Question

A stationary hydrogen atom of mass $M$ emits a photon corresponding to the longest wavelength of Balmer series. The recoil velocity acquired by the atom is
($R=$ Rydberg constant and $h=$ planck's constant)

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

Answer 1

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

Longest wavelength is emitted in Balmer series of the transition of electron takes place from $n_2=3$ to $n_1=2$
$\therefore$ Longest wavelength in Balmer series
$\frac{1}{{\lambda }_L}=R(\frac{1}{2^2}-\frac{1}{3^2})$ or
${\lambda }_L=\frac{36}{5R}$
The momentum is conserved.
So,
$\frac{h}{\lambda }=Mv$ or $v=\frac{h}{M\lambda }$
putting the value of $\lambda$, we have
$v=\frac{h}{M}\left(\frac{5}{36}R\right)$
$v=\frac{5}{36}\frac{Rh}{M}$