Solutions to Problems on Relativistic Energy (Serway et al., p.47) R. Egerton

34. (a) K = 5(m0 c^2), total energy E = K + (m0 c^2) = 6 (m0 c^2) = 6 (511 keV)

= 3.07 MeV = (3.06E6)(1.60E-19) Joule = 4.9E-13 Joule

(b) K = (gamma – 1)(m0 c^2), gamma = 1 + 5 (511 keV)/(511 keV) = 6

(1-v^2/c^2) = 1/36, v/c = (35/36)^1/2 = 0.986, v = 2.96E8 m/s

38. (a) K = 400 (M0 c^2) = (gamma – 1) (M0 c^2), gamma = 401

(1-v^2/c^2) = 1/(401)^2 = 6.22E-06, v/c = (1-6.22E-06)^(1/2) = 0.9999969

v = 2.998E08 m/s

(b) K = (400)(938.3 MeV) = 375 000 MeV

40. (a) K = (e)(50 000) = (1.60E-19)(50E03) = 8.01E-15 Joule = 50 keV

= (gamma-1)(m0 c^2), so gamma = 1 + (50 keV)/(511 keV) = 1.0978

(1-v^2/c^2) = 1/(1.0978^2) = 0.8297, v/c = (1-0.8297)^1/2 = 0.4127

v = 1.24E08 m/s

(b) Classical physics would give: (1/2)(m0 v^2) = 8.01E-15 Joule

v^2 = 1.759E16, v =1.326E8 m/s

42. Mass of products = (222.0175 + 4.0026) u = (226.0201) u