3 lectures: Nuclear Physics Particle Physics 1 Particle Physics 2. Nuclear and Particle Physics. Nuclear Physics Topics. Composition of Nucleus features of nuclei Nuclear Models nuclear energy Fission Fusion Summary. About Units. Energy - electron-volt
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Particle Physics 1
Particle Physics 2
Nuclear and Particle Physics
opposite of “cathode rays”
nucleus (1900 – 1920)
613C* 612C + n
Ea, g= Ei– Ef
F. A. Scott, Phys. Rev.48, 391 (1935)
Dear Radioactive Ladies and Gentlemen!
I have hit upon a desperate remedy to save…the law of conservation of energy.
…there could exist electrically neutral particles, which I will call neutrons, in the nuclei…
The continuous beta spectrum would then make sense with the assumption that in beta decay, in addition to the electron, a neutron is emitted such that the sum of the energies of neutron and electron is constant.
But so far I do not dare to publish anything about this idea, and trustfully turn first to you, dear radioactive ones, with the question of how likely it is to find experimental evidence for such a neutron…
I admit that my remedy may seem almost improbable because one probably would have seen those neutrons, if they exist, for a long time. But nothing ventured, nothing gained…
Thus, dear radioactive ones, scrutinize and judge.
Positron track going
upward through lead
ro = 1.2 x 10-15 m
Find the ratio of the radii for the following nuclei:
1H, 12C, 56Fe, 208Pb, 238U
1 : 2.89 : 3.83 : 5.92 : 6.20
energy units MeV/c2
Problem: Estimate the lowest possible energy of a neutron contained in a typical nucleus of radius 1.33×10-15 m.
E = p2/2m = (cp)2/2mc2
x p = h/2x (cp) = hc/2
(cp) = hc/(2 x) = hc/(2 r)
(cp) = 6.63x10-34 Js * 3x108 m/s / (2 * 1.33x10-15 m)
(cp) = 2.38x10-11 J = 148.6 MeV
E = p2/2m = (cp)2/2mc2 = (148.6 MeV)2/(2*940 MeV) = 11.7 MeV
At t = 1/, N is 1/e (0.368) of the original amount
In each potential well, the lowest energy states are occupied.
Because of the Coulomb repulsion the proton well is shallower than that of the neutron.
But the nuclear energy is minimized when the maximum energy level is about the same for protons and neutrons
Therefore, as Z increases we would expect nuclei to contain progressively more neutrons than protons.
U has A = 238, Z = 92
+ about 200 MeV energy
1H + 1H → 2H + e+ + n
e+ + e- → g + g
2H + 1H → 3He + g
3He + 3He → 4He + 1H + 1H
1 pp collision in 1022→ fusion!
7.65 MeV above 12C ground state