Semi-Empirical Mass Formula part II Quantum Terms [Sec. 4.2 Dunlap]

Inclusion of the Coulomb Term If these terms were all there were - then through the weak force (that converts protons into neutrons) the PROTONS WOULD CONVERT TO NEUTRONS to make the nucleus more tightly bound. There would be no more atoms as we know them What is it then that keeps N  Z THE ANSWER IS QUANTUM MECHANICS DISCRETE ENERGY LEVELS FOR NUCLEONS + PAULI EXCLUSION PRINCIPLE

The Asymmetry Term (a) Kinetic Energy Component UNDERSTANDING NUCLEON MOTION The amazing thing about the nuclear liquid is that although the N-N bonds keep the nucleon bound in the nucleus with NEAREST NEIGHBOR interactions – the NUCLEON MOVES SMOOTHLY THROUGHOUT THE WHOLE NUCLEAR VOLUME – ONLY BEING CONFINED BY THE NUCLEAR WALL. [How does this differ from a classical liquid?] Of course it is not right to show the motion a classical trajectory (as shown) – the motion is governed by WAVE (QUANTUM) MECHANICS – As shown in the upper diagram the nucleons move in well-defined quantum states – each having it’s own wavefunction and discrete energy. In the bulk it has N-N bonds in all directions, but at the surface it sees only INWARD N-N bonds

The Asymmetry Term (a) Kinetic Energy Component UNDERSTANDING NUCLEON MOTION Why is it that nucleons do not collide and have a rather “jerky” passage through the nucleus? The full mathematical answer is complicated (requiring many body theory), but the simple answer is that they travel as waves in discrete energy states – If they do “collide” – where are they to go. THE PAULI EXCLUSION PRINCIPLE forbids a nucleon from entering an already occupied state. Why do we show two nucleons per energy level. --- because there are two spin-states for each spatial wave.

The Asymmetry Term (a) Kinetic Energy Component There are two types of particles (Nucleons) in the nucleus – PROTONS NEUTRONS They each have their own potential well – why? Now we can see why N  Z. If we try and convert protons to neutrons we are going to SPEND ENERGY. For example to convert 2x PROTON  2 x NEUTRON It will cost 2E units of energy If we try and convert ANOTHER two protons it will cost us another 2 x 3E units of energy. We need to “dig” deeper and “ build” higher.

The Asymmetry Term (a) Kinetic Energy Component In enumerating the ENERGY COST incurred in transferring neutrons to protons we find we are dealing with a ARITHMETIC SERIES n=(N-Z)/4 is the number of neutron pairs transferred. Where: The sum of this series is simply:

The Asymmetry Term (a) Kinetic Energy Component From this point there is a long road, and a short road. The short road says that for a FERMI GAS (which is what we have inside the nucleus – fermion (spin ½ particles moving in a box – although we may also call it a FERMI LIQUID) the level spacing E varies as 1/A (i.e. 1/volume) so that the asymmetry energy becomes: which is exactly the form we are looking for, allowing us to write for the nuclear binding energy. NOTE: The new term is also a negative binding energy term

The Asymmetry Term (a) Kinetic Energy Component but the unfortunate thing about taking that “short road” is that we don’t know anything about the value of aA. To find something about aA we need to look at Fermi gas theory – which is conceptually easy – but mathematically messy. FERMI GAS THEORY In a Fermi Gas the particles (fermions) undergo linear motion with wavefunctions of the form: The density of such waves is limited and uniform

FERMI GAS THEORY In a Fermi Gas the particles (fermions) undergo linear motion with wavefunctions of the form:

The Asymmetry Term (a) Kinetic Energy Component FERMI GAS THEORY The density of such waves is limited and uniform For each particle type – NEUTRONS or PROTONS – at equilibrium – Z=N=A/2 We can populate “k-space” with A/2 particles and find the FERMI WAVENUMBER kF Number of spin states From which: The interesting thing here is that nuclear volume V is proportional to A Combining these results: By putting R0=1.2F gives: The same for ALL NUCLEI

The Asymmetry Term (a) Kinetic Energy Component The Nuclear Fermi Energy (EF) We can now work out the energy of nucleons at the Fermi surface. [Fastest moving nucleons] Which may readily be calculated: for ALL NUCLEI 33MeV 15MeV EF Approximately we know that the well depth of ALL NUCLEI is around 50MeV

The Asymmetry Term (a) Kinetic Energy Component The spacing between energy levels On a simple picture you might say – that 33MeV 15MeV EF E But a more accurate answer is required because the spacing are not even – we need E at EF . A/4 levels We must find dn/dE at the Fermi surface: Find dk in terms of dE so that Finally:

Asymmetry Energy (a) Kinetic Energy Component with We see a remarkable thing: And this result is independent of the nucleus BUT IS ONLY 50% OF THE EXPERIMENTAL RESULT

Asymmetry Energy (a) Potential Energy Component

Asymmetry Energy

Pairing Energy This term accounts for the fact that a pair of like nucleons is more strongly bound than is a pair of unlike nucleons. When like nucleons have equal and opposite spins they can exist in the same spatial quantum state This situation is shown in fig (a) in a kind of classical manner. The particles essentially spend more time in each other’s presence and the strong interaction between them is more effective – the binding is stronger. If on the other and the spins are not anti-parallel as shown in fig (b) then the pair only interact at two points on their (classical) orbits. The interaction energy is not so strong. There is no real “theory” for this effect, but clearly the bigger the nucleus the less important. Empirically.

Chart of the Nuclides Z=N Proton number * Unstable nuclei on either side of stable ones * Small Z tendency for Z=N * Large Z characterized by N>Z Neutron number

The number of unstable nuclei is around 2000 but is always increasing. * Tendency for even Z – even N to be the most stable nuclei * Even – Odd, and Odd – Even configurations are equally likely * Almost no Odd Z – Odd N are stable, and these are interesting small nuclei such as