An electron of mass 9 × 10⁻³¹ kg moves at 3 × 10⁷ m s⁻¹. Its kinetic energy, given 1 eV = 1.6 × 10⁻¹⁹ J, is approximately:

1. Kinetic energy: $E = \tfrac{1}{2}mv^2$. 2. Compute $v^2 = (3\times10^{7})^2 = 9\times10^{14}$ m²/s². 3. $E = \tfrac{1}{2} \times 9\times10^{-31} \times 9\times10^{14} = 4.05\times10^{-16}$ J. 4. Convert: $E = \dfrac{4.05\times10^{-16}}{1.6\times10^{-19}} \approx 2.5\times10^{3}$ eV. 5. So $E \approx 2.5$ keV. _Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", p.139_