Nuclear reaction problems (for honors and other interested parties)

Honors Section 1 try these; Honors Section 2 got these in class

Nuclear Collisions Problems

Data
1 eV = 1.6x10-19 J
1 Mev = 1.6 x10-13 J
q for a proton = 1.6x10-19 Coulombs
the mass of a proton is 1.67x10-27 kg
the mass of a neutron is 1.67x10-27 kg
D has one proton and one neutron
T has one proton and two neutrons,
He 3 has one neutron and two protons
He 4 has two neutrons and two protons,
Li has 3 protons and 3 neutrons
c = 3x108 m/s
see Appendix for energies produced by reactions

In collision between charged nuclei, the electrostatic potential energy due to the charges of the protons is given by

PEES = 9x109 q1q2/r where r is the distance between the two nuclei and q are the charges for each nucleus.

In order for the two charges to approach to where the strong forces take over they must get within 10-14 meters.
1) Calculate the kinetic energy the following reactions require.
2) Assuming the kinetic energy is shared equally by the two fuel nuclei, what are their velocities?
3) Using the energies for reactions from the Appendix find the decrease in mass between the products and the fuel
4) Find the velocities of the products (see Appendix for energies produced by reactions)


D +D -> T (1.01 MeV) + p ( 3.02 MeV) (50%)
D +D -> He3 (0.82 MeV) + n ( 2.45 MeV) (50%)
D +He3 -> He4 (3.6 MeV) + p (14.7 MeV)
He3+T -> He4 (4.8 MeV) + D ( 9.5 MeV) (43%)
He3+T -> He4 (0.5 MeV) + n ( 1.9 MeV) + p (11.9 MeV) (6%)

Appendix:
If you want to use fusion as a source of energy, the reaction chosen must satisfy several criteria. It must:
be exothermic. This one is obvious, but it limits the reactants to the low Z side of the curve of binding energy. It also makes helium He-4 the most common product because of its extraordinarily tight binding, although He3 and T also show up.
involve low Z nuclei. This is because the electrostatic repulsion must be overcome before the nuclei are close enough to fuse.
have two reactants. At anything less than stellar densities, three body collisions are too improbable.
have two or more products. This allows simultaneous conservation of energy and momentum without relying on the (weak!) electromagnetic force.
and conserve both protons and neutrons. The cross sections for the weak interaction are too small.
This makes the list of candidates pretty short. The most interesting reactions are the following.
1) D +T -> He4 (3.5 MeV) + n (14.1 MeV)
2) D +D -> T (1.01 MeV) + p ( 3.02 MeV) (50%)
3) D +D -> He3 (0.82 MeV) + n ( 2.45 MeV) (50%)
4) D +He3 -> He4 (3.6 MeV) + p (14.7 MeV)
5) T +T -> He4 + 2 n + 11.3 MeV
6) He3+He3 -> He4 + 2 p
7) He3+T -> He4 + p + n + 12.1 MeV (51%)
8) He3+T -> He4 (4.8 MeV) + D ( 9.5 MeV) (43%)
9) He3+T -> He4 (0.5 MeV) + n ( 1.9 MeV) + p (11.9 MeV) (6%)
10)D +Li6 -> 2 He4 + 22.4 MeV (11) p +Li6 -> He4 (1.7 MeV) + He3 ( 2.3 MeV)
11)He3+Li6 -> 2 He4 + p + 16.9 MeV
12)p +B11 -> 3 He4 + 8.7 MeV