Problem#1
Comparison of Energy Released per Gram of Fuel. (a) When gasoline is burned, it releases 1.3 x 108 J of energy per gallon (3.788 L). Given that the density of gasoline is 737 kg/m3 express the quantity of energy released in J/g of fuel. (b) During fission, when a neutron is absorbed by a 235U nucleus, about 200 MeV of energy is released for each nucleus that undergoes fission. Express this quantity in of fuel. (c) In the proton-proton chain that takes place in stars like our sun, the overall fusion reaction can be summarized as six protons fusing to form one 4He nucleus with two leftover protons and the liberation of 26.7 MeV of energy. The fuel is the six protons. Express the energy produced here in units J/g of of fuel. Notice the huge difference between the two forms of nuclear energy, on the one hand, and the chemical energy from gasoline, on the other. (d) Our sun produces energy at a measured rate of 3.86 x 1026 W. If its mass of 1.99 x 1030 kg were all gasoline, how long could it last before consuming all its fuel? (Historical note: Before the discovery of nuclear fusion and the vast amounts of energy it releases, scientists were confused. They knew that the earth was at least many millions of years old, but could not explain how the sun could survive that long if its energy came from chemical burning.)
Answer:
ρ = 737 kg/m3; 1 gal = 3.788 L = 3.788 x 10-3 m3.
The massa of a 235U nucleus is 235mp, then
(a) for 1 gallon, m = ρV = (737 kg/m3)(3.788 x 10-3 m3) = 2.79 kg = 2.79 x 10-3 g
So, (1.3 x 108 J)/(2.79 x 10-3 g) = 4.7 x 105 J/g
(b) 1 g contains (1.00 x 10-3 kg)/(235mp) = 2.55 x 1021 nuclei
So, (200 MeV/nucleus)(1.60 x 10-13 J/MeV)(2.55 x 1021 nuclei) = 8.2 x 1010 J/g
(c) A mass of 6mp produces 26.7 MeV.
(26.7MeV)(1.60 x 10-13 J/MeV)/(6mp) = 4.26 x 1014 J/kg = 4.26 x 1011 J/g
(d) The total energy available would be
(1.99 x 1030 kg)(4.7 x 107 J/kg) = 9.4 x 1037 J
Power = energy/t,
so, t = energy/power = (9.4 x 1037 J)/(3.86 x 1026 W) = 2.4 x 1011 s = 7600 yr
Problem#2
Use conservation of mass-energy to show that the energy released in alpha decay is positive whenever the mass of the original neutral atom is greater than the sum of the masses of the final neutral atom and the neutral atom 4He. (Hint: Let the parent nucleus have atomic number Z and nucleon number A. First write the reaction in terms of the nuclei and particles involved, and then add Z electron masses to both sides of the reaction and allot them as needed to arrive at neutral atoms.)
Answer:
Nuclei: AZXZ+ → A-4Z-2Y(Z – 2)+ + 42He2+
Add the mass of Z electrons to each side and we find:
∆m = m(AZX) – m(A-4Z-2Y) – m(42He)
where now we have the mass of the neutral atoms. So as long as the mass of the original neutral atom is greater than the sum of the neutral products masses, the decay can happen.
Problem#3
Use conservation of mass-energy to show that the energy released in β- decay is positive whenever the neutral atomic mass of the original atom is greater than that of the final atom. (See the hint in Problem 43.49.)
Answer:
Denote the reaction as AZX+ → A-4Z+1Y + e-.
The mass defect is related to the change in the neutral atomic masses by
[mX – Zme] – [mY – (Z + 1)me] – me = mX – mY
where mX and mY are the masses as tabulated in, for instance, Table (43.2).
Problem#4
Use conservation of mass-energy to show that the energy released in β+ decay is positive whenever the neutral atomic mass of the original atom is at least two electron masses greater than that of the final atom. (See the hint in Problem 43.49.)
Answer:
Nuclei: AZXZ+ → AZ-1Y(Z – 1)+ + β+
Adding (Z –1) electrons to both sides yields
AZX+ → AZ-1Y + β+
So in terms of masses:
∆m = m(AZX+) – m(AZ-1Y) – me = [m(AZX) – me] – m(AZ-1Y) – me = m(AZX) – m(AZ-1Y) – 2me
So the decay will occur as long as the original neutral mass is greater than the sum of the neutral product mass and two electron masses.
Problem#5
(a) Calculate the minimum energy required to remove one proton from the nucleus 12C6. This is called the proton-removal energy. (Hint: Find the difference between the mass of a 12C6 nucleus and the mass of a proton plus the mass of the nucleus formed when a proton is removed from 12C6.) (b) How does the proton-removal energy for 12C6 compare to the binding energy per nucleon for 12C6, calculated using Eq. (43.10)?
Answer:
The minimum energy to remove a proton from the nucleus is equal to the energy difference
between the two states of the nucleus (before and after proton removal)
12C6 → 1H1 + 11B5.
∆m = m(1H1) + m(11B5) – m(12C6)
The electron masses cancel when neutral atom masses are used
∆m = 1.007825 u + 11.009305 u – 12.000000 u = 0.01713 u
The energy equivalent of this mass increase is
(0.01713 u)(931.5 MeV/u) = 16.0 MeV
(b) We follow the same procedure as in part (a)
∆M = 6MH + 6Mn – 12M6
∆M = 6(1.007825 u) + 6(1.008665 u) – 12.000000 u = 0.09894 u.
Then, EB = (0.09894 u)(931.5 MeV/u) = 92.16 MeV and
EB/A = 92.16 MeV/12u = 7.68 MeV/u
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