## [ Be-12(p,n)B-12 ] the detector circuit

i joined a scattering experiment for $^{12}Be(p,n)^12B$. today the Beam-Line-Detector team was finished the BLD, and SHARAQ group is trimming the beam to the SHARAQ detector. i was worked on the neutron detector array, but i was absented for all preparation job due to a polarized proton target.

here is the detector circuit:

when the neutron hits on the plastics scintillator, it create s proton and further release photons. these photons is result of photoelectric effect and Compton scattering. the photons have different time on reaching the 2 ends of the scintillator due to the position. Thus, by analysis the timing, we can identify the position.

The Photo-Multipler Tube convert photon into electric signal. This signal go to a Splitter, the splitter simply copy the signal. one output is go to a Discriminator, which create a digitized signal. the PMT at the 2 ends will give different timing, and the Discriminator signals will be combined after the AND gate. If the neutron hit the scintrillator near either ends, the AND gate make sure that the later signal will be the trigger signal.

the trigger signal will enable the Charge-Digital-Converter (QDC) start integrate the delayed signal for 300ns. Since from the raw signal to the AND gate takes time, thus a delay will ensure a correct detection of the total energy for each signal. However, the signal will have exponential decay when traveling alone the PMT. the signal decay is:

$S = S_0 e^{- k x}$

For signal created at position r from the mid-point of the scintrillator , the 2 signals are:

$S_1 = S_0 e^{- k (L-r)}$

$S_2 =S_0 e^{-k ( L+r)}$

where L is the half length from the mid-point of the scintrillator. thus, if we use a geometric mean, we can get the original signal strength.

$S_0 = \sqrt{S_1 S_2} e^{k L}$

the trigger signal also combined with the discriminator signal for determination of the time. the trigger signal will define the t=0. by this, we can calculate back the location of the neutron. If we define the mid-point of the scintrillator be x=0. a neutron hits on position r will take different path and the time different for the scintrillator is:

$\Delta t_s = 2 r / c$

where c is the speed of light. we also have to add the time different due to the cable, thus, the total time different is:

$\Delta t = t_1 - t_2 = \frac{2r}{c} + t_0$

by fitting the statistic, the average of the position should be zero, thus the mean of Δt is the offset by the cable.

However, as we mention before, the signal will decay. since we are using Threshold Discriminator, different signal strength will give different time even for same start time. This is called Walk effect. For lower signal strength, the effect is stronger. and give larger Walk time. since the Walk time is always additional, the statistics of the QDC-TDC graph will has a tail at one side only.

In fact , there is a technique the tackle the Walk effect, which is by a Constant Fraction Discriminator. the principle is that, if we copy the signal and apply a negative fraction on 1 of them and summing up. the point of zero is always the same no matter the amplitude or the signal strength.

the Walk effect can be calculated by assuming the falling of the signal is just like a line with slope -m. with threshold -θ, and a decay ratio k. the time are:

$t_1 = \frac{\theta}{m}$

$t_2 = \frac{\theta}{k m} = \frac{1}{k} t_1$

$\Delta t = t_1-t_2 = \frac{k - 1}{k} \frac{\theta} {m}$

if we plot the signal strength versus time different, we have a curve in the form

$y = \frac{k-1}{k} \theta \frac{1}{x}$

From the equation, a smaller decay ratio, which corresponding to the location closer to the mid-point of the  scintrillator, the bigger the $1- \frac{1}{k}$ and give a border curve.

## Angular distribution of Neutrons from the Photo-Disintegration of the Deuteron

this paper was written on 1949. at that time, deuteron just discovered 20 years. this paper presents a method on detecting the diffraction cross section of the neutron from a disintegrated deuteron by gamma ray of energy 2.76MeV. and by this, they found the photo-magnetic to photo-electric cross section ration. the ratio is 0.295 ± 0.036.

the photo-electric dipole transition and photo-magnetic dipole transition can both be induced by the gamma ray. Photo carry 1 angular momentum, the absorption of photon will excited the spherical ground state $^1S$ into $^3P$. the 2 mechanisms of the disintegrations results 2 angular distributions of the neutrons. by examine the angular distribution, they find out the ratio.

the photo-magnetic cross section is isotropic and the photo-electric cross section is follow a of a $sin^2$ distribution. the average intensity of neutron detected on a angle is:

$I(\gamma ) = \int_{\gamma_1}^{\gamma_2} {(a + b sin^2(\gamma)) sin(\gamma) d\gamma } / \int_{\gamma_1}^{\gamma_2} {sin(\gamma) d\gamma }$

where a is the contribution from the photo-magnetic interaction and b is from photo-electric interaction. and $\gamma_1$ and $\gamma_2$ are the angle span by the finite size of the target and detector. the integration is straight forward and result is:

$I(\gamma) = a+b( 1 - 1/3 ( cos^2(\gamma_1) + cos(\gamma_1) cos(\gamma_2) + cos^2 ( \gamma_2) )$

and the author guided us to use the ration of 2 angle to find the ration of a and b. and the ration of a and b is related to the probability of the magnetic to the electric effect by

