When a hydrogen atom emits a photon during a transition from $n=4$ to $n=2,$ its recoil speed is about

The expression for the momentum of the photon,

$p=\frac{h}{\lambda }$

$mv=\frac{h}{\lambda }$

$v=\frac{h}{m\lambda }$

From the formula for the Bohr's for H-atom,

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Substitute the given values in the above equation.

$\frac{1}{\lambda }=\left(1.097\times {10}^7\right) \left(\frac{1}{2^2}-\frac{1}{4^2}\right)$

$\Rightarrow \lambda =4.86\times {10}^{-7} \mathrm{m}$

The recoil speed of photon is,

$v=\frac{h}{m\lambda }=\frac{6.62\times {10}^{-34}}{1.6\times {10}^{-27}\times 4.86\times {10}^{-7}}$

$\Rightarrow v=0.8513 \mathrm{m} {\mathrm{s}}^{-1}$