Difference between revisions of "Forest NucPhys I"

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Advanced Nuclear Physics
 
Advanced Nuclear Physics
  
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References:
+
:References:
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+
; Introductory Nuclear Physics: Kenneth S. Krane: ISBN 9780471805533
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Krane:
 
  
 
Catalog Description:
 
Catalog Description:
Line 28: Line 27:
  
 
=[[NucPhys_I_Syllabus]]=
 
=[[NucPhys_I_Syllabus]]=
−
[http://www.iac.isu.edu/mediawiki/index.php/NucPhys_I_Syllabus Click here for Syllabus]
+
[http://wiki.iac.isu.edu/index.php/NucPhys_I_Syllabus Click here for Syllabus]
  
 
=Introduction=
 
=Introduction=
−
The interaction of charged particles (electrons and positrons) by the exchange of photons is described by a fundamental theory known as Quantum ElectroDynamics.  QED has perturbative solutions which are limited in accuracy only by the order of the perturbation you have expanded to.  As a result the theory is quite useful in describing the interactions of electrons that are prevalent in Atomic physics.
+
The interaction of charged particles (electrons and positrons) through the exchange of photons is described by a fundamental theory known as Quantum ElectroDynamics(QED).  QED has perturbative solutions which are limited in accuracy only by the order of the perturbation you have expanded to.  As a result, the theory is quite useful in describing the interactions of electrons that are prevalent in Atomic physics.
  
  
−
Nuclear physics, however, encompasses the physics of describing not only the nucleus of an Atom but also the composition of the nucleons (protons and neutrons) which are the constituent of the nucleus. Quantum ChromoDynamic (QCD) is the fundamental theory designed to describe the interactions of the quarks and gluons inside a nucleon.  Unfortunately, QCD does not have a complete solution at this time.  At very high energies, QCD can be solved perturbatively.  This is an energy <math>E</math>  at which the strong coupling constant <math>\alpha_s</math> is less than unity where  
+
Nuclear physics describes how Atomic nuclei interact via the strong forces as well as how the strong force binds the constituents of a nucleus (protons and neutrons, a.k.a. nucleons).  Particle physics studies the interactions of fundamental particles, particles without substructure like quarks, photons, and electrons.  Both Nuclear and Particle physics rely on the "Standard Model", a field theory description of the strong, weak and electromagnetic forces.   Quantum ChromoDynamic (QCD) is one component to the Standard Model which represents the fundamental theory developed to describe the interactions of the quarks and gluons inside a nucleon, analogous to how QED describes the electromagnetic forces of electrons within the atom.  The electroweak and Higgs field are the remaining components to the Standard model.   
 +
Ideally, QCD is a field theory which could be used to describe how quarks interact to for nucleons and then describe how those nucleons interact to form a nucleus and eventually lead to a description of how the nucleus interacts with other nuclei.
 +
Unfortunately, QCD does not have a complete solution at this time.  At very high energies, QCD can be solved perturbatively.  This is an energy <math>E</math>  at which the strong coupling constant <math>\alpha_s</math> is less than unity where  
 
:<math>\alpha_s \approx \frac{1}{\beta_o \ln{\frac{E^2}{\Lambda^2_{QCD}}}}</math>
 
:<math>\alpha_s \approx \frac{1}{\beta_o \ln{\frac{E^2}{\Lambda^2_{QCD}}}}</math>
 
:<math>\Lambda_{QCD} \approx 200 MeV</math>
 
:<math>\Lambda_{QCD} \approx 200 MeV</math>
  
−
The "Standard Model" in physics is the grouping of QCD with Quantum ElectroWeak theory.  Quantum ElectroWeak theory is the combination of Quantum ElectroDynamics with the weak force; the exchange of photons, W-, and Z-bosons.
 
  
 
The objectives in this class will be to discuss the basic aspects of the nuclear phenomenological models used to describe the nucleus of an atom in the absence of a QCD solution.
 
The objectives in this class will be to discuss the basic aspects of the nuclear phenomenological models used to describe the nucleus of an atom in the absence of a QCD solution.
Line 109: Line 109:
  
 
==Nuclear Properties==
 
==Nuclear Properties==
 +
 +
[[NuclearProperties_Forest_NucPhys_I]]
  
 
The nucleus of an atom has such properties as spin, mangetic dipole and electric quadrupole moments.  Nuclides also  have stable and unstable states.  Unstable nuclides are characterized by their decay mode and half lives.
 
The nucleus of an atom has such properties as spin, mangetic dipole and electric quadrupole moments.  Nuclides also  have stable and unstable states.  Unstable nuclides are characterized by their decay mode and half lives.
Line 184: Line 186:
 
[http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_6/4_6_1.html National Physical Lab (UK)]
 
[http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_6/4_6_1.html National Physical Lab (UK)]
  
−
=Quantum Mechanics Review=
+
[http://ie.lbl.gov/education/isotopes.htm  Table of Isotopes at Lawrence Berkeley National Laboratory]
−
*Debroglie - wave particle duality
 
−
{| border="1"  |cellpadding="20" cellspacing="0
 
−
|-
 
−
| Particle || Wave
 
−
|-
 
−
| <math>E</math> || <math>\hbar \omega = h \nu</math>
 
−
|-
 
−
| <math>P</math> || <math>\hbar k = \frac{h}{\lambda}</math>
 
−
|}
 
−
* Heisenberg uncertainty relationship
 
−
:<math>\Delta x \Delta p_x \ge \frac{\hbar}{2}</math>
 
−
:<math>\Delta E \Delta t \ge \frac{\hbar}{2}</math>
 
−
:<math>\Delta \ell_z \Delta \phi \ge \frac{\hbar}{2}</math> where <math>\phi</math> characterizes the location of <math>\ell</math> in the x-y plane
 
−
 
 
−
* Energy conservation
 
−
:Classical: <math>\frac{p^2}{2m} + V(r) = E</math>
 
−
:Quantum (Shrodinger Equation): <math>-\frac{\hbar^2}{2m}\nabla^2 \Psi + V(r) \Psi = i\hbar \frac{\partial \Psi}{\partial t}</math>
 
−
: <math>\;\;\;\;\;\; p_x \rightarrow -i \hbar \frac{\partial}{\partial x} \;\;\; E \rightarrow i \hbar \frac{\partial}{\partial t}</math>
 
−
*Quantum interpretations
 
−
:E = energy eigenvalues
 
−
:<math>\Psi(x,t) = \psi(x)e^{-\omega t}</math> = eigenvectors <math>\omega=\frac{E}{\hbar}</math>
 
−
:<math>P = \int_{x_1}^{x_2}\Psi^*(x,t) \Psi(x,t)</math> = probability of finding the particle (wave packet) between <math>x_1</math> and <math>x_2</math>
 
−
:<math>\Psi^*</math> = complex conjugate <math>(i \rightarrow -i)</math>
 
−
:<math><f> = \int \Psi^* f \Psi dx =<\Psi^*| f| \Psi></math> = average (expectation) value of observable <math>f</math> after many measurements of <math>f</math>
 
−
:example: <math><p_x> =\int \Psi^* \left ( -i\hbar \frac{\partial}{\partial x}\right ) \Psi dx</math>
 
  
−
*Constraints on Quantum solutions
+
[http://www.nea.fr/janis/ French Nuclear data Java GUI]
−
#<math>\psi</math> is continuous accross a boundary : <math>\lim_{\epsilon \rightarrow 0} \left [ \psi(a+\epsilon) - \psi(a-\epsilon)\right ] =0</math> and <math>\lim_{\epsilon \rightarrow 0} \left [ \left(\frac{\partial \psi}{\partial x} \right )_{a+\epsilon} - \left(\frac{\partial \psi}{\partial x} \right )_{a-\epsilon}\right ] =0</math> ( if <math>V(x)</math> is infinite this second condition can be violated)
 
−
#the solution is normalized:<math>\int \psi^* \psi dx =<\psi^* | \psi >=1</math>
 
  
−
* Current conservation: the particle current density associated with the wave function \Psi is given by
+
=Quantum Mechanics Review=
−
:<math>j = \frac{\hbar}{2mi} \left ( \Psi^* \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^*}{\partial x}\right )</math>
 
−
== Schrodinger Equation ==
 
−
=== 1-D problems===
 
−
====Free particle====
 
  
−
If there is no potential field (V(x) =0) then the particle/wave packet is free.  The wave function is calculated using the time-dependent Schrodinger equation:
+
[[Quantum_Mechanics_Review_Forest_NucPhys_I]]
  
−
:<math>-\frac{\hbar^2}{2m} \nabla^2 \Psi(x,t) + 0 = i\hbar\frac{\partial \Psi(x,t)}{\partial t}</math>
+
= Nuclear Properties=
  
−
Using separation of variables Let:
+
[[NuclearProperties_Forest_NucPhys_I]]
−
: <math>\Psi(x,t) = \psi(x) f(t)</math>
 
  
−
Substituting we have
+
= The Nuclear Force=
  
−
:<math>-\frac{\hbar^2}{2m} f(t) \frac{d^2 \psi(x)}{d x^2}  = i \hbar \psi(x) \frac{d f(t)}{d t}</math>
+
[[NuclearForce_Forest_NucPhys_I]]
−
 
 
−
reorganizing you can move all functions of <math>x</math> on one side and <math>t</math> on the other suggesting that both sides equal some constant which we will call <math>E</math>
 
−
 
 
−
: <math>-\frac{\hbar^2}{2m} \frac{1}{\psi(x)} \frac{d^2 \psi(x)}{d x^2}  = i \hbar \frac{1}{f(t)} \frac{d f(t)}{d t} \equiv E</math>
 
