Use $m_{e} = 9.1 \times 1 0^{- 31} kg$ , $1 eV = 1.6 \times 1 0^{- 19} J$ and $E_{1} = - 13.6 eV$ . A free electron collides with a ground-state hydrogen atom and just excites it to the first excited state, losing all its kinetic energy. Find the speed of the electron before the collision, in units of $1 0^{6} m s^{- 1}$ (two decimals). Neglect atomic recoil.

Numerical answer type, answer in 1e6 m/s, tolerance 0.02.

Answer: 1.89 1e6 m/s

The required energy is

13.6 (1 - 1/4) = 10.2 eV = 1.632 e - 18 J

. Setting

\frac{1}{2} m_{e} v^{2} = 1.632 \times 1 0^{- 18}

gives

v = 2 K E / m_{e} = 1.89 \times 1 0^{6} m s^{- 1}