Question

A stationary hydrogen atom of mass $M$ emits a photon corresponding to the longest wavelength of Balmer series. The recoil velocity of the atom is

• A. $\frac{Rh}{M}$
• B. $\frac{Rh}{4M}$
• C. $\frac{3Rh}{4M}$
• D. $\frac{5Rh}{36M}$

Answer 1

The correct option is D $\frac{5Rh}{36M}$

Longest wavelength of Balmer series will occur for the shortest transition.
i.e transition from $n=3$ to $n=2$

Momentum of photon emitted
$p_{\text{photon}}=\frac{h}{\lambda }=Rh(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
Here, $n_1=2\&n_2=3$
$p_{\text{photon}}=Rh(\frac{1}{4}-\frac{1}{9})=Rh\left(\frac{9-4}{36}\right)$
$\Rightarrow p_{\text{photon}}=\frac{5Rh}{36}$

Using conservation of linear momentum,
$p_{\text{photon}}=p_{\text{atom}}$
$\frac{5Rh}{36}=Mv$
where $v$ is the recoil velocity of the atom.
$\Rightarrow v=\frac{5Rh}{36M}$