## The most tightly bound nucleus

which mean, it has most Binding Energy per nucleon. many will say it is 56Fe, which has 8.790MeV per nucleon.

But in a research by Richard Shurtleff and Edward Derringh from Wentworth Institute of Technology say it is 62Ni on 1988, which has 8.795MeV per nucleon.

so, the total different of the binding energy between 2 nucleus are about 300 keV, which is much larger then the contribution of the 2 extra electron in Ni then Fe.

Thus raise up a question on the nuclear fusion process inside stars. why the end product is not 62Ni but 56Fe?

there does not have any stable nucleus to bring 56Fe to 62Ni at the old star. this is the reason from them. and they concluded that, 56Fe is the end product is not due to just the binding energy, but also the environment.

## Deuteron

The deuteron is the nucleus that contains 1 proton and 1 neutron. The spin and isospin of proton and neutron are the same, both are equal to half.  It is the only stable state for 2 nucleons. Deuteron provides an unique aspect to study the inter nuclear force. The strong force are believed to be charge independent. Thus, the strong force can be more easily to study on deuteron due to the absent of other force or eliminate from the Coulomb force, which is understood very much.

The mass of deuteron is 1876.1244MeV. The binding energy is then 2.2245MeV. It was determined by the slow neutron capture of a proton. The emitted gamma ray is approximately equal to the binding energy and the deuteron mass was calculated.

Deuteron has no excited state. It is because any excitation will easily to make the system break apart.

The parity is positive from experiment. If we separate the deuteron wavefunction into 3 parts. The proton wavefunction, neutron wavefunction and the orbital wavefunction. Under the only force, the strong force in this system, proton and neutron are the same nucleon with different state. Thus, the parity are the same for proton and neutron. So, the product of these 2 wavefunction always has positive parity. The total parity then is solely given by the angular orbital.

Any orbital wave function can be represented by the spherical harmonic, $Y(l,m)$.

The parity transform is changing it to

$Y(l,m) \rightarrow (-1)^l Y(l,m)$

So, the experimental face of positive parity fixed the angular momentum must be even.

Ok, we just predicted the possible angular momentum from parity.

The experimental fact on spin is 1. Since J = L + S, and the value of J can take every integer from |L-S| to L + S. and L must be even. The spin of proton and neutron is 1/2. Thus the possible S is 0 or 1 ( we are using L-S coupling scheme ). J = 1 = L + S , that tell us S must be odd to give out 1 for an even L. Thus S=1. So, the only possible L is 0 and 1. Thus, the possible state of deuteron is (L,S) = (0,1) or (2,1). Therefore, a deuteron is a mixed state, if without any further argument.

Now, 2 out of 3 parts of the wave function symmetry were determined by symmetry argument. The isospin can now be fixed by the 2 fermions state must be antisymmetry. The spatial state symmetry is even by L = 0 or 2. And for the state (L , S) = ( 0, 1 ), the spin state is symmetric. Thus, the isospin must be antisymmetric. Since the algebra for isospin and spin are the same. We use T = 0 for the isospin. Thus a complete wavefunction is ( L , S , T ) = ( 0 , 1, 0 ). For the other possible state (L , S) = ( 2 , 1 ) , we can use same argument for isospin state. And for the degenerated state with Ms = +1, 0, -1. By the symmetry of the raising and lowering ladder operator, they all preserved the symmetry. Thus, the Ms = 0 state can only be the + state.

So, we now have 2 possible states of deuteron. If the hamiltonian is commute with L^2 and  S^2, both L and L is a good quantum number and those states are eigen state. And the deuteron ground state must be one of them.

## Stability of a nucleus ( Liquid Drop Model )

when look at the table of the nuclear world, why there are some nucleus more stable then the other? which mean, why some will decay while some are not?

OK, this basically the ultimate question that nuclear physics want to answer.

so, the very fundamental reason, no one know.

but in the elementary level, or by experimental fact and some assumption. we have Binding Energy to estimate or predict the stability of a nucleus. when the Binding Energy is larger then Zero, it must be unstable and will decay under conservation laws. if it is less then zero, it may be stable or not, it depends on whether it reach the bottom of energy level.

Binding Energy can also be though as the energy required to break the nucleus.

In liquid drop model, we imagine the nucleus is like a liquid. and nucleons inside just like liquid molecules. experiments show that nucleus is a spherical object. and it density is a constant. and the interaction range of nuclear force is short, few fm. thus, it likes a incompressible liquid drop. the radius of it is related to the mass number:

$R^3 = A$

the Binding Energy ( = $\Delta M(A,Z,N)$ = mass deficit) is given by theoretical assumption and experimental fact.