  
 +
= Nuclear Models=
  
−
Solving the temporal (t) part:
+
Given the basic elements of the nuclear potential from the last chapter, one may be tempted to construct the hamiltonian for a group of interacting nucleons in the form
  
−
: <math>\frac{1}{f(t)} \frac{d f(t)}{d t} = -\frac{i E}{\hbar} \Rightarrow f(t) = e^{-iEt/\hbar}</math> : just integrate this first order diff eq.
+
:<math>H = \sum_i^A T_i + \sum_i<j^A V_{ij}</math>
−
 
 
−
Solving the spatial (x) part:
 
−
: <math>\frac{1}{\psi(x)} \frac{d^2 \psi(x)}{d x^2}  =- \frac{2m E}{\hbar^2}</math>
 
−
 
 
−
Such second order differential equations have general solutions of
 
−
 
 
−
: <math>\psi(x) = A e^{ikx} + B e^{-ikx}</math> where <math>k^2 = \frac{2mE}{\hbar^2}</math>
 
−
 
 
−
Now put everything together
 
−
 
 
−
:<math>\Psi(x,t) = \psi(x) f(t) = \left (A e^{ikx} + B e^{-ikx} \right ) e^{-iEt/\hbar}</math>
 
−
:<math>= A e^{i(kx-\omega t)} + B e^{-i(kx-\omega t)}</math>
 
−
 
 
−
;Notice
 
−
:<math><\Psi(x,t) | \Psi(x,t)> = <\psi(x)| \psi(x)> \Rightarrow  </math>the wave function amplitude does not change with time
 
−
;also, if the operator for an observable A does not change in time, then
 
−
:<math><\Psi(x,t) | A | \Psi(x,t)> = <\psi(x)| A| \psi(x)> \Rightarrow  </math> even though particles are not stationary they are in a quantum state which does not change with time (unlike decays).
 
−
: the term of amplitude<math> A</math> represents a wave traveling in the +x direction while the second term represents a wave traveling in the -x direction. 
 
−
 
 
−
;Example
 
−
:consider a free particle traveling in the +x direction
 
−
: Then
 
−
: <math>\Psi(x,t) = A e^{i(kx-\omega t)}</math>
 
−
: if the particles are coming from a source at a rate of<math> j</math> particles/sec then
 
−
:<math>j = \frac{\hbar}{2mi} \left ( \Psi^* \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^*}{\partial x}\right )</math>
 
−
: <math>= \frac{\hbar}{2mi} \left ( A^*A[ik] - A A^* [-ik]\right ) = \frac{\hbar k}{m} \left ( A^*A\right )= \frac{\hbar k}{m} \left | A \right |^2</math>
 
−
: <math>\Rightarrow A = \sqrt{\frac{mj}{\hbar k}}</math>
 
−
 
 
−
====Step Potential====
 
−
Consider a 1-D quantum problem with the Step potential V(x) define below where <math>V_o >0</math>
 
−
:<math>V(x) =\left \{  {0 \;\;\;\; x <0 \atop V_o \;\;\;\; x>0} \right .</math>
 
−
 
 
−
 
 
−
Break these types of problems into regions according to how the potential is defined.  In this case there will be 2 regions
 
−
=====x<0=====
 
−
 
 
−
When x<0 then V =0 and we have a free particle system which has the solution given above.
 
−
 
 
−
:<math>\Psi_1(x,t) = \psi_1(x) f(t) = \left (A e^{ikx} + B e^{-ikx} \right ) e^{-iEt/\hbar}</math>
 
−
:<math>= A e^{i(kx-\omega t)} + B e^{-i(kx-\omega t)}</math> where <math>k^2 =\frac{2mE}{\hbar^2}</math> and  <math>\omega =\frac{E}{\hbar}</math>
 
−
 
 
−
=====x>0=====
 
−
:<math>-\frac{\hbar^2}{2m} \nabla^2  \Psi_2(x,t) + V_o = i\hbar\frac{\partial \Psi_2(x,t)}{\partial t}</math>
 
−
:<math>-\frac{\hbar^2}{2m} \frac{\partial^2 \Psi_2(x,t)}{\partial x^2} + V_o = i\hbar\frac{\partial \Psi_2(x,t)}{\partial t}</math>
 
−
 
 
−
separation of variables: <math>\Psi_2(x,t) = \psi_2(x)f(t)</math>
 
−
:<math>-\frac{\hbar^2}{2m} f(t)\frac{\partial^2 \psi_2(x)}{\partial x^2} + V_o\psi(x)f(t) = i\hbar \psi_2(x)\frac{\partial f(t)}{\partial t}</math>
 
−
:<math>-\frac{\hbar^2}{2m} \frac{1}{\psi_2(x)}\frac{\partial^2 \psi_2(x)}{\partial x^2} + V_o = i\hbar \frac{1}{f(t)}\frac{\partial f(t)}{\partial t} \equiv E =</math> Constant
 
−
 
 
−
The time dependent part of the problem is the same as the free particle solution.  Only the spatial part changes because the Potential is not time dependent.
 
−
 
 
−
:<math>-\frac{\hbar^2}{2m} \frac{1}{\psi_2(x)}\frac{\partial^2 \psi_2(x)}{\partial x^2}  =E - V_o = </math> Constant
 
−
:<math> \frac{\partial^2 \psi_2(x)}{\partial x^2}  =-\frac{2m}{\hbar^2}(E - V_o) \psi_2(x) </math>
 
−
 
 
−
If <math>E > V_o</math> then we have a wave that traverses the step potential partly reflected and partly transmitted, otherwise it will be reflected back and the part that is transmitted will tunnel through the barrier attenuated exponentially for x>0.
 
−
 
 
−
Here is how it works out mathematically
 
−
 
 
−
=====<math>E > V_o</math>=====
 
−
For the case where <math>E > V_o</math>:
 
−
:<math> \frac{\partial^2 \psi_2(x)}{\partial x^2}  =-\frac{2m}{\hbar^2}(E - V_o) \psi_2(x)  \equiv -k_2^2 \psi_2(x)< 0 </math>
 
−
:<math> \frac{\partial^2 \psi_2(x)}{\partial x^2}  \equiv -k_2^2 \psi_2(x)< 0 \Rightarrow</math>SHM solutions
 
−
 
 
−
The above Diff. Eq. is the same form as the free particle but with a different constant
 
−
 
 
−
;Let
 
−
:<math>\psi_2(x) = Ce^{ik_2x} + De^{-ik_2x}</math>
 
−
 
 
−
Now apply Boundary conditions:
 
−
:<math>\psi(x=0) = \psi_2(x=0)</math>
 
−
: <math>A + B = C + D : e^{\pm i 0} = 1</math>
 
−
and
 
−
:<math>\frac{\partial \psi}{\partial x}|_{x=0} = \frac{\partial \psi_2}{\partial x}|_{x=0}</math>
 
−
:<math>k(A-B) = k_2 (C-D)</math>
 
−
 
 
−
We now have a system of 2 equations and 4 unknowns which we can't solve.
 
−
 
 
−
;Notice
 
−
: The coefficient "D" in the above system represent the component of <math>\psi_2</math> represent a wave moving from the right towards x=0.    If we assume the free particle encountered this step potential by originating from the left side, then there is no way we can have a component of <math>\psi_2</math> moving to the left.  Therefore we set <math>D=0</math>.
 
−
 
 
−
:The coefficient A represent the incident plane wave on the barrier.  The remaining coefficients B and C represent the reflected and transmitted components of the traveling wave, respectively.
 
−
 
 
−
;Know our system of equations is:
 
−
: <math>A+B =C</math>
 
−
:<math>A -B = \frac{k_2}{k} C</math>
 
−
 
 
−
;If I assume that the coefficient A is known (I know what the amplitude of the incoming wave is) then I can solve the above system such that
 
−
: <math>A+B = C = (A-B)\frac{k}{k_2}</math>
 
−
:<math>B=A\frac{1-\frac{k_2}{k}}{1+\frac{k_2}{k}}</math>
 
−
similarly
 
−
: <math>C = A+B = A\left (1+ \frac{1-\frac{k_2}{k}}{1+\frac{k_2}{k}} \right ) = \frac{2A}{1+\frac{k_2}{k}}</math>
 
−
======Reflection (R) and Transmission (T) Coefficients======
 
−
 
 
−
:<math>R\equiv \frac{j_{reflected}}{j_{incident}} = \frac{|B|^2}{|A|^2} = \left ( \frac{1-\frac{k_2}{k}}{1+\frac{k_2}{k}} \right )^2</math>
 
−
 
 
−
:<math>T\equiv \frac{j_{transmitted}}{j_{incident}} = \frac{C^*ik_2C + Cik_2C^*}{A^*ikA + AikA^*} = \frac{k_2 |C|^2}{k |A|^2} =\frac{4 \frac{k_2}{k}}{\left ( 1 + \frac{k_2}{k} \right )^2} </math>
 
−
:<math>=1 - R =1-\left ( \frac{1-\frac{k_2}{k}}{1+\frac{k_2}{k}} \right )^2 =\frac{ \left( 1+\frac{k_2}{k} \right )^2 -  \left ( 1-\frac{k_2}{k}\right )^2}{\left ( 1+\frac{k_2}{k}\right )^2}=\frac{4 \frac{k_2}{k}}{\left ( 1 + \frac{k_2}{k} \right )^2} </math>
 
−
 
 
−
=====<math>E < V_o</math>=====
 
−
:<math> \frac{\partial^2 \psi_2(x)}{\partial x^2}  =-\frac{2m}{\hbar^2}(E - V_o) \psi_2(x) >0</math>
 
