Question

11. (2 pts) A photon with energy 44.8 eV is absorbed by a hydrogen atom in its ground state, knocking the

electron out. What is the speed of the electron as it leaves?

Answer 1

Answer: 3.3·10^6 m/s

Explanation: Ionization energy of H is 13.6 eV. So the kinetic energy of electron is 31,2 eV. Kinetic Energy of Electron is E = ½mv^2

1 eV = 1.60·10^-19 J and m = 9.1·10^-31 kg. Solving speed v = √( 2E / m)

v = √(2·31.2·1.60·10^-19 J /9.1·10^-31 kg) = 3.3·10^6 m/s