$\Delta M(A,Z,N) = - a_1 A + a_2 A^{\frac {2}{3} } + a_3 Z^2 / A^{ \frac{1}{3}} + a_4 (N-Z)^2 /A \pm a_5 A^{- \frac{3}{4} }$

the first 3 terms are theoretical assumption and the lat 2 terms are from experimental fact. All coefficients are given by experimental measurement.

The first term is the “volume energy” by the nuclear force, which is proportional to the number of nucleons.

the 2nd term is the “surface tension” from the “liquid”. we can see its dimension is area.  (why this term is + ? ) it  explained why smaller nucleus has less Binding energy.

the 3rd term is the Coulomb potential term.

the 4th term is the balance term.  if the number of neutron and proton is no balance,

the 5th term is the “Symmetry term“. for even-even of neutron and proton number, the nucleus is more stable, thus, we choose minus sign for it. for odd-odd combination, nucleus are more unstable, thus, plus sign for it. for other, like ood – even or even-odd combination, this term is zero.

the value of the coefficients are:

$a_1 \simeq 15.6 MeV$

$a_2 \simeq 16.8 MeV$

$a_3 \simeq 0.72 MeV$

$a_4 \simeq 23.3 MeV$

$a_5 \simeq 34 MeV$

The below plot is the Binding Energy per nucleon in  Z against N.

Lets use the liquid drop model and Binding Energy to look the β-decay. the β-decay conserved the mass number A. there are 2 β-decays.

$\beta_- : n \rightarrow p + e^- + \bar{\nu_e}$

$\beta_+ : p \rightarrow n + e^+ + \nu_e$

so, the β+ decay decrease the number of proton while β– decay increase the number of proton.

The below diagram show the β-decay for A = 22. we can see the 22Ne is stable, since no more β-decay can help to reach a lower energy level.

## 2p-2p decay of 8C and isospin-allowed 2p decay of the isobaric-analog state in 8B

this paper reports another 2 protons decay mode in 8C. They also discover an “enhancement” at small relative energy of 2 protons. They also reported that an isobaric analog state, 8C and 8B, have same 2-protons decay, which is not known before.

the 1st paragraph is a background and introduction. 2 protons decay is rare. lightest nucleus is 6Be and heaviest and well-studied is 45Fe. the decay time constant can be vary over 18 orders and the decay can be well treated by 3-body theory.

the 2nd paragraph describes the decay channel of 8C and 8B. it uses the Q-value to explain why the 2-protons decay is possible. it is because the 1-proton decay has negative binding energy, thus, it require external energy to make it decay. while 2-protons decay has positive binding energy, thus, the decay will automatic happen in order to bring the nucleus into lower energy state. it also consider the isospin, since the particle decay is govt by strong nuclear force, thus the isospin must be conserved. and this forbid of 1-proton decay.

it explains further on the concept of 2-protons decay and 2 1-protons decay. it argues that, in the 8C, the 2-protons decay is very short time, that is reflected on the large energy width, make the concept of 2 1-protons decay is a unmeasurable concept. however, for the 8Be, the life time is 7 zs (zepto-second $10^{-21}$), the 8Be moved 100 fm ( femto-meter \$latex 10^{-15} ), and this length can be detected and separate the 4-protons emission in to 2 2-protons decay.

the 3rd paragraph explain the experiment apparatus – detector.

the 4th paragraph explains the excitation energy spectrum for 8C, 6Be.

the 5th and 6th paragraphs explain the excitation energy spectrum for the 6Be form 8C decay. since the 2 steps 2 – protons decay has 4 protons. the identification for the correct pair of the decay is important. they compare the energy spectrum for 8C , 6Be and 6Be from decay to do so.

the 7th paragraph tells that they anaylsis the system of 2-protons and the remaining daughter particle, by moving to center of mass frame ( actually is center of momentum frame ) and using Jacobi T coordinate system, to simplify the analysis. the Jacobi T coordinate is nothing but treating the 2-protons the 2 protons are on the arm of the T, and the daughter particle is on the foot of the T.

## Informations we can extract

in scattering experiment,  the raw informations we can know or observe are only few things:

1. the number of particles counted at particular solid angle. ( when you have a unit sphere, the area on the surface is called solid angle)
2. The polarization (spin)
3. charge
4. energy
5. momentum (time of flight)

Since the number of particles counted is related to the intensity of the incident beam, the density of the target, the interaction and the differential cross section.

on the other hand, the number of particles counted should be related to intensity of incident beam, density of the target and interaction potential. Thus, the differential cross section is related to the interaction potential.

