a photon of $10.2eV$. Find the kinetic energy of the recoil atom. [Mass of proton, $m_p=1.67\times {10}^{-27}kg$ ]

$K=\frac{1}{2}m{v}^2(v=$ recoil speed of atom, $m=$ mass hydrogen atom)

        $\Rightarrow K=\frac{1}{2}m{\left(\frac{p}{m}\right)}^2=\frac{p^2}{2m}$

Now by conservation of linear momentum,p=Momentum of atom = Momentum of photon =$\frac{h\nu }{c}=\frac{10.2 \times 1.6\times {10}^{-19}}{3\times {10}^8}kg-m/s = 5.44\times {10}^{-27} kg -m/s$

Substituting the value of the momentum of atom we get

            $K=\frac{{\left(5.44\times {10}^{-27}\right)}^2}{2\times 1.67\times {10}^{-27}}=8.86\times {10}^{-27}J$