One of the most worrisome waste products of a nuclear reactor is plutonium 239 (

Question

One of the most worrisome 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 time later. The energy emitted in the plutonium decay is 8.40 ? 10-13 J and is entirely converted to kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus is 6.68 ? 10-27 kg, while that of the uranium is 3.92 ? 10-25 kg (note that the ratio of the masses is 4 to 235).

(a) Calculate the speeds of the two nuclei, assuming the plutonium nucleus is originally at rest.
m/s (He)
m/s (U)

(b) About how much kinetic energy does each nucleus carry away?
J (He)
J (U)

Explanation / Answer

energy emitted in the decay = KE of the He + KE of U

KE of He = 0.5*mHe*vHe^2 + 0.5*mU*vU^2 = E........(1)

before decaying the plutonium is at rest

initial momentum = Pi = 0

after decaying final momentum

Pf = mHe*vHe + mU*vU

from momentum conservation Pf = Pi

mU*vU =- mHe*vHe  

vU = -(mHe*vHe)/mU = (mHe/mU)*vHe = (4/235)*vHe =
-0.017*vHe................(2)

from 1 & 2

0.5*mHe*vHe^2 + 0.5*mU*(0.017^2)*vHe^2 = E

vHe^2 *((0.5*6.68*10^-27)+(0.5*(0.017^2)*3.92*10^-25)) = (8.4*10^-13)

vHe = 1.57*10^7 m/s

vU = -0.017*vHe = 2.67*10^5 m/s

------------

part(b)

for HE

KE(He) = 0.5*mHe*vHe^2 = 0.5*6.68*10^-27*(1.57*10^7)^2 = 8.23*10^-13 J    <------answer

for U

KE(U) = 0.5*mU*vU^2 = 0.5*3.92*10^-25*(2.67*10^5)^2 = 1.39*10^-14 J   <-------answer

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