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Consider the reaction $_{92}^{285} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow_{57}^{148} \mathrm{La}+_{35}^{87} \mathrm{Br}+_{0}^{1} \mathrm{n}.$ (a) Write the conservation of relativistic energy equation symbolically in terms of the rest energy and the kinetic energy, setting the initial total energy equal to the final total energy. (b) Using values from Appendix B, find the total mass of the initial particles. (c) Using the values given below, find the total mass of the particles after the reaction takes place. (d) Subtract the final particle mass from the initial particle mass. (e) Convert the answer to part (d) to MeV, obtaining the kinetic energy of the daughter particles. Neglect the kinetic energy of the reactants. Note: Lanthanum-148 has atomic mass 147.932 236 u; bromine - 87 has atomic mass 86.920 711 19 u.

a) $K E_{i}+m_{i} c^{2}=K E_{f}+m_{f} c^{2}$

b) 236.052588$u$

c) 235.861612$u$

d) 0.190976$u$

e)

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National University of Singapore

Gravitation