Plutonium-239 is the fissionable isotope produced in breeder reactors; it is als
ID: 993417 • Letter: P
Question
Plutonium-239 is the fissionable isotope produced in breeder reactors; it is also produced in ordinary nuclear plants and in weapons tests. It is an extremely poisonous substance with a half-life of 24,100 years. (a) Write an equation for the decay of 239Pu via alpha emission. (b) The atomic mass of 239Pu is 239.05216 u and that of 235U is 235.04393 u. Calculate the energy released per 239Pu atom, in MeV, in decaying via alpha emission. (c) What will be the initial activity, in disintegrations per second, of 1.00 g of 239Pu buried in a disposal site for radioactive wastes?
Explanation / Answer
1) Equation for the decay of 23994Pu via 42 emission.
23994Pu ---------> 23592U + 42.
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2) Energy released per Plutonium-239 atom on disintegration to U-235 and 1- particle.
Atomic mass of 239Pu = 239.05126 u
Atomic mass of 235U = 235.04393 u
Atomic mass of 4 = 4.00151 u
Total mass of 235U + 4 = 235.04393 + 4.00151 = 239.04544 u.
The mass defect in the radioactivity process is,
m = (239Pu) – (235U + 4)
m = (239.05126) – (239.04544)
m = 0.00582 u
Then binding energy (B.E.) in MeV is given as,
E = m x 931.5
E = 0.00582 x 931.5
E = 5.421 MeV
5.421 MeV energy released.
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3) Initial activity (A), in disintegrations per second (dps), of 1.00 g of 239Pu buried.
Formula,
A=N ………….. ( radioactivity constant)
( = 0.693/T½ )
A = (0.693/ T½) N ……………………….(1)
Where,
Half-life of Pu-239 T½ = 24100 = 24100 x365 days x 24 hrs x 60 min x 60 s = 7.6 x 1011 s.
N = number of Pu atoms in 1g Pu
N = moles of Pu x Avogadro’s number
N = (mass of Pu buried x molar mass of Pu-239) x Avogadro’s number
N=[1.0g / 239]X(6.02x1023mol-1)
N = 2.5 x1021 atoms
Put these values in eq. (1)
A = [0.693/(7.6 x 1011)] x (2.5 x1021)
A = 2.28 x 109 dps.
Initial activity of 239Pu is 2.29 x 109 dps.
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