The thermal energy generated when radiation from radionuclides is absorbed in ma
ID: 1547511 • Letter: T
Question
The thermal energy generated when radiation from radionuclides is absorbed in matter can serve as the basis for a small power source for use in satellites, remote weather stations, and other isolated locations. Such radionuclides are manufactured in abundance in nuclear reactors and may be separated chemically from the spent fuel. One suitable radionuclide is 238Pu (T1/2 = 87.7 y), which is an alpha emitter with Q = 5.50 MeV. At what rate is thermal energy generated in 3.90 kg of this material? W
Explanation / Answer
now we find the number of atoms present in 3.9 kg PU
the number of atoms=3900*6.023*10^23/238=98.7*10^23 atoms
the alpha emitter =5.5 MeV=5.5*1.6*10^-13=8.8*10^-13 J
the energy released by 3.9 kg 238 Pu=number of atoms*alpha emitter
=98.7*10^23*8.8*10^-13
=868.56*10^10 J
time T1/2=87.7*365*24*60*60=2.8*10^9
the rate of energy released =868.56*10^10/2.8*10^9
=3102 W
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