−
 
 
−
;Let : <math>k_3 \equiv \sqrt{\frac{2m}{\hbar^2}(V_o-E)}</math>
 
−
 
 
−
Then
 
−
 
 
−
:<math> \frac{\partial^2 \psi_3(x)}{\partial x^2}  =k_3 \psi_3(x) >0 \Rightarrow</math> exponential decay
 
−
 
 
−
;Assume solution:
 
−
: <math>\psi_3 = G e^{k_3x} + Fe^{-k_3x}</math>
 
−
 
 
−
;Recall the solution for x<0
 
−
:<math>\psi_1(x,t) = A e^{ikx} + B e^{-ikx}</math> where <math>k^2 =\frac{2mE}{\hbar^2}</math>
 
−
 
 
−
;Apply Boundary conditions
 
−
 
 
−
If <math>x \rightarrow \infty</math>
 
−
 
 
−
Then <math>e^{\infty} \rightarrow \infty \Rightarrow G =0</math>
 
−
 
 
−
: <math>\psi_3 =  Fe^{-k_3x}</math>
 
−
 
 
−
;Continuous conditions at x=0
 
−
:<math>A+B = F</math>
 
−
:<math>ik(A-B) = -k_3F</math>
 
−
 
 
−
Assuming A is known we have 2 equations and 2 unknowns again
 
−
 
 
−
:<math>A+B = \frac{ik}{-k_3} (A-B)</math>
 
−
:<math>B=-A\left(\frac{1+\frac{ik}{k_3}}{1-\frac{ik}{k_3}} \right) = A\left(\frac{1-\frac{ik_3}{k}}{1+\frac{ik_3}{k}} \right)</math>
 
−
: <math>F = A\left (1-\left(\frac{1-\frac{ik_3}{k}}{1+\frac{ik_3}{k}}\right) \right)=\frac{2A}{1+\frac{ik_3}{k}}</math>
 
−
======Reflection (R) and Transmission (T) Coefficients=======
 
−
 
 
−
:<math>R\equiv \frac{j_{reflected}}{j_{incident}} = \frac{|B|^2}{|A|^2} = \left(\frac{1-\frac{ik_3}{k}}{1+\frac{ik_3}{k}} \right)\left(\frac{1-\frac{ik_3}{k}}{1+\frac{ik_3}{k}} \right)^*</math>
 
−
:<math>=\left(\frac{1-\frac{ik_3}{k}}{1+\frac{ik_3}{k}} \right)\left(\frac{1-\frac{-ik_3}{k}}{1+\frac{-ik_3}{k}} \right) = 1</math>
 
−
:<math>T\equiv \frac{j_{transmitted}}{j_{incident}} = \frac{F^*k_3F - Fk_3F^*}{A^*ikA + AikA^*} =0 =1-R </math>
 
−
 
 
−
;Evanescent waves:: Waves like <math>\psi_3</math> which carry no current. There is a finite probability of penetrating the barrier (tunneling) but no net current is transmitted. A feature which separates Quantum mechanics from classical.
 
−
 
 
−
====Rectangular Barrier Potential====
 
−
Barrier potentials are 1-D step potentials of height (V_o > 0) which have a finite step width:
 
−
 
 
−
: <math>V(x) =0 \;\;\; x<0</math>
 
−
:<math>V(x) =\left \{  {V_o \;\;\;\; 0 \le x \le a \atop 0 \;\;\;\; x>a} \right .</math>
 
−
 
 
−
We now have 3 regions in space to solve the schrodinger equation
 
−
 
 
−
We now from the free particle solutions that on the left and right side of the barrier we should have
 
−
 
 
−
 
 
−
 
 
−
:<math>\psi_1 = = A e^{ikx} + B e^{-ikx)} \;\;\; x <0</math>
 
−
:<math>\psi_3 = = F e^{ikx} + G e^{-ikx)} \;\;\; x > a</math>
 
  
 
where
 
where
  
−
:<math>k^2= \frac{2mE}{\hbar^2}</math>
+
:<math>T_i</math> represent the kinetic energy of the ith nucleon
 +
:<math>V_{ij}</math> represents the potential energy between two nucleons.
  
−
But in the region <math>0 \le x \le a</math> we have the save type of problem as the step in which the solution depends on the Energy of the system with respect to the potential.  One solution for the <math>E>V_o</math> (oscilatory) system and one for the <math>E<V_o</math> (exponetial decay) system.
+
If you assume that the nuclear force is a two body force such that the force between any two nucleons doesn't change with the addition of more nucleons,
 +
Then you can solve the Schrodinger equation corresponding to the above Hamiltonian for A<5.
  
−
:<math>\psi_2 = \left \{  {= Ce^{ik_2x} + De^{-ik_2x} \;\;\;\; E> V_o  \atop = Ce^{k_3x} + De^{-k_3x} \;\;\;\; E < V_o } \right .</math>
+
For A< 8 there is a technique called Green's function monte carlo which reportedly finds solution that are nearly exact.
−
where
+
J. Carlson, Phys. Rev. C 36, 2026 - 2033 (1987), B. Pudliner, et. al., Phys. Rev. Lett. 74, 4396 - 4399 (1995)
−
:<math>k_2^2=\frac{2m(E-V_o)}{\hbar^2}\;\;\; k_3^2=\frac{2m(V_o-E)}{\hbar^2}</math>
 
  
−
=====<math>E > V_o</math>=====
+
==Shell Model==
−
For the case where <math>E > V_o</math>:
 
  
−
Before we set <math>D =0</math> because there wasn't a wave moving to the left towards the <math>x=0</math> interface.  The rectangular barrier though could have a wave reflect back form the <math>x=a</math> interface.
+
===Independent particle model===
  
−
;Apply Boundary conditions:
+
This part of the Shell model suggests that the properties of a nucleus with only one unpaired nucleon are determined by that one unpaired nucleon. The unpaired nucleon usually, though no necessarily, occupies the outer most shell as a valence nucleon.
−
:<math>\psi_1(x=0) = \psi_2(x=0)</math>
 
−
: <math>A + B = C + D : </math>
 
−
and
 
−
:<math>\psi_2(x=a) = \psi_3(x=a)</math>
 
−
: <math>Ce^{ik_2a} + De^{-ik_2a} = Fe^{ika} + Ge^{-ika} </math>
 
−
and
 
−
:<math>\frac{\partial \psi_1}{\partial x}|_{x=0} = \frac{\partial \psi_2}{\partial x}|_{x=0}</math>
 
−
:<math>k(A-B) = k_2 (C-D)</math>
 
−
and
 
−
:<math>\frac{\partial \psi_2}{\partial x}|_{x=a} = \frac{\partial \psi_3}{\partial x}|_{x=a}</math>
 
−
:<math>k_2(Ce^{ik_2a}-De^{-ik_2a}) = k (Fe^{ika}-Ge^{-ika})</math>
 
  
−
We now have a system of 4 equations
+
=== SN-130 Example===
−
and 6 unknowns (A,B,C, D, F and G).
 
  
−
But:  
+
The low lying excited energy states for Sn-130 taken from the  [http://ie.lbl.gov/ensdf/ LBL website] are given below.
  
−
:<math>G=0</math> : no source for wave moving to left when x>a
+
[[Image:Sn-130_LowLyingE_Levels.tiff | 200 px]]
  
−
If we treat <math>A</math> as being known (you know the incident wave amplitude) then we have 4 unknowns (B,C,D, and F) and the 4 equations:
 
  
−
: <math>A + B = C + D : </math>
+
The listing indicates that the ground state of Sn-130 is a spin 0 positive parity <math>(J^{\pi} = 0^+)</math> state.  The first excited state of this nucleus is 1.22 MeV above the ground state and has <math>(J^{\pi} = 2^+)</math>.  The next excited state is 1.95 MeV above the ground state and has <math>(J^{\pi} = 7^-)</math>.
−
:<math>k(A-B) = k_2 (C-D)</math>
 
−
: <math>Ce^{ik_2a} + De^{-ik_2a} = Fe^{ika}  : </math>
 
−
:<math>k_2(Ce^{ik_2a}-De^{-ik_2a}) = k Fe^{ika}</math>
 
  
 +
Let's see how well the shell model does at predicting these <math>(J^{\pi} )</math> states
  
−
======Transmission======
+
==Liquid Drop Model==
−
:<math>T \equiv \frac{|F|^2}{|A|^2}</math> = the transmission coefficient
 
  
−
To find the ration of F to A <br>
+
Bohr and Mottelson considered the nucleon in terms of its collective motion with vibrations and rotations that resembled a suspended drop of liquid.
−
#solve the last 2 equations for C & D in terms of F<br>
 
−
#solve the first 2 equations for A in terms  C and D<br>
 
−
# 3.)substitute your values for C and D from the last 2 equations so you have the ratio of B/A in terms of F/A
 
  
 +
== Electric Quadrupole Moment==
  
−
; 1.)solve the last 2 equations
+
[[Electric_QuadrupoleMoment_Forest_NuclPhys_I]]
−
: <math>Ce^{ik_2a} + De^{-ik_2a} = Fe^{ika}  : </math>
 
