Chapter 38 - Photons and Matter Waves - Problems - Page 1183: 51

$K = 4.33\times 10^{-6}~eV$

We can use Equation (38-20) to find the required kinetic energy: $K = \frac{h^2}{2m \lambda^2}$ $K = \frac{(6.63\times 10^{-34}~J~s)^2}{(2)(9.109\times 10^{-31}~kg)~(590\times 10^{-9}~m)^2}$ $K = 6.9314\times 10^{-25}~J$ $K = (6.9314\times 10^{-25}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $K = 4.33\times 10^{-6}~eV$