Problem#1
(a) Is the n → p + β- + v̅e decay energetically possible? If not, explain why not. If so, calculate the total energy released. (b) Is the decay n → n + β+ + ve energetically possible? If not, explain why not. If so, calculate the total energy released.
Answer:
Compare the total mass on each side of the reaction equation. Neglect the masses of the neutrino and antineutrino.
1 u is equivalent to 931.5 MeV.
(a) The energy released is the energy equivalent of
mn – mp – me = 8.40 x 10-4u = (8.40 x 10-4)(931.5 MeV) = 783 keV
(b) mn > mp , and the decay is not possible.
Β- and β+ particles have the same mass, equal to the mass of an electron.
Problem#2
What nuclide is produced in the following radioactive decays? (a) α decay of 239Pu94; (b) β– decay of 24Na11; (c) β+ decay of 15O8.
Answer:
In each case determine how the decay changes A and Z of the nucleus. The β+ and β- particles have charge but their nucleon number is A = 0.
(a) α-decay: Z increases by 2, A = N + Z decreases by 4 (an α particle is a 2He4 nucleus)
239Pu94 → 4He2 + 235U92
(b) β- decay: Z increases by 1, A = N + Z decreases by the same (an β- particle is an electron -1e0)
24Na11 → -1e0 + 24Mg12
(c) β+ decay: Z descreases by 1, A = N + Z decreases by the same (an β+ particle is a positron +1e0)
15O8 → +1e0 + 15N7
In each case the total charge and total number of nucleons for the decay products equals the
charge and number of nucleons for the parent nucleus; these two quantities are conserved in the decay.
Problem#3
238U decays spontaneously by emission to 234Th. Calculate (a) the total energy released by this process and (b) the recoil velocity of the 234Th nucleus. The atomic masses are 238.050788 u for 238U and 234.043601 u for 234Th.
Answer:
The mass defect is equal to the difference between the initial and final masses of the constituent
particles.
(a) The mass defect is 238.050788u – 234.043601u – 4.002603u = 0.004584u.
The energy released is (0.004584 u)(931.5 MeV/u) = 4.270 MeV.
(b) Take the ratio of the two kinetic energies, using the fact that K = p2/2m.
KTh/Kα = (pTh2/2mTh)/(pα2/2mα) = mα/mTh = 4/234
The kinetic energy of the Th is
KTh = 4KTh/(234 + 4) = (4/238)(4.270 MeV) = 0.07176 MeV = 1.148 x 10-14 J
Solving for v in the kinetic energy gives
K = ½ mv2
1.148 x 10-14 J = ½ (1.6605 x 10-27 kg)v2
v = 2.43 x 105 m/s
Problem#4
The atomic mass of 14C is 14.003242 u. Show that the β– decay of 14C is energetically possible, and calculate the energy released in the decay.
Answer:
Subtract the electron masses from the neutral atom mass to obtain the mass of each nucleus.
If β- decay of 14C is possible, then we are considering the decay
14C6 → 14N7 + β-
∆m = M(14C6) – M(4N7) – me
Δm = (14.003242u – 6(0.000549 u)) – (14.003074u – 7(0.000549u)) – 0.0005491u
Δm = +1.68 x 10-4u
So, E = Δmc2 = (+1.68 x 10-4u)(931.5 MeV/u) = 156 keV
Problem#5
What particle (α particle, electron, or positron) is emitted in the following radioactive decays? (a) 27Si14 → 27Al13; (b) 238U92 → 234Th90; (c) 74As33 → 74Se34.
Answer:
An α particle has charge +2e and nucleon number 4. An electron has charge -e and nucleon
number zero. A positron has charge +e and nucleon number zero.
(a) A proton changes to a neutron, so the emitted particle is a positron (β+).
(b) The number of nucleons in the nucleus decreases by 4 and the number of protons by 2, so the emitted particle is an alpha-particle.
(c) A neutron changes to a proton, so the emitted particle is an electron (β-).
We have considered the conservation laws. We have not determined if the decays are energetically allowed
Problem#6
(a) Calculate the energy released by the electron-capture decay of (see Example 43.7). (b) A negligible amount of this energy goes to the resulting atom as kinetic energy. About 90% of the time, the nucleus emits two successive gamma-ray photons after the electron-capture process, of energies and respectively, in decaying to its ground state. What is the energy of the neutrino emitted in this case?
Answer:
The energy released is the energy equivalent of the difference in the masses of the original
atom and the final atom produced in the capture. Apply conservation of energy to the decay products.
1 u is equivalent to 931.5 MeV.
(a) As in the example, (0.000897 u)(931.5 MeV u) = 0.836 MeV.
(b) 0.836 MeV – 0.122 MeV – 0.014 MeV = 0.700 MeV.
Problem#7
Tritium (1H3) is an unstable isotope of hydrogen; its mass, including one electron, is 3.016049 u. (a) Show that tritium must be unstable with respect to beta decay because the decay products (3He2 plus an emitted electron) have less total mass than the tritium. (b) Determine the total kinetic energy (in MeV) of the decay products, taking care to account for the electron masses correctly.
Answer:
In beta decay an electron, e-, is emitted by the nucleus. The beta decay reaction is
3H1 → e- + 3He2. If neutral atom masses are used, 3H1 includes one electron and 2He3 includes two electrons.
One electron mass cancels and the other electron mass in 32He represents the emitted electron. Or, we can subtract the electron masses and use the nuclear masses. The atomic mass of 2He3 is 3.016029 u.
(a) The mass of the 3H1 nucleus is 3 016049u – 0.000549u = 3.015500u The mass of the 2He3 nucleus is 3 016029u – 2(0.000549 u) = 3.014931 u. The nuclear mass of 2He3 plus the mass of the
emitted electron is 3.014931u – 0.000549u = 3.015480 u. This is slightly less than the nuclear mass for 1H3, so the decay is energetically allowed.
(b) The mass decrease in the decay is 3.015500u – 3.015480u = 2.0 x 10-5u. Note that this can also be
calculated as (3H1) – m(3He2) where atomic masses are used. The energy released is
(2.0 x 10-5u)(931.5 MeV/u) = 0.019 MeV.
The total kinetic energy of the decay products is 0.019 MeV, or 19 keV.
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