−
:<math>Ce^{ik_2a}-De^{-ik_2a} = \frac{k}{k_2} Fe^{ika}</math>
 
−
for C and D
 
−
:<math>2Ce^{ik_2a} =Fe^{ika}\left ( 1+\frac{k}{k_2} \right)</math>
 
−
:<math>2De^{-ik_2a} =Fe^{ika}\left ( 1-\frac{k}{k_2} \right)</math>
 
  
−
;2.) solve the first 2 equations for B in terms of C & D
 
  
−
: <math>A + B = C + D : </math>
 
−
:<math>A-B = \frac{k_2}{k} (C-D)</math>
 
  
−
for A in terms of C and D
+
=Nuclear Decay =
  
−
:<math>2B=C \left ( 1- \frac{k_2}{k}\right ) + D\left ( 1+ \frac{k_2}{k}\right )</math>
+
[[Nuclear_Decay_Forest_NucPhys_I]]
−
:<math>=\frac{F}{2}e^{i(k-k_2)a}\left ( 1+\frac{k}{k_2}\right ) \left ( 1- \frac{k_2}{k}\right ) +\frac{F}{2}e^{i(k+k_2)a}\left ( 1-\frac{k}{k_2} \right)\left ( 1+ \frac{k_2}{k}\right )</math>
 
−
:<math>=\frac{F}{2}e^{i(k-k_2)a}\left (  \frac{k}{k_2}-\frac{k_2}{k}\right ) +\frac{F}{2}e^{i(k+k_2)a}\left ( - \frac{k}{k_2}+\frac{k_2}{k}\right )</math>
 
−
:<math>\Rightarrow \frac{B}{A} = \frac{Fe^{ika}}{4A}\left[ \left ( e^{-ik_2a} -e^{ik_2a}\right ) \frac{k}{k_2} +\left ( -e^{-ik_2a} -e^{ik_2a}\right ) \frac{k_2}{k}  \right ]</math>
 
−
:<math>= -\frac{Fe^{ika}}{4A} \left [ 2i\sin(k_2a) \frac{k}{k_2} -2 i\sin(k_2a)\frac{k_2}{k}\right ]</math>
 
  
 +
=Nuclear  Reactions=
  
−
;3.) Find Reflection Coeff in terms of Transmission Coeff
+
[[Forest_NucPhys_I_Nuclear_Reactions]]
−
:<math>\frac{B}{A}=- \frac{F}{A}\frac{ie^{ika}\sin(k_2a)}{2} \left [ \frac{k^2-k_2^2}{kk_2} \right ]</math>
 
  
−
: <math>T +R = \frac{|F|^2}{|A|^2} + \frac{|B|^2}{|A|^2} = \frac{|F|^2}{|A|^2} + \frac{F^*}{A^*}\frac{-ie^{-ika}\sin(k_2a)}{2} \left [ \frac{k^2-k_2^2}{kk_2} \right ]\frac{F}{A}\frac{ie^{ika}\sin(k_2a)}{2} \left [ \frac{k^2-k_2^2}{kk_2} \right ] = 1</math>
+
=Electro Magnetic Interactions=
−
:<math>\Rightarrow \frac{|F|^2}{|A|^2} \left (1 + \frac{\sin^2(k_2a)}{4}\left [ \frac{k^2-k_2^2}{kk_2} \right ]^2 \right) = 1</math>
 
  
−
or
 
  
−
:<math>T = \frac{|F|^2}{|A|^2} = \frac{1}{\left (1 + \frac{\sin^2(k_2a)}{4}\left [ \frac{k^2-k_2^2}{kk_2} \right ]^2 \right) }</math>
+
[[TF_DerivationOfCoulombForce]]
  
−
since
+
=Weak Interactions=
−
:<math>k^2= \frac{2mE}{\hbar^2} \;\;k_2^2= \frac{2m(E-V_o)}{\hbar^2}</math>
 
  
−
Then
+
Neutrino
  
−
:<math>T = \frac{|F|^2}{|A|^2} = \frac{1}{\left (1 + \frac{\sin^2(k_2a)}{4}\left [ \frac{V_o^2}{E(E-V_o)} \right ] \right) }</math>
+
:<math>\nu_e + {40 \atop 18 }Ar_{22} \rightarrow {40 \atop 19 }K^*_{21} + e^- </math>
  
 +
The min neutrino energy needed for this reaction assuming the electron energy is ignorable:
  
−
=== 3-D problems===
+
:<math>\Delta M = \left [ m\left({40 \atop 19 }K\right)- m\left({40 \atop 18 }Ar\right) \right] c^2 </math>
 +
: <math>= \left [ 39.96399848 - 39.9623831225 \right] 931.502 \frac{\mbox{MeV}}{\mbox {u}}  </math>
 +
:<math>= 0.00162 931.502 \frac{\mbox{MeV}}{\mbox {u}} = 1.505 MeV</math>
  
−
==== Infinite Spherical Well ====
+
A reaction to the ground state of potassium is forbidden in this charged current electron neutrino reaction resulting in one or more gammas being emitted  as the Potassium nucleus de-excited to the ground state.  There are about 98 observed excited states of the K-40 nucleus.
  
−
What is the solution to Schrodinger's equation for a potential V which only depends on the radial distance (r) from the origin of a coordinate system?
+
Detecting this signal in DUNE means you need to see and electron and at least one gamma in coincidence. 
  
−
:<math>V =\left \{  {0 \;\;\;\; r<a \atop \infty  \;\;\;\; r>a} \right .</math>
+
Potassium 40 decay
  
−
Such a potential lends itself to the use of a Spherical coordinate system in which the schrodinger equation has the form
+
89% of the time
 +
:<math>{40 \atop 19 }K_{21} \rightarrow {40 \atop 20 }Ca_{20} + \beta (E_{max} = 1.3 MeV) + \bar{\nu}</math>
  
−
:<math>\hat{H}\psi(r,\theta,\phi) = E\psi(r,\theta,\phi)</math>
+
10.72% of the time
−
: <math>-\frac{\hbar^2}{2m}\nabla^2 \psi(r,\theta,\phi)+V\psi(r,\theta,\phi) = E\psi(r,\theta,\phi)</math>
 
  
−
In spherical coordinates
+
:<math>{40 \atop 19 }K_{21} \rightarrow {40 \atop 18 }Ar_{22} + \gamma (E = 1.46 MeV) + \bar{\nu}</math>
  
−
:<math>\nabla^2 = \frac{1}{r} \frac{\partial^2}{\partial r^2} r + \frac{1}{r^2} \left ( \frac{1}{\sin(\theta)} \frac{\partial}{\partial \theta} \sin(\theta)\frac{\partial}{\partial \theta} + \frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial \phi^2}\right )</math>
+
0.001% of the time
  
−
;Note
+
:<math>{40 \atop 19 }K_{21} \rightarrow {40 \atop 20 }Ca_{20} + e^+ + \nu</math>
−
:<math>\frac{1}{r} \frac{\partial^2}{\partial r^2}  r = \frac{1}{r} \frac{\partial}{\partial r} r \left ( \frac{1}{r} \frac{\partial}{\partial r} \right )  = \left (\frac{1}{r} \frac{\partial}{\partial r}  \right)^2 \equiv -\left ( \frac{\hat{p}_r}{\hbar} \right )^2</math>
 
−
: <math>\left ( \frac{1}{\sin(\theta)} \frac{\partial}{\partial \theta} \sin(\theta)\frac{\partial}{\partial \theta} + \frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial \phi^2}\right ) \equiv -\frac{\hat{L}^2}{\hbar^2}</math>
 
  
−
so
+
Potassium has half life of <math>1.251×10^9</math> years.
  
−
:<math>\hat{H}\psi(r,\theta,\phi) = \left ( \frac{\hat{p}_r^2}{2m} + \frac{\hat{L}^2}{2mr^2} + V \right ) \psi(r,\theta,\phi)= E\psi(r,\theta,\phi)</math>
+
Potassium-Argon dating measures the amount of argon trapped in rock.  Originally molten rock has no argon.  After molten rock solidifies, the argon from decaying Potassium is trapped.    In primordial material, Argon-39 is dominant.  Argon-40 dominates in the earths atmosphere.
  
−
Using separation of variables:
 
  
−
:<math>\psi(r,\theta,\phi) \equiv R(r) \Theta(\theta) \Phi(\phi)</math>
+
Argon-36,38,40 are stable.
  
−
which we can also write as
+
Argon-39 is the longest lived isotope with a half life of 269 years
  
−
:<math>\psi(r,\theta,\phi) \equiv R(r) Y_{l,m}(\theta, \phi)</math>
+
Argon-39 decay:
  
−
where
+
:<math>{39 \atop 18 }Ar_{21} \rightarrow {39 \atop 19 }K_{20} + e^- + \bar{\nu}</math>
  
−
:<math>Y_{l,m}(\theta, \phi) \equiv \Theta(\theta) \Phi(\phi)</math>
+
=Strong Interaction=
  
−
Substitute
+
=Applications=
  
−
: <math>\frac{1}{2mR(r)} \hat{p}_r^2 R(r)+ \frac{1}{2mr^2Y_{l,m}} \hat{L}^2 Y_{l,m}= E-V</math>
+
=Homework problems=
  
−
=====V=0 =====
+
[[NucPhys_I_HomeworkProblems]]
−
 
 
−
We have a constant on the right hand side so the left hand side must also be constant
 
−
 
 
−
:<math>\frac{1}{2mr^2Y_{l,m}} \hat{L}^2 Y_{l,m} = \frac{l(l+1)\hbar^2}{2mr^2} =</math>a "centrifugal" barrier which keeps particles away from r=0
 
−
 
 
−
substituting
 
−
:<math>\frac{1}{2mR(r)} \hat{p}_r ^2R(r) + \frac{l(l+1)\hbar^2}{2mr^2} = E-V</math>
 
