The most tightly bound nucleus

Leave a comment

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.


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 }


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]

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.

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
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

Special Relativity I

Leave a comment

i just state the formula and the usage of it.

the basic equation is

E^2 = (p c)^2 + (m c^2)^2

where E is total energy, p is momentum

here we can see the advantage of using MeV as unit of mass. the equation is now further simplified by using MeV/c as momentum unit.

E^2 = p^2 + m^2

which is Pythagorean theorem!

the speed of the particle is from the formula

\beta = \frac {v}{c} = \frac {p}{E}

For example, proton mass is 940 MeV/c2, if we say an proton is moving at 94MeV, or a 94 MeV proton. we mean, the KINETIC Energy (K.E.) of proton is 94MeV. The total energy is

E = m + K.E.

Thus, a 94 MeV proton is moving at 41.7% of light speed. by using a right-angle triangle of base 10, side 11, and the hight is Square-Root 21.

another way around is, a proton at 90% speed of light, how much K.E. it has? which is around 3000 MeV or 3 GeV [Giga eV]

Size, Energy and Unit

Leave a comment

As we know, atom is very small. Lets get some idea how small it is.

Let us transform ourself, like Alice in Wonderland, i guess Alice is just 100 times smaller. but this time, we goes much much smaller, become same same as an atom. Hydrogen atom, which is the smallest one, is about 1 meter hight, and Uranium is about 2 meters. and Alice, now becomes mush better then the earth and she can touch the moon easily.

how large is proton in this scale? it is just a diameter of our hair! and electron is still much smaller, so far, we believe that it is just a point, no size. so no matter how powerful your microscope, you never see it! which also means, electron has no internal structure. Thus, you may ask the structure of proton. um… we talk this later.

When someone want to talk about physics, he has to bring up ENERGY. that make his sound professional. and we are going to do the same. However, our unit of energy is different from day usage, like Calorie, or kilo-Watt-hour (kWh). we use  MAGA-ELECTRON-VOLT (MeV). before we get some idea about these units. we should understand kilo [k] = 1000, maga [M] = 1000 X 1000.

1 Calorie, which is the energy require the raise 1 degree of 1 kg water. in a cold winter, say 10 degree. if you drink tap water of 350ml, then you lose about 9 Calorie. if you drink 350ml water at 60 degree. then, you gain, 17.5 Calorie. according to this websiteIf your body mass is 64 kg and you walk at a speed of 5.63 km/hr then you will burn approximately 4.6 Calories/minute – if you walk for one hour you will burn 60 × 4.6 = 276 Calories. well, not much help for the hot water.

1 kWh is equal to 3.6 mage joule. so, what is joule? joule is another energy unit, 1 Calorie = 4.2 kilo joule. Thus, 1 kWh = 860 Calorie = about 3 hour walk. next time when you check your electric bill, you can imagine how far you have to walk to consume this energy.

as you may see, 1 Cal can raise 1 kg water up 1 degree. how many water molecules in 1 kg of water? It is 3 × 1025 !! the world population is just 6 × 109. if this number is money, then every one is a billion billionaire!! So, you can see, each molecule only share very very little among of energy. in order to save some zero in front of 1. we use a very small unit. maga-electron-volt.

according to Einstein Energy-Mass equation: E = m c2, we found that a proton is about 940 MeV/ c2. which is to say, if we want to make a proton out of no where, we have to at least give 940MeV. um, pretty small compare to our daily life. But it is very big in nuclear physics.

from this, we can know electron mass is about 0.5 MeV/ c2.

the reason why we use “energy” unit on “mass”, is not just because Einstein tell us we can do so, but it has a practical reason. We always deal with relativity and using MeV as a unit of mass bring huge convenient, both on calculation and imagination.


Older Entries