Answer to Question #236833 in Quantum Mechanics for Zeedan

Question

Due to collision of an of energy 22 eV with a hydrogen atom, the atom exited to the higher energy state and motion of electron is retarded. Immediately a photon of wavelength 1216 Å is emitted. What is the velocity of the electron after the collision?

P.S. Answer is 2.037 x 10^6 m/s

Answer 1

$mv^2/2=22\cdot e-hc/\lambda\to v=\sqrt{2\cdot(22\cdot e-hc/\lambda)/m}=$

$=\sqrt{2\cdot(22\cdot 1.6\cdot10^{-19}-6.626\cdot10^{-34}\cdot10^8/1216\cdot10^{-10})/(9.1\cdot10^{-31})}\approx2.6\ (Mm/s)$