−
 
 
−
 
 
−
In the region where V=0
 
−
 
 
−
:<math>\frac{1}{\hbar^2R(r)}\hat{p}_r^2 R(r) + \frac{l(l+1)}{r^2} = \frac{2m}{\hbar^2}E</math>
 
−
 
 
−
The Radial equation becomes
 
−
 
 
−
:<math>\left ( \frac{\hat{p}_r^2}{\hbar^2}+ \frac{l(l+1)}{r^2} \right ) R(r)= \left ( \frac{-1}{r} \frac{\partial^2}{\partial r^2} r + \frac{l(l+1)}{r^2} \right ) R(r)=\frac{2mE}{\hbar^2}R(r)</math>
 
−
 
 
−
Let
 
−
 
 
−
:<math>k^2 = \frac{2mE}{\hbar^2}</math>
 
−
 
 
−
Then we have the "spherical Bessel"differential equation with the solutions:
 
−
 
 
−
:<math>j_l(kr) = \left (\frac{-r}{k} \right ) ^l \left (\frac{1}{r} \frac{d}{dr} \right )^l j_o (kr)</math>
 
−
 
 
−
where
 
−
 
 
−
:<math>j_o(kr) = \frac{sin(kr)}{kr}</math>
 
−
 
 
−
=====<math>Y_{l,m}</math> and <math> j_l</math> Table =====
 
−
{| border="1"  |cellpadding="20" cellspacing="0
 
−
|-
 
−
| <math>l</math> ||<math> m_l </math> ||<math> j_l</math> || <math>Y_{l,m_l}</math>
 
−
|-
 
−
| 0 || 0 || <math>\frac{\sin(kr)}{kr} = \frac{1}{2ikr} \left ( e^{ikr}-e^{-ikr} \right)</math> || <math>\sqrt{\frac{1}{4 \pi}}</math>
 
−
|-
 
−
| 1 || 0 || <math>\frac{\sin(kr)}{(kr)^2} -\frac{\cos(kr)}{kr}</math> || <math>\sqrt{\frac{3}{4 \pi}}\cos(\theta)</math>
 
−
|-
 
−
|  || <math>\pm</math> 1 ||  || <math>\mp \sqrt{\frac{3}{8 \pi}}\sin(\theta)e^{\pm i \phi}</math>
 
−
|-
 
−
| 2 || 0 || <math>\left ( \frac{3}{(kr)^3} - \frac{1}{kr} \right )\sin(kr) -\frac{3\cos(kr)}{(kr)^2}</math> || <math>\sqrt{\frac{5}{16 \pi}}(3\cos^2(\theta)-1)</math>
 
−
|-
 
−
|  || <math>\pm</math> 1 ||  || <math>\mp \sqrt{\frac{15}{8 \pi}}\sin(\theta)\cos(\theta)e^{\pm i \phi}</math>
 
−
|-
 
−
|  || <math>\pm</math> 2 ||  || <math>\mp \sqrt{\frac{15}{32 \pi}}\sin^2(\theta)e^{\pm 2 i \phi}</math>
 
−
|}
 
−
 
 
−
[[Image:SphericalBesselFunctions.jpg]][[Image:SphereicalHamronics_Ylm.jpg]]
 
−
 
 
−
The general solution for the 3-D spherical infinite potential well problem is
 
−
 
 
−
:<math>\psi_{k,l,m}(r,\theta,\phi) = j_l(kr) Y_{l,m}(\theta, \phi)</math> = eigen function(s)
 
−
 
 
−
where
 
−
 
 
−
:<math>k,l,m</math> are quantization number and <math>E_k = \frac{\hbar^2 k^2}{2m} =</math> quantum energy level = eigen state(s)
 
−
 
 
−
=====Energy Levels =====
 
−
 
 
−
To find the Energy eignevalues we need to know the value for "k".  We apply the boundary condition
 
−
 
 
−
:<math>j_l(kr)= 0 </math> at <math>r=a</math>
 
−
 
 
−
to determine the "nodes"  of <math>j_l</math>; ie value of <math>ka</math> so if you tell me the size of the well then I can tell you the value of k which will satisfy the boundary conditions.  This means that "k" is not a "real" quantum number in the sense that it takes on integral values.
 
−
 
 
−
We simple label states with an integer <math>(n)</math> representing the <math>n^{th}</math> zero crossing via:
 
−
 
 
−
: <math>| n,l> = j_l(ka) Y_{l,m_l}</math>
 
−
 
 
−
 
 
−
For example:
 
−
 
 
−
;In the <math> l =0</math> case
 
−
:<math>j_o(ka) =\frac{sin(ka)}{ka} = 0 </math>when <math>(ka) = \pi, 2\pi, 3\pi, 4\pi, ...</math>
 
−
:You arbitrarily label these state as <math>n=1 \Rightarrow (ka) =\pi \;\;\;\; k = \pi/a \;\;\;\;\; E_0=\frac{\hbar^2 (\pi)^2}{2ma^2}, n=2 \Rightarrow (ka) = 2\pi  </math>
 
−
:<math>|1,0> = j_o(\pi r/a) Y_{0,0}  \;\;\; E=E_0</math>
 
−
:<math>|2,0> = j_o(2\pi r/a) Y_{0,0};\;\; E=2^2E_0 = 4E_0</math>
 
−
:<math>|3,0> = j_o(3\pi r/a) Y_{0,0};\;\; E=3^2E_0=9E_0</math>
 
−
:<math>|4,0> = j_o(4\pi r/a) Y_{0,0};\;\; E=4^2E_0=16E_0</math>
 
−
 
 
−
;In the <math> l =1</math> case
 
−
:<math>|1,1> = j_1(4.49  r/a) Y_{1,m_l}\;\;\; E=\left(\frac{4.49}{\pi}\right )^2E_0=2.04E_0</math>
 
−
:<math>|2,1> = j_1(7.73 r/a) Y_{1,m_l}\;\;\; E=\left(\frac{7.73}{\pi}\right )^2E_0=6.05E_0</math>
 
−
:<math>|3,1> = j_1(10.9 r/a) Y_{1,m_l}\;\;\; E=\left(\frac{10.9}{\pi}\right )^2E_0=12.04E_0</math>
 
−
:<math>|4,1> = j_1(14.07 r/a) Y_{1,m_l}\;\;\; E=\left(\frac{14.07}{\pi}\right )^2E_0=20.1E_0</math>
 
−
 
 
−
;Notice
 
−
:The angular momentum is degenerate for each level making the degeneracy for each energy <math>= 2l+1</math>
 
−
 
 
−
[[Image:EnergyLevel3-DInfinitePotentialWell.jpg]]
 
−
 
 
−
==== Simple Harmonic Oscillator ====
 
−
 
 
−
The potential for a Simple Harmonic Oscillator (SHM) is:
 
−
:<math>V(r) = \frac{1}{2} kr^2</math>
 
−
 
 
−
This potential is does not depend on any angles.  It's a central potential.  Our solutions for Y_{l,m} from the 3-D infinite well potential will work for the SHM potential as well!  All we need to do is solve the radial differential equation:
 
−
 
 
−
:<math>\frac{1}{R(r)}\hat{p}_r^2 R(r) + \frac{l(l+1)}{r^2} = \frac{2m}{\hbar^2}\left ( E - \frac{1}{2} kr^2 \right )</math>
 
−
:<math>\left ( \frac{-1}{r} \frac{\partial^2}{\partial r^2} r + \frac{l(l+1)}{r^2} \right ) R(r)= \frac{2m}{\hbar^2}\left ( E - \frac{1}{2} kr^2 \right )R(r)</math>
 
−
 
 
−
or
 
−
:  <math>\frac{\partial^2}{\partial r^2} R(r) +\frac{2}{r} \frac{\partial}{\partial r} R(r) + \left ( \frac{2m}{\hbar^2}\left ( E - \frac{1}{2} kr^2 \right)  -\frac{l(l+1)}{r^2}\right ) R(r)= 0</math>
 
−
 
 
−
When solving the 1-D harmonic oscillator solutions were found which are of the form
 
−
:<math>\psi_x(x) = A_n e^{-x^2/2} H_n(x)</math>
 
−
 
 
−
where
 
−
 
 
−
:<math>H_n(x) = (-1)^ne^{x^2} \frac{d^n}{dx^n} e^{-x^2}</math>
 
  
−
If you construct the solution
 
−
:<math>\psi(x,y,z) = \psi(x) \psi(y) \psi(z) \sim e^{x^2/2+y^2/2+z^2/2} f(x,y,z) \sim e^{r^2/2} f(x,y,z)</math>
 
  
−
Assume R(r) may be written as
+
=Midterm Exam Topics list=
  
−
:<math>R(r) = G(r) e^{-r^2/2}</math>
+
Basically everything before section 5.3 (The Nuclear Force).  Section 5.3 and below is not included on the midterm.
  
