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Unformatted text preview: rgy E released by the four alpha decays is found from the
difference in mass ∆m between the parent nucleus, thorium 228 Th , and the products of the
90
decays, which are four alpha particles and the “greatgreatgranddaughter” nucleus. After
calculating the mass difference in atomic mass units (u), we will find the equivalent amount
of released energy, given that 1 u of mass is equivalent to 931.5 MeV of energy.
SOLUTION
a. Four successive α decays reduce the atomic mass number A of the parent nucleus by
4 × 4 = 16 , and the atomic number Z by 4 × 2 = 8 . Therefore, the overall reaction process
may be summarized as ( A
A−16
4
Z P → Z −8 D + 4 2 He ) Substituting A = 228, Z = 90 and the appropriate chemical symbols into this reaction
process, we obtain ( 228
212
4
90Th → 82 Pb + 4 2 He Therefore, the product of the four α decays is lead
b. The mass of thorium 228
90Th is 212
82 Pb )
. 228.028 715 u, the mass of lead 212
82 Pb is 211.991 871 u (see Appendix F at the back of the book), and the mass of a single α particle is 4.002 063 u.
Following the reaction process above, the mass difference ∆m is ∆m = 228.028 715 u −
14 244
4
3
Mass of parent 211.991871 u
14 244
4
3 − 4(4.002 603 u) = 0.026 432 u
14 244
4
3 Mass of
greatgreatgrandaughter Mass of 4 α
particles The energy equivalent of this mass difference, then, is 931.5 MeV E = ( 0.026 432 u ) = 24.62 MeV
1u CHAPTER 32 IONIZING RADIATION, NUCLEAR ENERGY, AND
ELEMENTARY PARTICLES
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (c) The biologically equivalent dose (in rems) is the absorbed dose (in rads) times the relative
biological effectiveness. (See Equation 32.4.)
2. (d) Equation 32.3 defines the relative biological effectiveness as the dose of 200keV Xrays
that produces a certain biological effect divided by the dose of radiation that produces the
same biological effect. Thus, since RBEA = 2 × RBEB, it takes a smaller dose of A to
produce the same biological effect as B, smaller by a factor of 2.
3. 9.0 × 10−3 rem
4. (b) Since electric charge must be conserved, we know that the number of protons must be the
same before and after the reaction takes place. Therefore, Z + 7 = 6 + 1, so Z = 0. We also
know that the total number of nucleons must be conserved, so the total number before and
after the reaction takes place must be the same. Therefore, A + 14 = 14 + 1, so A = 1. With a
1
single uncharged nucleon in the nucleus, the unknown species must be a neutron 0 n .
5. (a) The symbol for the α particle is 4 He , and the symbol for the proton p is 1 H . Therefore,
1
2
there are 2 + 13 = 15 protons present before the reaction takes place and 15 + 1 = 16 protons
present after the reaction takes place, which violates the conservation of electric charge.
There are 27 + 4 = 31 nucleons present before the reaction takes place and 1 + 31 = 32
nucleons present after the reaction takes place, which violates the conservation of nucleon
number.
6. (e) The compound nucleus is 236
92 U for any X and Y. Since the total number of nucleons is conserved, it follows that 1 + 235 = AX + AY + η, where η is the number of neutrons 1
0n produced by the reaction. Therefore, greater values of η lead to smaller values for AX and
AY. Since electric charge also is conserved, it follows that 0 + 92 = ZX + ZY + η(0).
Therefore, ZX + ZY = 92 for any X and Y. Chapter 32 Answers to Focus on Concepts Questions 1625 7. (b) In a fission reactor each fission event, on average, must produce at least one neutron.
Otherwise there would be no neutrons to cause additional fission events, and it would not be
possible to establish a controlled chain reaction. If the fission products of the reaction each
have a binding energy per nucleon that is less than the binding energy per nucleon of the
starting nucleus, the reaction does not produce energy. It only produces energy when the
fission products have, on average, a binding energy per nucleon that is greater than the
binding energy per nucleon of the starting nucleus.
8. (a) In order for a fusion reaction to be potentially energyproducing, the binding energy per
nucleon of the starting nuclei must be smaller than the binding energy per nucleon of the
product nucleus. Since the maximum binding energy per nucleon occurs at a nucleon
number of about 60, the starting nuclei with nucleon numbers of 30 in reaction I have the
smaller binding energy per nucleon, as required.
9. (c) The energy produced by a fusion reaction is the mass defect ∆m (in u) for the reaction
times 931.5 MeV/u, since an energy of 931.5 MeV is equivalent to 1 u. The mass defect is
the total mass of the initial nuclei minus the total mass of the product nuclei. Thus, to obtain
the ranking, we need only calculate the mass defect for each reaction from the given masses
and rank the defects in descending order. The results are: reaction I (∆m = 0.0035 u),
reaction II (∆m = 0.0138 u), reaction III (∆m = 0.0053 u).
10. (d) The quark theory explains the electric charge that each hadron carries. An antipr...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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