## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $K = 0.00512~MeV$ (b) $K = 9.39~MeV$ (c) $K = 37.6~MeV$
(a) We can find the kinetic energy when $\gamma = 1.01$: $K = (\gamma-1) ~mc^2$ $K = (1.01-1) ~(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$ $K = 8.1981\times 10^{-16}~J$ $K = (8.1981\times 10^{-16}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $K = 5.12\times 10^{3}~eV$ $K = 0.00512\times 10^{6}~eV$ $K = 0.00512~MeV$ (b) We can find the kinetic energy when $\gamma = 1.01$: $K = (\gamma-1) ~mc^2$ $K = (1.01-1) ~(1.67\times 10^{-27}~kg)(3.0\times 10^8~m/s)^2$ $K = 1.503\times 10^{-12}~J$ $K = (1.503\times 10^{-12}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $K = 9.39\times 10^{6}~eV$ $K = 9.39~MeV$ (c) We can find the kinetic energy when $\gamma = 1.01$: $K = (\gamma-1) ~mc^2$ $K = (1.01-1)(4)(1.67\times 10^{-27}~kg)(3.0\times 10^8~m/s)^2$ $K = 6.012\times 10^{-12}~J$ $K = (6.012\times 10^{-12}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $K = 3.76\times 10^{7}~eV$ $K = 37.6\times 10^{6}~eV$ $K = 37.6~MeV$