−
substituting this into the differential equation gives
+
Topics of emphasis:
  
−
:  <math>\frac{\partial^2 G}{\partial r^2}  + \left ( \frac{2}{r} -  \alpha r\right ) \frac{\partial G}{\partial r}  + \left (  \lambda - \beta - \frac{l(l+1)}{r^2}\right ) G(r)= 0</math>
+
#1-D Schrodinger Equation based problems involving discrete potentials ( wells, steps) and continuous potentials (simple Harmonic, coulomb).
 +
#Calculating form factors given the density of a nucleus
 +
#Determining binding and nucleon mass separation energies
 +
# <math>\vec{I}</math>, <math>\vec{\ell}</math>, and <math>\vec{s}</math> angular momentum operations
 +
#Calculating scattering rates given the cross-section and a description of the experimental apparatus
  
 +
Formulas given on test
  
−
The above differential equation can be solved using a power series solution
+
Schrodinger Time independent 1-D equation
  
−
:<math>G = \sum_i^\infty a_ir^i</math>
+
: <math>\left ( \frac{- \hbar^2}{2m} \frac{ \part^2}{\part x^2} + V \right ) \psi = E \psi</math>
  
−
After performing the power series solution; ie find a recurrance relation for the coefficents a_i after substituting into the differential equation  and require the coefficent of each power of r to vanish.
+
Particle Current Density
  
−
You arrive at a soultion of the form
+
:<math>j = \frac{ \hbar}{2im} \left ( \Psi^* \frac{\part \Psi}{\part x} -  \Psi \frac{\part \Psi^*}{\part x}\right )</math>
  
−
:<math>\psi(r,\theta,\phi) \equiv R(r) Y_{l,m}(\theta, \phi) = e^{-r^2/2} G_{l,n} Y_{l,m}(\theta, \phi) </math>
+
Form Factor
  
−
where
+
If the density has no <math>\theta</math> or <math>\phi</math> dependence
  
−
:<math>\ G_{l,n} =</math> polynomial in <math>r</math> of degree <math>n</math> in which the lowest term in <math>r</math> is <math>r^l</math>
+
:<math>F(q) = \frac{4 \pi}{q} \int \sin(qr) \rho(r) rdr</math>
  
 +
Coulomb energy difference between point nucleus and one with uniform charge distribution
  
−
these polynomials are solutions to the differential equation
+
:<math>\Delta E = \frac{2}{5} \frac{Z^4e^2}{4 \pi \epsilon_0} \frac{R^2}{a_0^3}</math>
  
−
:  <math>r^2\frac{\partial^2 G}{\partial r^2}  + 2\left ( r-r^3\right ) \frac{\partial G}{\partial r}  + \left (  2nr^2- l(l+1)\right ) G(r)= 0</math>
+
Nucleus Binding Energy
  
−
if you do the variable substitution
+
: <math>B(^A_ZX_N) = \left [ Z m(^1H) + Nm_n -m(^AX) \right ]c^2</math>
  
−
:<math>t = r^2</math>
+
Neutron Separation energy
  
−
you get
+
:<math>S_n = B(^A_ZX_N) - B(^{A-1}_ZX_{N-1})</math>
  
−
:<math>t\frac{\partial^2 S}{\partial t^2}  + \left ( l + \frac{3}{2} -t \right )  \frac{\partial S}{\partial t}  + k S= 0</math>
+
Proton Separation energy
  
−
the above differential equation is called the "associated" Laguerre differential equation with the Laguerre polynomials as its solutions.
+
:<math>S_p = B(^A_ZX_N) - B(^{A-1}_{Z-1}X_N)</math>
  
−
The following table gives you the Radial wave functions for a few SHO states:
+
Semiempirical Mass Formula
  
−
{| border="1"  |cellpadding="20" cellspacing="0
+
:<math>M(Z,A) = Z m(^1H) + Nm_n - B(^A_ZX_N)/c^2</math>
−
|-
 
−
| <math>n</math> ||<math> l</math> ||<math>E_n (\hbar \omega_o )</math> || <math>R(r)</math>
 
−
|-
 
−
| 0 || 0 || <math>\frac{3}{2} </math> || <math>= \frac{2 \alpha^{3/2}}{\pi^{1/4}}  e^{- \alpha^2 r^2/2}</math>
 
−
|-
 
−
| 1|| 1 || <math>\frac{5}{2} </math> || <math>= \frac{2\sqrt{2} \alpha^{3/2}}{\sqrt{3} \pi^{1/4}} \alpha r e^{- \alpha^2 r^2/2}</math>
 
−
|-
 
−
| 2 || 0 || <math>\frac{7}{2} </math> || <math>= \frac{2\sqrt{2} \alpha^{3/2}}{\sqrt{3} \pi^{1/4}} (\frac{3}{2} -\alpha^2 r^2 e^{- \alpha^2 r^2/2}</math>
 
−
|-
 
−
| 2 || 2 ||  || <math>= \frac{4 \alpha^{3/2}}{\sqrt{15} \pi^{1/4}} \alpha^2 r^2  e^{- \alpha^2 r^2/2}</math>
 
−
|-
 
−
| 3 || 1 ||  <math>\frac{9}{2} </math>  || <math>= \frac{4 \alpha^{3/2}}{\sqrt{15} \pi^{1/4}} (\frac{5}{2} \alpha r  -\alpha^3 r^3  e^{- \alpha^2 r^2/2}</math>
 
−
|-
 
−
|}
 
−
 
 
−
:Note
 
−
:Again there is a degeneracy of <math>2l+1</math> for each <math>l</math>
 
−
: Again E is independent of <math>l</math> (central or constant potentials)
 
−
: if <math>n</math> is odd <math>l</math> is odd and if <math>n</math> is even<math> l</math> is even
 
−
: multiple values of <math>l</math> occur for a give <math>n</math> such that <math>l \le n</math>
 
−
: The degeneracy is <math>\frac{1}{2} (n+1) (n+2)</math> because of the above  points
 
−
 
 
−
==== The Coulomb Potential for the Hydrogen like atom====
 
−
 
 
−
The Coulomb potential is defined as :
 
−
:<math>V(r) = -\frac{Ze^2}{4 \pi \epsilon_0 r} </math>
 
  
 
where  
 
where  
−
:<math>Z =</math> atomic number
 
−
:<math>e =</math> charge of an electron
 
−
:<math>\epsilon_0=</math> permittivity of free space = <math>8.85 \times 10^{-12} Coul^2/(N m^2)</math>
 
−
 
−
This potential is does not depend on any angles.  It's a central potential.  Our solutions for Y_{l,m} from the 3-D infinite well potential will work for the Coulomb  potential as well!  All we need to do is solve the radial differential equation:
 
−
 
−
:<math>\frac{1}{R(r)}\hat{p}_r^2 R(r) + \frac{l(l+1)}{r^2} = \frac{2m}{\hbar^2}\left ( E + \frac{k}{r}  \right )</math>
 
−
:<math>\left ( \frac{-1}{r} \frac{\partial^2}{\partial r^2} r + \frac{l(l+1)}{r^2} \right ) R(r)= \frac{2m}{\hbar^2}\left ( E + \frac{k}{r}  \right )R(r)</math>
 
−
 
−
or
 
−
:  <math>\frac{1}{r}\frac{\partial^2}{\partial r^2} rR(r)  + \left ( \frac{2m}{\hbar^2}\left ( E + \frac{k}{r}  \right)  -\frac{l(l+1)}{r^2}\right ) R(r)= 0</math>
 
−
 
−
====== Radial Equation ======
 
  
−
Use the change of variable to alter the differential equiation
+
:<math>B(^A_ZX_N) = = \alpha_V A - \alpha_S A^{2/3} - \alpha_C \frac{Z(Z-1)}{A^{1/3}} - \alpha_{sym} \frac{(A-2Z)^2}{A} - \delta</math>
  
−
Let
+
{| border="1"  |cellpadding="20" cellspacing="0  
−
 
 
−
:<math>G(r) \equiv r R(r)</math>
 
−
 
 
−
Then the differential equation becomes:
 
−
 
 
−
:  <math>\frac{\partial^2}{\partial r^2} G(r)  + \left ( \frac{2m}{\hbar^2}\left ( E + \frac{k}{r}  \right)  -\frac{l(l+1)}{r^2}\right ) G(r)= 0</math>
 
−
 
 
−
Consider the case where <math>|V| > |E| \Rightarrow</math> (Bound states)
 
−
 
 
−
Bound state also imply that the eigen energies are negative
 
−
 
 
−
:<math>E = - |E|</math>
 
−
 
 
−
Let
 
−
:<math>\kappa^2 \equiv \frac{2m |E|}{\hbar^2}</math>
 
−
:<math>\rho \equiv 2 \kappa r</math>
 
−
: <math>\lambda \equiv \left ( \frac{Ze^2 m }{\kappa \hbar^2} \right ) = Z \sqrt{\frac{\mathcal{R}}{|E|}}</math>
 
−
:<math>\mathcal{R} = \frac{me^4}{2\hbar^2} = \frac{\hbar^2}{2m a_o^2} =1.09737316 \times 10^7\frac{1}{m} =</math> Rydberg's constant
 
−
: <math>a_o = \frac{\hbar^2}{me^2} = 5.291772108 \times 10^{-11} m= 52918 fm =</math> Bohr Radius
 
−
 
 
−
 
 
−
:  <math>\frac{\partial^2}{\partial \rho^2} G(r)  - \frac{l(l+1)}{\rho^2}G(r) +  \left ( \frac{\lambda}{\rho} -\frac{1}{4}  \right)  G(r)= 0</math>
 
−
 
 
−
====== Boundary conditions ======
 
−
 
 
−
* if <math>\rho</math> is large then the diff equation looks like
 
−
 
 
−
:  <math>\frac{\partial^2}{\partial \rho^2} G(r)  -\frac{1}{4}  G(r)= 0</math>
 
−
: <math>\Rightarrow G(r) \sim Ae{-\rho/2} + B e^{\rho/2}</math>
 
−
 
 
−
To keep<math> G(r \rightarrow \infty )</math> finite at large <math>\rho</math> you need to have B=0
 
−
 
 
−
* if \rho is very small ( particle close to the origin) then the diff equation looks like
 
−
 
 
−
: <math>\frac{\partial^2}{\partial \rho^2} G(r)  - \frac{l(l+1)}{\rho^2}  G(r)= 0</math>
 
