Q. 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 $= 1.6 \times 10^{- 27}kg$)
A) $10\ ms^{- 1}$
B) $2 \times 10^{- 2}\ ms^{- 1}$
C) $4\ ms^{- 1}$
D) $8 \times 10^{2}\ ms^{- 1}$
Explanation
(c) The Hydrogen atom before the transition was at rest. Therefore from conservation of momentum
$p_{H - atom} = p_{photon} = \frac{E_{radiated}}{c} = \frac{13.6\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)eV}{c}$
$1.6 \times 10^{- 27} \times v = \frac{13.6\left( \frac{1}{1^{2}} - \frac{1}{5^{2}} \right) \times 1.6 \times 10^{- 19}}{3 \times 10^{8}}$
$\Rightarrow v = 4.352\ m/s = 4m/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.