Question

(a) Write the decay equation for the a decay of 228 *Th. (Enter your answer in the form ^x.) chemPad Help x. Greek (b) What energy (in MeV) is released in this decay? The mass of the daughter nuclide is 224.00679 u. (Assume 1 u = 931.5 MeV/c2. Give your answer to at least 2 decimal places.) MeV (c) Assuming the residual nucleus is formed in its ground state, how much energy (in MeV) goes to the a particle? (Give your answer to at least 2 decimal places.) MeV

Answer 1

° () d- decay of nucleus ņX is written as AX Z -A4 10 " He y + Z 2 Z-2. 11 where He is 2 12 228 Th is 10 the alpha karticle decay equation for d-decay of Th Ra + 4 He 1 228 224 2 90 228 224 3 6 Th. Rata (6) Energy released in the decay is given s ley Q-value of the reaction which is B = mass of Parent nucleus - Mass of daugter nucliu, td particles SM / 228 Th) - 1/224 Ra) - M(2) ] c2 Q = 1(228.02874)–(224.00679) – 14.00200) Juc? a = [0.01935 X 9316] Meu Hoa 18.024 Men z

10 U mass of ( 228 Th) 228.02874 u Mass of (224 Ral = 224.00679 mass of (He), a particle = 4:00 2604 and 1 u = 931.5 MeV/c2 'to energy released in alpha decay of 28th 12 18.02 Mev 12 2 as 4 since (C) Energy given to alpha particle appears kinetic energy residual nudluis is heavy its kinetie energy is negligible and since it is uni gerbelnd state , no no energy is spend on that. Energy released in the decay is shared by o bath alpha particle and residual neeellis. 5

10 11 12 I valuee [M (Reactant) Mlproduct]c? = kinetic energy of product - R. E of reactant so o= R. E(X) + K. EL 204 Ra) – K. El 228 Thi = P + pra ama 2 Mra at rest before decay, so rom momentum conservation total momenton should be conserved. Therefore PatPrazo & Pd -Prazp 1 228 Th in after decay 2 2- P² F [1+ Ma 7 5 &M RO MRA 6 Ź KEQ t 0 = P .Ima qM2 Q K.Ed) { Meat Me MRA KETI = 410 [A24 19 4 A-4J KEⓇ) [ 224.Jx(18:0245) mer A 228 K-E (2) = 17.71 mer