Problem#1
The polonium isotope 210Po84 has atomic mass 209.982857 u. Other atomic masses are 206Pb82, 205.974449 u; 209Bi83, 208.980383 u; 210Bi83, 209.984105 u; 2019Po84 208.982416 u; and 210At85 209.987131 u. (a) Show that the alpha decay of 210Po84 is energetically possible, and find the energy of the emitted α particle. (b) Is 210Po84 energetically stable with respect to emission of a proton? Why or why not? (c) Is 210Po84 energetically stable with respect to emission of a neutron? Why or why not? (d) Is 210Po84 energetically stable with respect to β- decay? Why or why not? (e) Is 210Po84 energetically stable with respect to β+ decay? Why or why not?
Answer:
Calculate the mass change in the decay. If the mass decreases the decay is energetically allowed.
(a) m210Po – mPb – mHe = 0.00581u, or Q = 5.41 MeV.
The energy of the alpha particle is (206/210) times this, or 5.30 MeV
(b) m210Po – mBi – mH = –0.00535u < 0, so the decay is not possible.
(c) m210Po – m209Po – mn = –0.00822u < 0, so the decay is not possible.
(d) mAt > m210Po, so the decay is not possible.
(e) mBi + 2me > m210Po, so the decay is not possible.
Problem#2
Irradiating Ourselves! The radiocarbon in our bodies is one of the naturally occurring sources of radiation. Let’s see how large a dose we receive. C14 decays via β- emission, and 18% of our body’s mass is carbon. (a) Write out the decay scheme of carbon-14 and show the end product. (A neutrino is also produced.) (b) Neglecting the effects of the neutrino, how much kinetic energy (in MeV) is released per decay? The atomic mass of 14C is 14.003242 u. (c) How many grams of carbon are there in a 75-kg person? How many decays per second does this carbon produce? (Hint: Use data from Example 43.9.) (d) Assuming that all the energy released in these decays is absorbed by the body, how
many MeV/s and J/s does the 14C release in this person’s body? (e) Consult Table 43.3 and use the largest appropriate RBE for the particles involved. What radiation dose does the person give himself in a year, in Gy, rad, Sv, and rem?
Answer:
The amount of kinetic energy released is the energy equivalent of the mass change
in the decay. me = 0.0005486 u and the atomic mass of 14N7 is 14.003074 u. The energy equivalent of
1u is 931.5 MeV. 14C has a half-life of T1/2 = 5730 yr = 1.81 x 1011 s. The RBE for an electron is 1.0.
(a) 14C6 → e- + 14N7 + v̅e
(b) The mass decrease is ∆M = m(14C6) – [me + m(14N7)]. Use nuclear masses, to avoid difficulty in accounting for atomic electrons.
The nuclear mass of 14C6 is 14.003242u – 6me = 13.999950u.
The nuclear mass of 14N7 IS 14.003074u – 7me = 13.999234u.
∆M = 13.999950u – 13.999234u – 0.000549u = 1.67 x 10-4u.
The energy equivalent of ∆M is 0.156 MeV.
(c) The mass of carbon is (0.18)(75 kg) = 13.5 kg. From Example 43.9, the activity due to 1 g of carbon in a living organism is 0.255 Bq. The number of decay/s due to 13.5 kg of carbon is
(13.5 x 103 g)(0.255 Bq/g) = 3.4 x 103 decays/s.
(d) Each decay releases 0.156 MeV so 3.4 x 103 decays/s releases 530 MeV/s = 8.5 x 10-11 J/s.
(e) The total energy absorbed in 1 year is (8.5 x 10-11 J/s)(3.156 x107 s) = 2.7 x 10-3 J.
The absorbed dose is (2.7 x 10-3 J)/75kg = 3.6 x 10-5 J/kg = 36 µGy = 3.6 mrad.
With RBE 1.0, = the equivalent dose is 36 µSv = 3.6 mrem.
Problem#3
Pion Radiation Therapy. A neutral pion (π0) has a mass of 264 times the electron mass and decays with a lifetime of 8.4 x 10-17 s to two photons. Such pions are used in the radiation treatment of some cancers. (a) Find the energy and wavelength of these photons. In which part of the electromagnetic spectrum do they lie? What is the RBE for these photons? (b) If you want to deliver a dose of 200 rem (which is typical) in a single treatment to 25 g of tumor tissue, how many π0 mesons are needed?
Answer:
mπ = 264me = 2.40 x 10-28 kg. The total energy of the two photons equals the rest mass energy mπc2 of the pion.
(a) Eph = ½ mπc2 = ½ (2.40 x 10-28 kg)(3.00 x 108 m/s)2 = 1.08 x 10-11 J = 67.5 MeV
Eph = hc/λ so λ = hc/Eph = (1.24 x 10-6 eV.m)/(67.5 x 106 eV) = 1.84 x 10-14 m = 18.4 fm
These are gamma ray photons, so they have RBE = 1.0.
(b) Each pion delivers 2(1.08 x 10-11 J) = 2.16 x 10-11 J
The absorbed dose is 200 rad 2.00 Gy = 2.00 J/kg
The energy deposited is (25 x 10-3 kg)(2.00 J/kg) = 0.050 J
The number of π0 mesons needed is 0.050 J/(2.16 x 10-11 J/mesons) = 2.3 x 109 mesons
Note that charge is conserved in the decay since the pion is neutral. If the pion is initially at
rest the photons must have equal momenta in opposite directions so the two photons have the same λ and are emitted in opposite directions. The photons also have equal energies since they have the same momentum and E = pc.
Problem#4
Gold, 179Au79 undergoes β- decay to an excited state of 198Hg80. If the excited state decays by emission of a γ photon with energy 0.412 MeV, what is the maximum kinetic energy of the electron emitted in the decay? This maximum occurs when the antineutrino has negligible energy. (The recoil energy of the 198Hg80 nucleus can be ignored. The masses of the neutral atoms in their ground states are 197.968225 u for 179Au79 and 197.966752 u for 198Hg80.)
Answer:
Find the energy equivalent of the mass decrease. Part of the released energy appears as the emitted photon and the rest as kinetic energy of the electron.
179Au79 → 198Hg80 + -1e0
The mass change is 197.968225 u – 197.966752 u = 1.473 x 10-3u
(The neutral atom masses include 79 electrons before the decay and 80 electrons after the decay. This one additional electron in the product accounts correctly for the electron emitted by the nucleus.) The total energy released in the decay is (1.473 10 u)(931.5 MeV/u) = 1.372 MeV. This energy is divided between the energy of the emitted photon and the kinetic energy of the β- particle. Thus the β- particle has kinetic energy equal to 1.372 MeV – 0.412 MeV = 0.960 MeV.
Problem#5
Calculate the mass defect for the β+ decay of 11C6. Is this decay energetically possible? Why or why not? The atomic mass of 11C6 is 11.011434 u.
Answer:
Nuclei: AZXZ+ → AZ-1Y(Z – 1)+ + β+
Adding (Z –1) electrons to both sides yields
6C11 → 115B + β+
So in terms of masses:
∆m = m(6C1) – m(115B) – 2me =
∆m = 0.00103u. Decay is energetically possible.
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