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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?

Respuesta :

keijosalonen

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