Question

The recoil speed of hydrogen atom after it emits a photon in going from $n=2$ state to $n=1$ state is nearly [Take $R_{\infty }=1.1\times {10}^7{m}^{-1}$ and $h=6.63\times {10}^{-34}Js$]

• A. $1.5m{s}^{-1}$
• B. $3.3m{s}^{-1}$
• C. $4.5m{s}^{-1}$
• D. $6.6m{s}^{-1}$

Answer 1

Answer: (B)

$3.3m{s}^{-1}$

Potential energy of the photon released is:

$\frac{1}{2}(\frac{3}{4}\times 13.6\times 1.602\times {10}^{-19})=\frac{p\times 3\times {10}^8}{2}\Rightarrow p=5.45\times {10}^{-27}$

is the momentum of photn which is also equal to hydrogen

Mass of hydrogen is $m_H=1.67\times {10}^{-27}kg$
Therefore, recoil velocity is $v=\frac{5.45\times {10}^{-27}}{1.67\times {10}^{-27}}=3.26m/s$