Problem#1
At the beginning of Section 43.7 the equation of a fission process is given in which 235U is struck by a neutron and undergoes fission to produce 144Ba, 89Kr, and three neutrons. The measured masses of these isotopes are 235.043930u (235U), 143.922953 u (144Ba), 88.917630 u (89Kr) and 1.0086649 u (neutron). (a) Calculate the energy (in MeV) released by each fission reaction. (b) Calculate the energy released per gram of 235U in MeV/g.Answer:
The energy released is the energy equivalent of the mass decrease. 1 u is equivalent to 931.5 MeV. The mass of one 235U nucleus is 235mp.
(a) 235U92 + 1n0 → 144Ba56 + 8
We can use atomic masses since the same number of electrons are included on each side of the reaction equation and the electron masses cancel. The mass decrease is
∆m = m(235U92) + m(1n0) – m(144Ba56) – m(89Kr36) – 3m(1n0)
∆m = 235.043930u + 1.0086649 u – 143.922953 u – 88.917630 u – 3(1.0086649 u)
∆m = 0.1860 u
The energy released is
E = ∆mc2 = 0.1860(931.5 MeV/u) = 173.3 MeV
(b) The number of 235U nuclei in 1g is (1.00 x 10-3 kg)/(235mp) = 2.55 x 1021.
The energy released per gram is (173.3 MeV/nucleus)(2.55 x 1021 nuclei/g) = 4.42 x 1023 MeV/g
Problem#2
Consider the nuclear reaction
28Si14 + γ → 24Mg12 + X
where X is a nuclide. (a) What are Z and A for the nuclide X? (b) Ignoring the effects of recoil, what minimum energy must the photon have for this reaction to occur? The mass of a 28Si14 atom is 27.976927 u, and the mass of a 24Mg12 atom is 23.985042 u.
Answer:
The charge and the nucleon number are conserved. The energy of the photon must be at least
as large as the energy equivalent of the mass increase in the reaction.
(a) A + 24 = 28 so, A = 4. Z + 12 = 14 so, Z = 2. Then, X is an α particle.
(b) –∆m = m(24Mg12) + m(α) – m(28Si14)
–∆m = 23.985042 u + 4.002603 u – 27.976927 u = 0.010718 u.
Eγ = (–∆m)c2 = 0.010718 u(931.5 MeV/u) = 9.984 MeV
Problem#3
The second reaction in the proton-proton chain (see Fig. 43.16) produces a 3He2 nucleus. A 3He2 nucleus produced in this way can combine with a nucleus:
3He2 + 4He2→ 7Be4 + γ
Calculate the energy liberated in this process. (This is shared between the energy of the photon and the recoil kinetic energy of the beryllium nucleus.) The mass of a 7Be4 atom is 7.016929 u.
Answer:
The energy released is the energy equivalent of the mass decrease that occurs in the reaction.
The energy liberated will be
m(3He2) + m(4He2) – m(7Be4)
= (3.016029 u + 4.002603 u – 7.016929 u)(931.5 MeV/u) = 1.586 MeV
Using neutral atom masses includes four electrons on each side of the reaction equation and the result is the same as if nuclear masses had been used.
Problem#4
Consider the nuclear reaction
4He2 + 7Li3 → X + 1n0
where X is a nuclide. (a) What are Z and A for the nuclide X? (b) Is energy absorbed or liberated? How much?
Answer:
Charge and the number of nucleons are conserved in the reaction. The energy absorbed or released is determined by the mass change in the reaction.
(a) Z = 3 + 2 – 0 = 5 and A = 4 + 7 – 1 = 10.
The nuclide is a boron nucleus, and m(He) + m(Li) – m(n) – m(B) = –3.00 x 10-3 u, and
So E = –3.00 x 10-3 u(931.5 MeV/u) = 2.79 MeV of energy is absorbed.
Problem#5
In a 100.0 cm3 sample of water, 0.015% of the molecules are D2O. Compute the energy in joules that is liberated if all the deuterium nuclei in the sample undergo the fusion reaction of
Example 43.13.
Answer:
First find the number of deuterium nuclei in the water. Each fusion event requires two of them, and each such event releases 4.03 MeV of energy.
The molecular mass of water is 18.015 x 10-3 kg/mol. m = ρV so the 100.0 cm3 sample has a mass of m = (1000 kg/m3)(100.0 x 10-6 m3) = 0.100 kg.
The sample contains 5.551 moles and (5.551 mol)(6.022 x 1023 molecules/mol) = 3.343 x 1024 molecules.
The number of D2O molecules is 5.014 x 1020. Each molecule contains the two deuterons needed for one fusion reaction. Therefore, the energy liberated is
(5.014 x 1020)(4.03 x 106 eV) = 2.021 x 1027 eV = 3.14 x 108 J = 314 MJ
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