Answer to Question #160165 in Atomic and Nuclear Physics for joel

Question

Using the information on atomic masses given below, show that a nucleus of uranium 238 can disintegrate with the emission of an alpha particle according to the reaction (238_92U) → (234_90Th) + (4_2He). Calculate
(a) the total energy released in the disintegration
(b) the kinetic energy of the alpha particle, the nucleus being at rest before disintegration. Mass of (238_92U) = 23812492 a.m.u. Mass of (234_90Th) = 23411650 a.m.u. Mass of (4_2He) = 400387 a.m.u where 1 a.m.u is equivalent to 930 MeV.

Answer 1

a)

$$Q=(M_U-M_{Th}-M_{He})c^2\\Q=(238.05079−234.04363−4.00260)930\\Q=4.25\ MeV$$

b)

$$K_{He}=\frac{4}{238}Q\\K_{He}=\frac{4}{238}4250=71.4\ keV$$