The polarization can be measured by 2nd scattering of known polarization target. or directly from a polarized primary target. or by a polarized beam.

For a nucleus there are 12 properties, and we can group them into 2, 1 is static properties, another is dynamic properties.

Static properties

intrinsic:

1. mass
3. spin
4. parity

extrinsic:

1. relative abundance
2. decay half-live
3. magnetic dipole

Dynamic properties

1. decay modes
2. reaction mode
3. cross section
4. excited state

Of course, the final goal of nuclear physics is find the Hamiltonian for governing the motion of nuclear matter. Thus, we can base on the intrinsic static properties, to deduced the extrinsic and dynamic properties.

think about in atomic physics, we know the potential, the spin, then we can give out every things, like the radius, parity, decay half-live, cross section, etc….

so, the nuclear Hamiltonian is the KEY to open the door of understanding of nuclear matter.

## Hydrogen Atom (Bohr Model)

OK, here is a little off track. But that is what i were learning and learned. like to share in here. and understand the concept of hydrogen is very helpful to understand the nuclear, because many ideas in nuclear physics are borrow from it, like “shell”.

The interesting thing is about the energy level of Hydrogen atom. the most simple atomic system. it only contains a proton at the center, um.. almost center, and an electron moving around. well, this is the “picture”. the fact is, there is no “trajectory” or locus for the electron, so technically, it is hard to say it is moving!

why i suddenly do that is because, many text books said it is easy to calculate the energy level and spectrum for it. Moreover, many famous physicists said it is easy. like Feynman, Dirac, Landau, Pauli, etc… OK, lets check how easy it is.

anyway, we follow the usual say in every text book. we put the Coulomb potential in the Schrödinger equation, change the coordinate to spherical. that is better and easy for calculation because the coulomb potential is spherical symmetric. by that mean, the momentum operator (any one don’t know what is OPERATOR, the simplest explanation is : it is a function of function.) automatically separated into 2 parts : radial and angular part. The angular part can be so simple that it is the Spherical harmonic.

Thus the solution of the “wave function” of the electron, which is also the probability distribution of  the electron location, contains 2 parts as well. the radial part is not so trivial, but the angular part is so easy. and it is just $Y(l,m)$.

if we denote the angular momentum as L, and the z component of it is Lz, thus we have,

$L^2 Y(l,m) = l(l+1) \hbar^2 Y(l,m)$

$L_z Y(l,m) = m \hbar Y(l,m)$

as every quadratic operator, there are “ladder” operator for “up” and “down”.

$L_\pm Y(l,m) =\hbar \sqrt{l(l+1) - m(m\pm 1)} Y(l,m \pm 1)$

which means, the UP operator is increase the z-component by 1, the constant there does not brother us.

it is truly easy to find out the exact form of the $Y(l,m)$ by using the ladder operator. as we know, The z component of the a VECTOR must have some maximum. so, there exist an $Y(l,m)$ such that

$L_+ Y(l,m) =0$

since there is no more higher z-component.

by solve this equation, we can find out the exact form of $Y(l,m)$ and sub this in to L2, we can know$Max(m) = l$. and apply the DOWN operator, we can fins out all $Y(l,m)$, and the normalization constant is easy to find by the normalization condition in spherical coordinate, the normalization factor is $sin(\theta)$, instead of 1 in rectangular coordinate.

$\int_0^\pi \int_0^{2 \pi} Y^*(l',m') Y(l,m) sin(\theta) d\theta d \psi = \delta_{l' l} \delta_{m' m}$

more on here

## Differential Cross Section

In nuclear physics, cross section is a raw data from experiment. Or more precisely differential cross section, which is some angle of the cross section, coz we cannot measure every scatter angle and the differential cross section gives us more detail on how the scattering going on.

The differential cross section (d.s.c.) is the square of the scattering amplitude of the scatter spherical wave, which is the Fourier transform of the density.

$d.s.c = |f(\theta)|^2 = Fourier ( \rho (r), \Delta p , r )$

Where the angle θ come from the momentum change. So, sometime we will see the graph is plotted against momentum change instead of angle.

By measuring the yield of different angle. Yield is the intensity of scattered particle. We can plot a graph of the Form factor, and then find out the density of the nuclear or particle.

However, the density is not in usual meaning, it depends on what kind of particle we are using as detector. For example, if we use electron, which is carry elected charge, than it can feel the coulomb potential by the proton and it reflected on the “density”, so we can think it is kind of charge density.

Another cross section is the total cross section, which is sum over the d.s.c. in all angle. Thus, the plot always is against energy. This plot give us the spectrum of the particle, like excitation energy, different energy levels.