Consider the reaction 235 92 U + 1 0 n 148 57 La + 87 35 Br + 1 0 n. (a) Write t

Question

Consider the reaction 23592U + 10n 14857La + 8735Br + 10n.

(a) Write the conservation of relativistic energy equation symbolically in terms of the rest energy and the kinetic energy, setting the initial total energy to the final total energy. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) Using values given above, find the total mass of the initial particles.
u

(c) Using the values given above, find the total mass of the particles after the reaction takes place.
u

(d) Subtract the final particle mass from the initial particle mass.
u

(e) Convert the answer to part (d) to MeV, obtaining the kinetic energy of the daughter particles, neglecting the kinetic energy of the reactants.
Mev

Explanation / Answer

Mass of 235U = 235.043923u

mass of n = 1.008665u

mass of 148La =147.932236u

mass of 87Br = 86.9207119u

(a) K =Qi-Qf = [Dmc^2]

(b) total mass of initial partciles

mi = (235.043923u) +(10*1.008665u) = 245.1305734u

(c) totla mass of final particles

mf = (147.932236u)+(86.9207119u)+(10*1.008665u) = 244.9395979u

(d) Dm =mf -mi = (244.9395979u) -(245.1305734u)

Dm = -0.1909755u

(e) 1uc^2 =931 MeV

Dmc^2 = (-0.1909755uc^2)(931MeV)= -177.8 MeV

K = 177.8 MeV

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