−
 
 
−
The general solution for this type of Diff Eq is
 
−
 
 
−
:<math>G(r) = \frac{A}{\rho^l} + B \rho^{l+1}</math>
 
−
 
 
−
where A =0 so <math>G(r \rightarrow 0)</math> is finite
 
−
 
 
−
A general solution is formed using a linear combination of these asymptotic solutions
 
−
 
 
−
:<math>G(r) = e^{-\rho/2} \rho^{l+1} F(\rho)</math>
 
−
 
 
−
where
 
−
 
 
−
:<math>F(\rho) = \sum_{i=0}^{\infty} C_i \rho^i</math>
 
−
 
 
−
substitute this power series solution into the differential equation gives
 
−
 
 
−
: <math>\rho \frac{d^2 F}{d \rho^2} + (2l + 2 - \rho) \frac{d F}{d \rho} - (l +1 - \lambda)F =0</math>
 
−
 
 
−
which is again the associated Laguerre differential equation with a general series solution containing functions of the form
 
−
 
 
−
:<math>F(\rho) \sim e^{\rho}</math>
 
−
 
 
−
with the recurrance relation
 
−
 
 
−
:<math>C_{i+1} = \frac{i+l+1 - \lambda}{(i+1)(i+2l+2)} C_i</math>
 
−
 
 
−
 
 
−
notice that
 
−
 
 
−
:<math>G(r) = e^{-\rho/2} \rho^{l+1} e^{\rho}</math>
 
−
 
 
−
now diverges for large <math>\rho</math>.
 
−
 
 
−
To keep the solution from diverging as well we need to truncate the coefficients<math> C_{i+1}</math> at some <math> i_{max}</math> by setting the coefficient to zero when
 
−
 
 
−
:<math>i_{max} = \lambda -l -1</math>
 
−
 
 
−
This value of <math>\lambda</math> for the truncations identifies a quantum state according to the integer <math>\lambda</math> which truncates the solution and gives us our energy eigenvalues
 
−
 
 
−
:<math>\lambda^2 = \frac{Z^2 \mathcal{R}}{|E|}</math>
 
−
 
 
−
or since \lambda is just a dummy variable
 
−
 
 
−
:<math>E_n = - |E_n| = -\frac{Z^2 \mathcal{R}}{n^2}</math>
 
−
 
 
−
====== Coulomb Eigenfunctions and Eigenvalues ======
 
−
 
 
−
 
 
−
{| border="1"  |cellpadding="20" cellspacing="0  
 
−
|-
 
−
| <math>n</math> ||<math> l</math> || Spec Not. ||<math>E_n (-\frac{Z^2 \mathcal{R}}{n^2} )= 13.6 eV</math> || <math>R(r)</math>
 
 
|-
 
|-
−
| 1 || 0 || 1S || <math>\frac{1}{1} </math> || <math>=2 \left ( \frac{Z}{a_o} \right)^{3/2}  e^{- Z r/a_o}</math>
+
| Parameter  || Krane
 
|-
 
|-
−
|2|| 1 || 2S || <math>\frac{1}{4} </math> || <math>= \left ( \frac{Z}{2a_o} \right)^{3/2}  (2 - Zr/a_o) e^{- Z r/2a_o}</math>
+
| <math>\alpha_V</math> || 15.5
 
|-
 
|-
−
| 2 || 1 || 2P || <math>\frac{1}{4} </math> || <math>= \left ( \frac{Z}{2a_o} \right)^{3/2} \frac{Zr}{\sqrt{3}a_o} e^{- Z r/2a_o}</math>
+
| <math>\alpha_S</math> || 16.8
 
|-
 
|-
−
| 3 || 0 || 3S  || <math>\frac{1}{9} </math> || <math>= \left ( \frac{Z}{3a_o} \right)^{3/2} \left [ 2 - \frac{4Zr}{3 a_o} + 4 \left (\frac{Zr}{3\sqrt{3} a_o} \right)^2 \right ] e^{- Z r/3a_o}</math>
+
| <math>\alpha_C</math> || 0.72
 
|-
 
|-
−
| 3 || 1 ||  3P || <math>\frac{1}{9} </math> || <math>= \frac{4\sqrt{2}}{9}  \left ( \frac{Z}{3a_o} \right)^{3/2} \left ( \frac{Zr}{a_o} \right) \left ( 1-\frac{Zr}{6a_o} \right ) e^{- Z r/3a_o}</math>
+
| <math>\alpha_{sym}</math> || 23
 
|-
 
|-
−
| 3 || 2 || 3D || <math>\frac{1}{9} </math> || <math>= \frac{2\sqrt{2}}{27\sqrt{5}} \left ( \frac{Z}{3a_o} \right)^{3/2}  \left ( \frac{Zr}{a_o} \right)^2 e^{- Z r/3a_o}</math>
+
| <math>\alpha_p</math> || <math>\pm</math>34
 
|-
 
|-
 
|}
 
|}
  
 +
:<math>g_{\ell} =\left \{  {1 \;\;\;\; proton  \atop 0 \;\;\;\; neutron} \right .</math>
 +
:<math>g_{s} =\left \{  {5.5856912 \pm 0.0000022 \;\;\;\; proton  \atop -3.8260837 \pm 0.0000018 \;\;\;\; neutron} \right .</math>
  
−
The SHO and Coulomb schrodinger equations have Laguerre polynomial solutions for the radial part with the SHO solution polynomials of <math>r^2</math> and the Coulomb solution polynomials linear in <math>r</math>.  The number of degenerate quantum states differs though, the SHO  has 10 degenerate states while the Coulomb potential has 9 states.
 
−
 
−
== Angular Momentum ==
 
−
 
−
 
−
As you may have noticed in the quantum solution to the coulomb potential (Hydrogen Atom) problem above, the quantum number <math>\ell</math> plays a big role in the identification of quantum states.    In atomic physics the states S,P,D,F,... are labeled according to the value of <math>\ell</math>.  Perhaps the best part is that as long as there is no angular dependence to the potential, you can reused the spherical harmonics as the angular component to the wave function for your problem.  Furthermore, the angular momentum is a constant of motion because the potential is without angular dependence (central potential), just like the classical case. 
 
−
 
−
The mean angular momentum for a given quantum state is given as
 
−
 
−
:<math><\ell^2> - \hbar^2 \ell (\ell+1)</math>
 
−
 
−
since <math>\ell</math> has its origin in
 
−
 
−
:<math>\vec{\ell} = \vec{r} \times \vec{p}</math>
 
−
 
−
and the uncertainty principle has
 
−
 
−
:<math>\Delta x \Delta p_x \ge \frac{\hbar}{2}</math>  we expect that the uncertainty principle will also impact <math>\ell</math> such that
 
−
 
−
:<math>\Delta \ell_z \Delta \phi \ge \frac{\hbar}{2}</math>
 
−
 
−
where <math>\phi</math> characterizes the location of <math>\ell</math> in the x-y plane.
 
−
 
−
or in other words, once we determine one component of <math>\ell</math> (ie: <math>\ell_x</math> ) we are unable to determine the remaining components ( <math>\ell_y</math> and <math>\ell_z</math> ).
 
−
 
−
As a result, the convention used is to define quantum states in terms of <math> \ell_z</math>  such that
 
−
 
−
:<math><\ell_z> = \hbar m_{\ell}</math>
 
−
 
−
This means the <math>m_{\ell}</math> represent the projection of \ell along the axis of quantization (z-axis).
 
−
 
−
;Notice
 
−
:  <math>m_{\ell}  < \sqrt{\ell(\ell+1}</math> : if <math>m_{\ell} = \ell</math> then we would know \<math>ell_x</math> and <math>\ell_y</math>.
 
  
−
=== Intrinsic angular Momentum (Spin) ===
+
=Final=
 +
1.) Calculate the magnetic moment of a proton assuming that it may be described as a neutron with a positive pion <math>(\pi^+)</math> in an  <math>\ell =1</math>  state.
  
−
The Stern Gerlach experiment showed us that electrons have an intrinsic angular momentum or spin which affects their trajectory through an inhomogeneous magnetic field.  This prperty of a particle has no classical analog.  Spin is treated in the same way as angular momentum, namely
 
  
−
:<math><s^2> = \hbar^2 s(s+1)</math>
+
2.) Show that the phase shift (<math>\delta_0</math>) for the scattering of a neutron by a proton can be given by the equation
−
:<math><s_z> = \hbar m_s = \pm \hbar \frac{1}{2}</math>
 
  
−
;Note
+
:<math>\delta_0 = \tan^{-1} \left( \frac{\tan(k_1R) - \frac{k_1}{k_2} \tan(k_2R)}{ \tan(k_1R) \tan(k_2R)+\frac{k_1}{k_2}}\right )</math>
−
: Nucleons like electrons are also spin \frac{1}{2} objects.
 
−
 
 
−
=== Total angular momentum ===
 
−
 
 
−
The total angular momentum of a quantum mechanical system is defined as
 
−
 
 
−
:<math>\vec{j} = \vec{\ell} + \vec{s}</math>
 
−
 
 
−
such that \vec{j} behaves quantum mechanically jusst like its constituents such that
 
−
 
 
−
:<math><j^2> = \hbar^2 j(j+1)</math>
 
−
:<math><j_z> - <\ell_z + s_z> = \hbar m_j</math>
 
  
 
where  
 
where  
  
−
:<math>m_j = m+{\ell} \pm \frac{1}{2}</math>
+
:<math>k_1^2 = \frac{2m(V+E)}{\hbar^2}</math>
−
 
+
:<math>k_2^2 = \frac{2mE)}{\hbar^2}</math>
−
In spectroscopic notation where <math>\ell</math> is labeled by s,p,d,f,g,...
 
