Mass excess is given by deltaM(Z,A) = M(Z,A)-A. The masses can therefore be found with M(Z,A) = delta M (Z,A ) + A by using 1 u = 931.5 MeV we get:
- M(n) = 1.008665 u
- M(1H) = 1.007825 u
- M(4He) = 4.002603 u
- M(56Fe) = 55.934938 u
- M(142Ce) = 141.909244 u
- M(238U) = 238.050788 u
- The most stable nuclei is 56Fe. It has the highest binding energy compared to the number of nucleons.
- 1.00 kg 2H is 496.5 mol by fusion 248,2 mol 4He is created which is 993.6 g. The difference in mass is equal to the energy 1000 g-993.6 g = 6.36 g. The energy is then 5.7 1014 J, 3.56 1027 MeV or 1.59 108 kWh.
- 1.00 kg 233U fission to 92Rb, 138Cs and three neutrons. The mass difference is 0.785 g. Which gives 7,1 1013 J, 4.40 1026 MeV or 1.96 107 kWh.
- The energy from the fission is distributed in different ways: most of it goes to the kinetic energy of for the fission products. Other parts goes to the kinetic energy of the neutrinos and the neutrons, and some of it goes to “prompt” gamma-rays and beta/gamma rays from the fission products.
binding energy B per nucleon:
It is 0.00868 u per nucleon
and 8.26 MeV per nucleon
times bigger binding energy for the nucleus.
- M(n) = 939.6 MeV.
- M(e) = 0.511 MeV.
- M("u") = 931.5 MeV.
- 8.55 MeV/nucleon.
- 8.79 MeV/nucleon.
- 7.86 MeV/nucleon.
6: The energy
M(235U) - (M(131Xe + M(101Ru) + 3M(n)) =
(ΔM(235U) + 235) - (Δ(131Xe) + 131) - (ΔM(101Ru) + 101) - 3(ΔM(m) + 1)
= ΔM(235U) - ΔM(131Xe) - ΔM(101Ru) - 3ΔM(n)
=40.916 + 88.421 + 87.952 - 3 8.071 = 193.08 Mev
Energy usage per 10 km:
1 L = 700 g. Mm(C8H18) = 114 g/mol, this means that 700 g is 6.14 mol.
1 g 235U=4.25
If all of the atoms fission and assuming 100% efficiency this will give 5.12
ΔG = ΔGp'r'o'd'u'c't - ΔGr'e'a'c't'a'n't = - 237 k'J / m'o'l. Which means that for each H2O molecule it is generated 237kJ/NA = 2.456 10-6MeV. Comparing to 0.0303 u = 23.2 MeV realesed by fusion to helium. We see that the nuclear reactions functions on a scale 10^6 times bigger than chemical.
fusion of 2H gives 5.75
In this problem it is usful to keep in mind the binding energy equation:
Most isobars (A= constant, Z varies) will have binding energy when plotted as a function of Z creating a parable. The last part of the equation in the formula for binding energy delta expresses the energy gained when the proton and/or the neutron are in a couple. A nucleus with a even amount of both will have a posetive delta, higher binding energy, a unpaired nucleus either proton or neutron will have a delta of zero, a nucleus with a unpaired neutron and proton will have a negative delta. Therefore for odd-nuclei the binding energy is independent of which nucleon is unpaired. The binding energy as a function of z will have a minimum value for a certain stable nucleus. For even nuclei with a given A value they will have a even-even, or a odd-odd konfiguration when Z varies. This will create two curves where the odd-odd nuclei will have a lower binding energy than the even-even nuclei. Odd-odd nuclei will therefore have more possibilities when decaying to a level with higher binding energy, either with adecay or a decay. Both of these mechanisms will lead to a more stable nucleus.
11: answered above