# Solutions 4

**1: **

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
^{56}Fe. It has the highest binding energy compared to the number of nucleons. - 1.00 kg
^{2}H is 496.5 mol by fusion 248,2 mol^{4}He 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 10^{14}J, 3.56 10^{27}MeV or 1.59 10^{8}kWh. - 1.00 kg
^{233}U fission to^{92}Rb,^{138}Cs and three neutrons. The mass difference is 0.785 g. Which gives 7,1 10^{13}J, 4.40 10^{26}MeV or 1.96 10^{7}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.

**2: **

binding energy B per nucleon:

It is 0.00868 u per nucleon

and 8.26 MeV per nucleon

**3:**

**4:**

- M(n) = 939.6 MeV.
- M(e) = 0.511 MeV.

- M("u") = 931.5 MeV.

**5:**

- 8.55 MeV/nucleon.

- 8.79 MeV/nucleon.

- 7.86 MeV/nucleon.

**6:** The energy

M(^{235}U) - (M(^{131}Xe + M(^{101}Ru) + 3M(n)) =

(ΔM(^{235}U) + 235) - (Δ(^{131}Xe) + 131) - (ΔM(^{101}Ru) + 101) - 3(ΔM(m) + 1)

= ΔM(^{235}U) - ΔM(^{131}Xe) - ΔM(^{101}Ru) - 3ΔM(n)

=40.916 + 88.421 + 87.952 - 3 8.071 = 193.08 Mev

**7:**

Energy usage per 10 km:

1 L = 700 g. M_{m}(C_{8}H_{18}) = 114 g/mol, this means that 700 g is 6.14 mol.

6.14 mol

1 g ^{235}U=4.25 10^{-3} mol = 2.56 10^{21} atoms.

If all of the atoms fission and assuming 100% efficiency this will give 5.12^{29} eV = 8.22 10^{7}kJ. Which will last for 135190 km.

**8:**