One of the waste products of a nuclear reactor is plutonium-239 (^239Pu). This n
ID: 2075239 • Letter: O
Question
One of the waste products of a nuclear reactor is plutonium-239 (^239Pu). This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus (4He +^235U). the latter of which is also radioactive and will itself decay some tune later. The energy emitted in the plutonium decay is 8.40 times 10^-27 J and is entirely converted into the kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus is 6.68 times 10^-27 kg. while that of the uranium is 3.92 times 10^-25 kg (note that the ratio of the masses is 4 to 235). Calculate the speed of the He. in meters per second, assuming the plutonium nucleus is originally at rest. Calculate the speed of the U. in meters per second, assuming the plutonium nucleus is originally at rest. How much kinetic energy, in joules, does the He nucleus carry away? How much kinetic energy, in joules, does the U nucleus carry away?Explanation / Answer
(A) Applying momentum conservation for this decay process,
initial = final momentum of system
0 = (6.68 x 10^-27) vHe - (3.92 x 10^-25 ) vU
vU = 0.017 vHe
Applying energy conservation,
(6.68 x 10^-27) vHe^2 /2 + (3.92 x 10^-25) vU^2 / 2 = 8.40 x 10^-13
(6.68 x 10^-27) vHe^2 + (3.92 x 10^-25) (0.017 vHe)^2 = 1.68 x 10^-12
6.79 x 10^-27 vHe^2 = 1.68 x 10^-12
vHe = 1.57 x 10^7 m/s
(B) VU = 0.017 vHe = 2.68 x 10^5 m/s
(C) K_He = mHe vHe^2 /2
= 8.26 x 10^-13 J
(D) KE_U = K_tot - K_He = 0.14 x 10^-13 J
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