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Homework Statement
Fission Reaction: Pu-239 + gamma --> Sr-92 + ? + 3n
Q-Value = ? (Answer in MeV, correct to 5 significant figures)
Atomic Masses (u)
Alpha Particle = 4.00150618
He-4 = 4.0026032497
Pu-239 = 239.0521565
Sr-92 = 91.911030
n (neutron) = 1.00866501
Homework Equations
Q-Value = Q = E = Δmc2 = (mi - mf)c2
The Attempt at a Solution
Q = (239.0521565 + gamma) - (91.911030 + 4.0026032497 + (3*1.00866501)) * c2
= (239.0521565 + gamma) - (98.939628)
Q = gamma = -140.1125285 * 931.5020
= -130515.1
= - 130520 MeV (correct to 5 significant figures)
Two Questions
1) I've seen examples where He-4 = 4.0026032497 is used in this calculation instead of the Alpha Particle = 4.00150618. Why is this? I know He-4 is missing it's two electrons, therefore it's mass is the equivalent of the alpha particle, but how come He-4 atomic mass ≠ alpha particle. And why should I use the atomic mass of He-4 in this calculation instead of the atomic mass of the alpha particle?
2) Is Q negative because it represents the amount of energy which must be provided for the reaction to occur? I.e. A gamma ray with 130520 MeV must be provided to split a Pu-239 atom into Sr-92 + He-4 + 3n