— Physical Chemistry Chemistry Question
Fission
Consider the following fission reactions of 235U by thermal neutrons:
235 94 141 92 38 (...)U + n Sr + Xe + (.....)
235 141 92 56U + n Ba + (.....) + 3 n
Identify the missing species and numbers.
Model Answer
235 94 141 92 38 54U + n Sr + Xe + 2 n
92235 141 92 56 36U + n Ba + Kr + 3 n
Consider the first of the above reactions. The unstable fission fragments undergo successive -decays giving Zr and Ce. Write down the net nuclear reaction and calculate the total energy released in MeV. You are given the following data on atomic masses :
m( 235U) = 235.0493 u
m( 94Zr) = 93.9063 u
m( 140Ce) = 139.9054 u
mn = 1.00866 u
1u = 931.5 Me V/c 2
Model Answer
The net nuclear reaction is
235 94 140 92 40 58
-U + n Zr + Ce + 2 n + 6 e + (Q)
The energy released is
Q = c 2
where the small energy of the initial thermal neutron has been ignored. (mN denotes the nuclear mass.) Now
mN( 235U) = m( 235U) – 92 me
ignoring the small electronic binding energies compared to rest mass energies.
Similarly for other nuclear masses.
Q = c 2
Using the given data,
Q = 213.3 MeV
A sample of natural uranium metal with a mass of 1 kg was put in a nuclear research reactor. When the total energy released reached 1 Mega Watt Day (MWd), it was removed from the reactor. What would be the percentage abundance of 235U in the uranium metal at that time, if it is 0.72 % in natural uranium. Your result in 7.2 above may be taken to be the average energy released per fission. Assume that all the energy is due to fission of 235U only.
Model Answer
1 MWd = 10 6 J s–1 × 24 × 3600 s = 8.64×10 10 J
Number of atoms of 235U fissioned = (8.64 × 10^10) / (213.3 × 1.60 × 10^-13) = 2.53 × 10^21
Mass of 235U fissioned = (2.53 × 10^21 × 235) / (6.02 × 10^23) = 0.99 g
Mass of 235U in 1 kg uranium removed from the reactor = 7.2 – 0.99 = 6.2 g
Abundance of 235U is 0.62 %