Question

The recoil speed of a hydrogen atom after it emits a photon in going from $n=5$ state to $n=1$ state will be :-

• A. $4.17m/s$
• B. $2.19m/s$
• C. $3.25m/s$
• D. $4.34m/s$

Answer 1

Answer: (A)

$(\Delta E)$ Releases when photon going from $n=5$ to
$n=\Delta E=(13.6-0.54)eV=13.06eV$

$P_i=P_f$ (By linear momentum conservation)
$0=\frac{h}{\lambda }-Mv=V_{\text{Recoil}}=\frac{h}{\lambda M}\dots$(i)
$\&\Delta E=\frac{hc}{\lambda }=\frac{hc}{\lambda M}\times M\Rightarrow Mc{V}_{\text{ Recoil }}$
$V_{\text{Recoil}}=\frac{\Delta E}{Mc}=\frac{13.06\times 1.6\times 10^{-19}}{1.67\times 10^{-27}\times 3\times 10^8}$
$=4.17m/sec$