A 232Th (thorium) nucleus at rest decays to a 228Ra (radium) nucleus with the emission of an alpha particle. The total kinetic energy of the decay fragments is 6.54 × 10?13 J. An alpha particle has 1.76% of the mass of a 228Ra nucleus. Calculate the kinetic energy of (a) the recoiling 228Ra nucleus and (b) the alpha particle.

Solution 102P Step 1: Initially the 232Th nucleus was at rest and decays into a 228Ra nucleus and an alpha particle. By the conservation of momentum, m V = m V --------------------(1) Ra Ra But it is given that, m 0.0176 m Ra -----------------(2) By putting this value in equation (1), m Ra Ra = 0.0176 m Ra V V = 0.0176 V ----------------------(3) Ra Step 2: The total kinetic energy would be, 1 2 1 2 1 2 2 K.E tot m2 RaV Ra + 2 V = [2 VRa Ra + 0.0176 m Ra V ] 1 2 V Ra = m2 Ra [V Ra + 0.0176 × 0.0176] 1 2 VRa2 1 2 1.0176 1 2 = m2 Ra [V Ra + 0.0176] = 2 Ra V Ra × 0.0176= 2 Ra V Ra × 57.81.