Example 27-6 Alpha Decay of 238U
The uranium isotope 238U undergoes alpha decay to 234Th. The binding energy per nucleon is 7.570 MeV in
7.597 MeV in 234Th. Find the energy released in the process 238U S 234Th + a.
Set Up
238
We’ll use the same principle that we used in
Example 27-4 (Section 27-4) to find the energy
released in fission: The released energy equals
the total binding energy of the nuclei present
after the decay, minus the binding energy of the
original (parent) nucleus. As in that example,
we’ll find the binding energy for each nucleus
by multiplying the binding energy per nucleon
times the number of nucleons A.
(energy released)
Solve
From Example 27-3 (Section 27-3), the binding energy of a 4He
nucleus (an alpha particle) is
U and
= (total binding energy of daughter plus alpha particle)
– (binding energy of original 238U nucleus)
EB(4He) = 28.297 MeV
The binding energy of a 234Th nucleus (A = 234) is
EB(234Th) = (234)(7.597 MeV) = 1777.7 MeV
and the binding energy of a 238U nucleus (A = 238) is
EB(238U) = (238)(7.570 MeV) = 1801.7 MeV
The released energy is then
Ereleased = EB(234Th) + EB(4He) 2 EB(238U)
= 1777.7 MeV + 28.297 MeV 2 1801.7 MeV
= 4.3 MeV
Reflect
The alpha particles emitted by radioactive isotopes with long half-lives, such as 238U, tend to have kinetic energies in the
4 to 5 MeV range.
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