Question

When a Hydrogen atom, initially at rest emits a photon resulting in transition n = 5 to n = 1 its recoil speed in m/s is [Take $h=6.6\times {10}^{-34};m_p=1.6\times {10}^{-27}kg;R={10}^7{m}^{-1}$]

Answer 1

$\frac{1}{\lambda }=R(\frac{1}{m^2}-\frac{1}{n^2})\Rightarrow \frac{1}{\lambda }=R(1-\frac{1}{25})=\frac{24}{25}R$
$P=\frac{h}{\lambda }\Rightarrow P=\frac{h}{25}\times 24R$
By conservation of momentum, the momentum of Hydrogen atom is $-P$ i.e. $P$ in the opposite direction. Therefore,
$mv=P$
$v=\frac{P}{m}$
$v=\frac{24}{25}\frac{hR}{m}=\frac{24}{25}\times \frac{6.6\times {10}^{-34}\times {10}^7}{1.6\times {10}^{-27}}=3.96m/s$