05 | February | 2011 | Nuclear Physics 101

Stability of a nucleus ( Liquid Drop Model )

February 5, 2011

GoLuckyRyan Basic, Porperties, theory beta, Binding Energy, Coulomb, decay, neutron Leave a comment

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.

Projection theorem

February 5, 2011

GoLuckyRyan advance, Spherical Stuffs, theory angular momentum, Bohr, g-factor, magneton, orbital, spin, spin-orbital, Wigner-Eckart, Zeeman Leave a comment

The simplest way to say is:

a operator can be projected on another one, for example, The orbital angular momentum cab be projected on the total angular momentum.

$L = L\cdot J \frac {J}{j(j+1)}$

a simple application is on the Zeeman effect on spin-orbital coupling. the Hamiltonian is:

$H_B = - \mu \cdot B = - ( \mu_l L + \mu_s S ) \cdot B$

by the Wigner-Eckart theorem:

$L = L\cdot J \frac {J}{j(j+1)}$

$S = S\cdot J \frac {J}{j(j+1)}$

then the Hamiltonian becomes:

$H_B = - \frac{1}{j(j+1)} ( \mu_l (L \cdot J) + \mu_s (S \cdot J) ) J\cdot B$

and introduce the Bohr Magneton and g-factor:

$H_B = - g \mu_B J \cdot B$

$g = - \frac{1}{j(j+1)} ( g_l (L \cdot J) + g_s (S \cdot J) )$

Review on angular momentum adding.

February 5, 2011

GoLuckyRyan Basic, Spherical Stuffs, theory angular momentum Leave a comment

for general angular momentum state, it is a “ugly”combination of the orbital state and the spin state:

$\left|l,m_l,s,m_s \right> = \left|l,m_l\right>\bigotimes \left|s,m_s \right>$

the above combination only possible when:

$[L_i,S_j]=0$

OK, we start with basic result.

$L = ( L_x,L_y,L_z)$

$L^2 \left|l,m_l \right> = l(l+1) \left |s,m_s \right>$

$L_z \left|l,m_l \right> = m_l \left |s,m_s \right>$

$[L_x,L_y]= i L_z$

$L_{\pm} \left|l,m_l\right> = \sqrt{ l(l+1) - m_l(m_l\pm1)} \left|l,m_l\pm 1 \right>$

the above are basic properties shared by spin angular momentum operator.

thus, we have follow result:

$[L_z, L_{\pm}]=\pm L_{\pm}$

$[L^2,L_{\pm}]=0$

and the total angular momentum operator is

$J= L\bigotimes 1 + 1 \bigotimes S$

and the state is

$\left| j, m_j \right> = \left| l, m_l, s, m_s \right>$

$J_z \left| j, m_j \right> = m_j \left| j, m_j \right>$

by $J_z= L_z\bigotimes 1 + 1 \bigotimes S_z$

$m_j =m_l+m_s$

by same on x component and y component:

$J_{\pm} = L_{\pm} + S_{\pm}$

and by $[L_i,S_j]=0$

$[J_x,J_y ] = i J_z$

$J^2 = L^2 + S^2 + 2 L \cdot S$

To find the possible $j$ , we can use the highest state withe $J_{\pm}$ and $J_z$ . Since the highest state is always symmetric. we have to find the other orthogonal states, which are anti-symmetric.

We can express $L \cdot S$ as

$\displaystyle L \cdot S = (L_x, L_y, L_z) \cdot (S_x, S_y, S_z) = L_x S_x + L_y S_y + L_z S_z$

Using the ladder operator $L_x = (L_+ + L_-)/2, L_y = (L_+ - L_-)/2i$ ,

$\displaystyle L \cdot S = \frac{1}{2}( L_+ S_- + L_- S_+) + L_z S_z$

Thus,

$\displaystyle J^2 = L^2 + S^2 + 2 L_z S_z + L_+ S_- + L_- S_+$

The expectation value of $J^2$ for $m=l, m_s = s$ is

$\displaystyle \left<JM|J^2 |JM\right> = l(l+1) + s(s+1) + 2 l s = (l+s)(l+s+1)$

The last step used $\left<JM| L_\pm S_\mp | JM\right> = 0$ for $j_{max}$ case. So, the max $J$ is $l+s$ . From minimum value of $J$ , the $m, m_s$ must be cancel each other as much as possible. Set $M = l - s$ , And the state $\left|JM\right>$ can be

$\displaystyle \left|JM\right> = \sum_{m,m_s} C^{JM}_{lmSm_s} \left|lmSm_s\right>$

But wait, the use of Chebsch-Gordan coefficient requires the knowledge of $J$ .

Since the total number of states for $L$ and $S$ are $(2l+1)(2s+1)$ , the total number of states should be the same after coupling, so, the total number of states for $J$ is