Deuteron

Leave a comment

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.

Magnetic field strength and Proton escape energy

Leave a comment

The paper has talked about the strong magnetic field will trap the low energy proton. So, how is the field strength and the proton energy relationship?

the proton moving radius can be formulated by:

R = \frac {m v}{e B}

and according the special relativity

v = c \frac{p c}{E} = c \sqrt { 1- \left( \frac {m c^2}{E} \right) ^2 }

Thus,

R = \frac { m c} { e B } \sqrt { 1- \left(\frac {m c^2}{E} \right)^2 }

Sub all the constant

R = 3.129738 \frac{1}{B} \sqrt { 1- \left(\frac {938 MeV }{E} \right)^2 }[m][T]

Thus, for slow proton, say 50MeV, the radius is

R = 983.02 / B [mm][T]

even for 1MeV

R = 144.40 / B [mm][T]

but for 1keV

R = 4.57 /B [mm][T]

[Pol. p target] Changing to the small setup

Leave a comment

Motivation

In order to convert the NMR single to the absolute magnetic field strength of the polarization. the thermal polarization ( polarization at thermal equilibrium ) should be measure and used to calibrate the NMR signal.

the thermal polarization is given by the Boltzmann statistic. The excited and the ground state population is:

\frac{ N_{\uparrow}} { N_{\downarrow} } = Exp \left( - 2 \frac { \mu_p B } {k_B T } \right)

the thermal polarization is the ratio of the different of spin-up and spin-down to the total spin.

P_{thermal} (B , T ) = tanh \left( \frac{ \mu_p B }{ k_B T } \right)

where \mu=p is the proton magnetic moment.

\mu_p = g_p \mu_N = g_p \frac{e \hbar }{2 m_p} = 1.410606662 \times 10^{-26} J T^{-1}

k_B is the Boltzmann constant.

k_B = 1.3806504 \times 10^{-23} J K^{-1}

The proton magnetic moment is small, the thermal polarization can be approximated as a linear function:

P_{thermal} ( B, T) = \frac{\mu_p B}{ k_B T}

since our polarization is on T = -5 ^oC  and B = 0.05 T , thus, the thermal polarization is:

P_{thermal} (0.05, 268.15) = 1.90509 \times 10^{-7}

which is very small to be detected, or to say, the signal is smaller then the noise level.

the small system has a better sensitivity, down to 10^{-7}.

Set-Up

TO-do

  • Connect the Static field power and water cooling system
  • connect the Controler Unit
  • connect the magnetic field sweeping
  • Test Run

First experiment of 6He with a polarized proton target

Leave a comment

DOI : 10.1140/epjad/i2005-06-110-5

this paper reported a first spin polarized proton solid target under low magnetic field ( 0.08 T ) and hight temperature ( 100K )

the introduction overview the motivation of a solid target.

  • a polarized gas target is ready on many nuclear experiment.
  • on the radioactive beam ( IR beam ), the flux of a typical IR beam is small, since it is produced by 2nd scattering.
  • a solid target has highest density of solid.
    • most solid target can only be polarized on low temperature ( to avoid environmental interaction to reduced the polarization )
      • increase the experimental difficult, since a low temperature should be applied by a cold buffer gas.
    • high field ( the low gyromagnetic  ratio ).
      • high magnetic field make low energy scattered proton cannot get out from the magnetic field and not able to detect.
  • a solid target can be polarized at high temperature and low magnetic field is very useful

the material on use is a crystal of naphthalene doped with pentacene.

the procedure of polarizing the proton is :

  1. use optical pumping the polarize the electron of pentacene
    • the population of the energy states are independent of temperature and magnetic field.
  2. by Dynamic Nuclear Polarization (DNP) method  to transfer  the polarization of the electron to the proton.
    • if the polarization transfer is 100% and the relaxation time is very long. the expected polarization of proton will be 72.8%

The DNP method is archived under a constant microwave frequency with a sweeping magnetic field. when the magnetic field and  microwave frequency is coupled. the polarization transfer will take place.

the next paragraph talks about the apparatus’s size and dimension, in order to fit the scattering experiment requirements.

the polarization measurement is on a scattering experiment with 6He at 71 MeV per nucleons. By measuring the polarization asymmetry \epsilon , which is related to the yield. and it also equal to the polarization of the target P_t  times the analyzing power A_y .

\epsilon = P_t \times A_y

with a reasonable guess of the target polarization. the analyzing power of  6He was found.

the reason why the polarization-asymmetry is not equal to the analyzing power is that, the target is not 100% polarized, where the analyzing power is defined. when the polarization of the target is 100%, both are the same.

in the analysis part. it used optical model and Wood-Saxon central potential to simulate the result. And compare the result from 6He to 6Li at same energy. the root mean square of 6Li is larger then 6He. it suggest the d-α core of 6Li may responsible for that.

they cannot go further discussion due to the uncertainly on the polarization of the target.

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

Leave a comment

DOI: 10.1103/PhysRevC.82.041304

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.

Mass of particles and nucleus

Leave a comment

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

Spin

Leave a comment

( this is just a draft, not organized )

Spin is a intrinsics property of elementary particle, such as electron, proton, and even photon. Intrinsics means it is a built-in property, like mass, charge. Which extrinsic properties are speed, momentum.

Spin is a vector or tensor quality while charge and mass are scaler.

Spin can react with magnetic field, like charge reacts with electric field or mass react with force produce acceleration. Thus, spin is like a bar-magnet inside particle, counter part of charge.

The magnitude of spin is half integer or integer of reduced Planck’s constant \hbar . Particles with half integer of spin are classified as Fermion, and those with integer spin are Boson. they follow different statistic while interact together, thus, this creates different physics for different group.

we are not going to the mathematic description this time.

the effect of spin causes the magnetic moment, that’s why it react with magnetic field. the other thing that creates magnetic moment is angular momentum for charge particle, like electron orbiting around nucleus. So, both spin and angular momentum can be imagined as a little magnet, thus, they can interact, in physics, we call the interaction between spin and angular momentum is coupling. for example, spin-orbital coupling, spin-spin coupling, etc..

when the spin interact with external magnetic field, it will precess around the magnetic field with Larmor frequency. and the direction of the spin while undergoes procession can only be certain angle. for spin half, like electron or proton. there are only 2 directions, and we called it up and down.

Older Entries