$a/b = 2/3 \tau$

. and the photo-magnetic to photo-electric cross section ratio is:

$\tau/(\tau+1)$

the detector was described in detail on 4 paragraphs. basically, it is a cylindrical linear detector base on the reaction $B^{10} ( n,\alpha)Li^7$. it was surrounded by paraffin to slow down fast nuetrons.

on the target, which is heavy water, $D_2 O$, they use an extraordinary copper toriod or donut shape container. it is based on 3 principles:

• The internal scattering of neutron
• Departure from point source
• The angular opening of the γ – ray source

they place the γ – ray source along the axis of the toriod, move it along to create different scattering angle.

they tested the internal scattering of the inside the toriod and found that it is nothing, the toriod shape does not have significant internal scattering.

they test the reflection of neutron form surrounding, base on the deviation from the inverse-square law. and finally, they hang up there equipment about 27meters from the ground and 30 meters from buildings walls. (their apparatus’s size is around 2 meters. They measured 45, 60, 75 and 90 degree intensity with 5 degree angular opening for each.

## 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.

## Mass of particles and nucleus

in Nuclear physics, the particle we deal with are so small and so light, if we use standard unit, then there will be many zero and we will lost in the zeros. for example, the electron has mass:

Mass( electron ) = 9.11 × 10-31 kilograms
Mass( proton ) = 1.67 × 10-27 kilograms

see? as the special relativity give us a translation tool – E = m c^2, thus, we can use MeV to talk about mass.

Mass ( electron ) = 0.511 MeV
Mass ( proton ) = 938.3 MeV

thus, we can see, Proton is roughtly 2000 times heavier then electron ( 1000 : 0.5 ).

Mass( neutron ) = 939.6 MeV

neutron is just 1.3 MeV heavier then proton.

The nucleus is formed by proton and neutron. so, in simple thought, an nucleus with Z proton and ( A-Z ) neutron should have mass

Z x Mass( proton ) + ( A – Z ) x Mass ( neutron ) = Mass ( A, Z )

where A is the atomic mass number, which is equal the number of nucleons in the nucleus, and Z is the proton number.

However, scientists found that it is not true.

Z x Mass( proton ) + ( A – Z ) x Mass ( neutron ) > Mass ( A, Z )

Some of the mass is missing! But that is explained why nucleus will not break down automatically. since it need extra energy to break it down.

we called the mass different is Mass Deficit. or Binding energy.

Mass Deficit = Mass( A, Z) – Mass ( proton + neutron )

some one may think that the binding energy is the energy for holding the nucleus together. in order to hold the nucleus, some mass was converted into the energy to holding it. this is INCORRECT. the correct argument is, the binding energy is th energy require to break it down.

think about a simple 2 bodies system, like sun and earth. at far far away, when both of them are at rest, the total energy is Mass( sun) + Mass ( earth ) + Potential energy.

when the earth moves toward to sun, the potential energy converted to the Kinetic energy, so the earth moving faster and faster. but, in order to stay in the orbit, some K.E. must be lost so that it does not have enough ( or the same) energy to run away. Thus, the total energy of the system is lesser then the total mass.

another analogy is electron orbit. when an electron was captured by an atom, it radiate energy in order to stay in some energy level. thus, the total energy of the system again less then the total mass.

any any case, the mass of the sun and earth and electron does not change, but the potential changes to negative, thus it makes to total energy lesser.

similar idea hold for nucleus, but the potential of it are great different, because there are a Coulomb Barrier. Thus, in order to make a nucleus. we have to put so many K.E. to again this barrier, then the resultant nucleus release the Mass Deficit energy and also the input K.E..

 a scratch on the nuclear potential. there are a Coulomb Barrie. ( by wolframalpha.com)
When the nucleus is radioactive and undergoes decay. this mean, it Mass deficit is positive. thus, it will automatically break down to another nucleus until it mass deficit is negative again. during this process, the emitted particle carry K.E. which is from the potential. Not the mass for one nucleons.
Remember, Mass( nucleus ) = Mass ( protons + neutrons ) + Potential

## magic number

we knew that for some atoms are more stable that others. like He, Ne, Ar, etc, which are belonged to noble gas. the reason for they are non-reactive is, there outer most electron shell is filled out.

similar things happened in nuclei. in the shell model of nuclei, protons and neutrons just like the electrons in atom. if the outer most shell of proton or neutron is filled out, the nuclei is very stable. and we called this number of proton or neutron be MAGIC NUMBER.

the first magic number is 2. nuclei with 2 protons is more stable then others. However, if only have 2 protons, with out neutron, it is very unstable because of coulomb force. and 2 neutron also unstable, if without a proton. the only stable 2 nucleons state is deuteron. If there are 2 protons and 2 neutrons, we called this double magic number, and this nuclei, which is He is very very stable.

the list of magic number is 2, 8, 20, 28, 50, 82, 126 in theory prediction.

however, when the nuclei become heavier and heavier, the stability of nuclei in the magic number lost. to understand this. we have to know that the magic number is come from the large spin-orbital coupling term in the Hamiltonian of the nuclei. and recent research suggest that, the spin-orbital coupling may change by the number of nucleons.