Physics - Modern Physics Question with Solution | TestHub

$\left(\Delta E\right)$ Releases when photon going from $n=5$ to $n=\Delta E$

$=\left(13.6-0.54\right)eV=13.06eV$

$P_i=P_f$ (By linear momentum conservation)

$0=\frac{h}{\lambda }-Mv=V_{Recoil}=\frac{h}{\lambda M}...\left(1\right)$

$\&\Delta E=\frac{hc}{\lambda }=\frac{hc}{\lambda M}\times M\Rightarrow Mc{V}_{Recoil}$

$V_{Recoil}=\frac{\Delta E}{Mc}=\frac{13.06\times 1.6\times 10^{-19}}{1.67\times 10^{-27}\times 3\times 10^8}=4.17m{\text{s}}^{-1}$