Question

A hydrogen atom emits a photon corresponding to an electron transition from $n=5$ to $n=1$. The recoil speed of hydrogen atom is almost
(mass of proton $\approx 1.6 \times 10^{-27} \,kg$ )

• A. $ 10\,ms^{-1} $
• B. $2 \times 10^{-2} \,ms ^{-1}$
• C. $4\, m s^{-1}$
• D. $8 \times 10^{2} \,m s^{-1}$

Answer 1

Answer: (C)

The hydrogen atom before the transition was at rest. Therefore, from conservation of momentum
$P_{\text{H-atom}}=P_{\text{photon}}=\frac{E_{\text {radiated }}}{c}=\frac{13.6\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] eV }{c}$
$1.6 \times 10^{-27} \times v$
$=\frac{13.6\left[\frac{1}{1^{2}}-\frac{1}{5^{2}}\right] \times 1.6 \times 10^{-19}}{3 \times 10^{6}}$
$\Rightarrow v=4.352\, m / s =\approx 4\, m / s$