Problem#1
Consider the nuclear reaction
2H1 + 14N7 → X + 10B5
where X is a nuclide. (a) What are Z and A for the nuclide X? (b) Calculate the reaction energy Q (in MeV). (c) If the 2H1 nucleus is incident on a stationary 14N7 nucleus, what minimum kinetic energy must it have for the reaction to occur?
Answer:
(a) Determine X by balancing the charge and nucleon number on the two sides of the reaction equation.
X must have A = 2 + 14 – 10 = 6 and Z = 1 + 7 – 5 = 3. Thus, X is 6Li3 and the reaction is
2H1 + 14N7 → 6Li3 + 10B5
(b) The neutral atoms on each side of the reaction equation have a total of 8 electrons, so the electron masses cancel when neutral atom masses are used. The neutral atom masses are found in Table 43.2.
mass of 2H1 + 14N7 is 2.014102u + 14.003074u = 16.017176 u
mass of 6Li3 + 10B5 is 6.015121u + 10.012937u = 16.028058 u
The mass increases, so energy is absorbed by the reaction. The Q value is (16.017176u – 16.028058 u)(931 5 MeV/u) = –10.14 MeV
(c) The available energy in the collision, the kinetic energy Kcm in the center of mass reference frame, is related to the kinetic energy K of the bombarding particle by Eq. (43.24).
The kinetic energy that must be available to cause the reaction is 10.14 MeV. Thus Kcm = 10.14 MeV. The mass M of the stationary target (14N7) is 14 u. 14 7 M = The mass m of the colliding particle (2H1) 12 is 2u. Then by Eq. (43.24) the minimum kinetic energy K that the 2H1 must have is
K = (M + m)Kcm/M = (14u + 2u)(10.14 MeV)/14u = 11.59 MeV
Problem#2
Energy from Nuclear Fusion. Calculate the energy released in the fusion reaction
3He2 + 2H1 → 4He2 + 1H1
Answer:
The energy released is the energy equivalent of the mass decrease that occurs in the reaction
m(3He2) + m(2H1) – m(4He2) – m(1H1) = 3.016029u + 2.014102u – 4.002603u – 1.007825u = 0.0197u
because 1u = 931.5 MeV, so the energy released is
E = 0.0197 x 931.5 MeV = 18.4 MeV
Problem#3
Consider the nuclear reaction
2H1 + 9Be4 → X + 4He2
where X is a nuclide. (a) What are the values of Z and A for the nuclide X? (b) How much energy is liberated? (c) Estimate the threshold energy for this reaction.
Answer:
(a) Determine X by balancing the charge and the nucleon number on the two sides of the reaction equation.
X must have A = 2 + 9 – 4 = 7 and Z = 1 + 4 – 2 = 3. Thus, X is 7Li3 and the reaction is
2H1 + 9Be4 → 7Li3 + 4He2
(b) If we use the neutral atom masses then there are the same number of electrons (five) in the
reactants as in the products. Their masses cancel, so we get the same mass defect whether we use nuclear masses or neutral atom masses. The neutral atoms masses are given in Table 43.2
mass of 2H1 + 9Be4 is 2.014102u + 9.012182 u = 11.26284 u
mass of 7Li3 + 4He2 is 7.016003 u + 4.002603u = 11.018606 u
The mass descreases, so corresponds to an energy release of (11.26284 u – 11.018606 u)(931 5 MeV/u) = 7.152 MeV
(b) The radius RBe of the 9Be4 nucleus is RBe = (1.2 x 10-15m)(9)1/3 = 2.5 x 10-15 m
The radius RH of the 2H1 nucleus is RH = (1.2 x 10-15m)(2)1/3 = 1.5 x 10-15 m
The nuclei touch when their center-to-center separation is
R = RBe + RH = 4.0 x 10-15 m
The Coulomb potential energy of the two reactant nuclei at this separation is
U = kq1q2/r = k(e)(4e)/r
U = (8.988 x 109‑ N.m2/C2)(4)(1.602 x 10-19 C)2/[(4.0 x 10-15 m)(1.602 x 10-19 J/eV)] = 1.4 MeV
This is an estimate of the threshold energy for this reaction.
Problem#4
The United States uses 1.0 x 1020 J of electrical energy per year. If all this energy came from the fission of 235U which releases 200 MeV per fission event, (a) how many kilograms of 235U would be used per year and (b) how many kilograms of uranium would have to be mined per year to provide that much 235U? (Recall that only 0.70% of naturally occurring uranium is 235U.)
Answer:
0.7% of naturally occurring uranium is the isotope 235U. The mass of one 235U nucleus is about
235mp.
(a) The number of fissions needed is
(1.0 x 1019 J)/[(200 x 106 eV)(1.60 x 10-19 J/eV)] = 3.13 x 1029.
The mass of 235U required is (3.13 x 1029)(235mp) = 1.23 x 105 kg
(b) (1.23 x 105 kg)/(0.7 x 10-2) = 1.76 x 107 kg
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