Question

A stationary hydrogen atom emits photon corresponding to the first time of Lyman series. If $R$ is the Rydberg's constant and $m$ is the mass of the atom, then the velocity acquired by the atom is (neglect energy absorbed by the photon).

• A. $\frac{4m}{Rh}$
• B. $\frac{Rh}{4m}$
• C. $\frac{3Rh}{4m}$
• D. $\frac{4m}{3Rh}$

Answer 1

Answer: (C)

$\frac{3Rh}{4m}$

For first line of Lyman series,
$\frac{1}{\lambda }=R(\frac{1}{1^2}-\frac{1}{2^2})=R(1-\frac{1}{4})=\frac{3}{4}R$
Again we have momentum,
$p=mv$ or $\frac{h}{\lambda }=mv$ or $v=\frac{h}{m\lambda }$
Putting the value of $\lambda$, we have
$v=\frac{h}{m}\left(\frac{3}{4}R\right)=\frac{3hR}{4m}$.