−
the value of j is added as a subsript
 
−
 
 
−
;for example
 
−
: <math>1S_{1/2}  = \ell=0</math> state with <math>m_s = + 1/2</math>
 
−
:<math>2P_{3/2} = \ell = 1</math> with <math>m_s = + 1/2</math>
 
−
:<math>2P_{1/2} = \ell =1</math> with <math>m_s = -1/2</math>
 
−
 
 
−
In Atomic systems, the electrons in light element atoms interact strongly according to their angular momentum with their spin playing a small role (you can use separation of variables to have <math>\psi = \psi(r)\psi(s)\psi(\ell)</math> .  In heavy atoms, the spin-orbit (<math>j</math>) interactions are as strong as the individual <math>\ell</math> and <math>s</math> interactions.  In his case the total angluar momentum (<math>j</math>) of each constituent is coupled to some <math> j_{tot}</math>, you construct <math>\psi = \psi(r) \psi(j)</math>.  When there is a very strong external magnetic field, <math>\ell</math> and <math>s</math> are even more decoupled.
 
−
 
 
−
===Wave Function Symmetry===
 
−
 
 
−
 
 
−
# All particles with integral spins (0,1,2) ( photons, dueteron,...) have symmetric total wave functions (\psi = \psi_A(r_1) \psi_B(r_2) = \psi_A(r_2) \psi_B(r_1))
 
−
#All particles with half-integral spins (1/2, 3/2, 5/2, ...) ( electrons, nucleons,...) have anti-symmetric wave functions.
 
−
 
 
−
==Parity==
 
−
 
 
−
Parity is a principle in physics which when conserved means that the results of an experiment don;t change if you perform the experiment "in a mirror".  Or in otehr words if you alter the experiment such that
 
−
:<math>\vec{r} \rightarrow - \vec{r}</math>
 
−
:<math>\mathcal{\hat{P}} \vec{r}=-\vec{r}</math>
 
−
the system is unchanged.
 
−
If
 
−
<math>\mathcal{\hat{P}} V(\vec{r})= V(-\vec{r}) =V(\vec{r})</math>
 
−
 
 
−
Then the potential (V(r))  is believed to conserve parity.
 
−
 
 
−
and
 
−
 
 
−
:<math>|\psi(\vec{r})|^2 = |\psi(-\vec{r})|^2</math>
 
−
::<math>\rightarrow \psi(-\vec{r}) = \pm \psi(\vec{r})</math>
 
−
 
 
−
;Positive parity
 
−
: <math>\mathcal{\hat{P}} \psi(\vec{r})= \psi(-\vec{r}) = \psi(\vec{r})</math>
 
−
 
 
−
; Negative parity
 
−
: <math>\mathcal{\hat{P}} \psi(\vec{r})=\psi(-\vec{r}) = -\psi(\vec{r})</math>
 
−
 
 
−
;Note:
 
−
:<math>\mathcal{\hat{P}} Y_{\ell,m}(\theta,\phi) = Y_{\ell,m}(\pi - \theta, \phi + \pi) = (-1)^{\ell} Y_{\ell,m}(\theta,\phi)</math>
 
−
:Thus is<math> \ell</math> is even then <math>Y_{\ell,m}</math> is Positive parity, if <math>\ell</math> is odd then <math>Y_{\ell,m}</math> is negative parity.
 
−
 
 
−
===3-D SHO ===
 
−
 
 
−
The Radial wave functions (R_{n,\ell}) of the 3-D SHO oscillator problem can be either positive or negative parity.
 
−
 
 
−
:<math>\mathcal{\hat{P}} R_{0,0} =\mathcal{\hat{P}} ( \frac{2 \alpha^{3/2}}{\pi^{1/4}}  e^{- \alpha^2 r^2/2}) = +R_{0,0}</math>
 
−
::Thus <math>\mathcal{\hat{P}} R_{0,0}Y_{0,0} = +R_{0,0}Y_{0,0}</math>
 
−
 
 
−
:<math>\mathcal{\hat{P}} R_{1,1} =\mathcal{\hat{P}} \left ( \frac{2\sqrt{2} \alpha^{3/2}}{\sqrt{3} \pi^{1/4}} \alpha r e^{- \alpha^2 r^2/2} \right )= -R_{1,1}</math>  
 
−
::Thus<math>\mathcal{\hat{P}} R_{1,1}Y_{1,m} = -R_{1,1}(-1)^1Y_{1,m} =R_{1,1}Y_{1,m}</math>
 
−
 
 
−
The conclusion is that the total wave function <math>\psi=R_{n,\ell}Y_{\ell,m}</math> is positive under parity.
 
−
 
 
−
===Parity Violation===
 
  
−
In 1957, Chien-Shiung Wu announced her experimental result that beta emission from Co-60 had a preferred direction.  In that experiment an external field was used to align the total angular momentum of the Co-60 source either towards or away from a scintillator used to detect <math>\beta</math> particles.  She reported seeing that only 30% of the <math>\beta</math> particles came out along the direction of the B-field (Co-60 spin direction).  In a mirror, the total angular momentum of the Co-60 source would point in the same direction as before
+
V = 36.7 MeV R = 2.1 fm
−
(<math>\vec{\ell} = \vec{r} \times \vec{p} = \vec{(-r)} \times \vec{(-p)})</math> while the momentum vector <math>(\vec{p} = -\vec{(-p)})</math> of the emitted <math>\beta</math> particles would change sign and hence direction.
 
  
−
;Consequence of the experimental observation
+
3.)
−
:The Weak interaction does not conserve parity
 
−
:Parity Violation for the Strong or E&M force has not been observed
 
  
−
==Transitions==
+
a.) Write the reaction equations for the following processes.  Show all reaction products.
  
−
For a stable particle its wave function will be in a stationary state that is static in time.  This means that if I measure the average energy of this state I will see no fluctuation because the state is stationary.
+
i.)<math>{226 \atop\; }Ra \; \; \; \alpha \mbox{- decays}</math>
  
−
In other words
+
ii.)<math>{110 \atop\; }In \; \; \; \beta^+  \mbox {-decays}</math>
  
−
:<math>\Delta E = \sqrt {<E^2> - <E>^2} = 0</math>
+
iii.)<math>{36 \atop\; }Ar (2^{nd} 0^{+})  \mbox{internal conversion}</math>
  
−
The implication of this using the uncertainty principle is that
+
iv.)<math>{12 \atop\; }C (2^+) \; \; \; \gamma \mbox{- decays}</math>
  
−
:<math>\Delta E \Delta t \ge \hbar/2 \Rightarrow t \rightarrow \infty</math>
+
b.) Determine the Q-values for the first two reactions above.
  
−
==Dirac Equation ==
 
  
−
= Nuclear Properties=
+
4.) Find the Quadrupole moment of <math>{209 \atop\; }Bi(9/2^-)</math> using the shell model and compare to the experimental value of -0.37 barns.
−
==Nuclear Radius ==
 
−
==Binding Energy ==
 
−
== Angular Momentum and Parity ==
 
  
  
−
= The Nuclear Force=
+
5.) Find <math>\frac{\mu}{\mu_{NM}}</math>, using the shell model, for the following nuclei
−
==Yukawa Potential ==
 
  
−
= Nuclear Models=
+
a.)<math> {75 \atop\; }Ge</math>
  
−
==Shell Model==
+
b.) <math>{87 \atop\; }Sr</math>
  
−
=Nuclear Decay and Reactions=
+
c.) <math>{91 \atop\; }Zr</math>
  
−
== Alpha Decay ==
+
d.) <math>{47 \atop\; }Sc</math>
  
−
==Beta Decay ==
+
6.) Use the shell model to predict the ground state spin and parity of the following nuclei:
  
−
==Gamma Decay ==
+
a.) <math>{7 \atop\; }Li</math>
  
−
=Electro Magnetic Interactions=
+
b.)<math> {11 \atop\; }B</math>
  
−
=Weak Interactions=
+
c.) <math>{15 \atop\; }C</math>
  
 +
d.)<math> {17 \atop\; }F</math>
  
−
=Strong Interaction=
+
7.) Tabulate the possible <math>m</math> states for a nucleus three quadrupole phonon state (<math>\lambda = 3</math> ).  Show that the permitted resultant states are <math>0^+</math>, <math>2^+</math>, <math>3^+</math>, <math>4^+</math>, and <math>6^+</math>.
−
 
 
−
=Applications=
 
−
 
 
−
=Homework problems=
 
−
 
 
−
[[NucPhys_I_HomeworkProblems]]
 

Latest revision as of 01:07, 18 December 2015

Advanced Nuclear Physics

References:
Introductory Nuclear Physics
Kenneth S. Krane: ISBN 9780471805533

Catalog Description:

PHYS 609 Advanced Nuclear Physics 3 credits. Nucleon-nucleon interaction, bulk nuclear structure, microscopic models of nuclear structure, collective models of nuclear structure, nuclear decays and reactions, electromagnetic interactions, weak interactions, strong interactions, nucleon structure, nuclear applications, current topics in nuclear physics. PREREQ: PHYS 624 OR PERMISSION OF INSTRUCTOR.

PHYS 624-625 Quantum Mechanics 3 credits. Schrodinger wave equation, stationary state solution; operators and matrices; perturbation theory, non-degenerate and degenerate cases; WKB approximation, non-harmonic oscillator, etc.; collision problems. Born approximation, method of partial waves. PHYS 624 is a PREREQ for 625. PREREQ: PHYS g561-g562, PHYS 621 OR PERMISSION OF INSTRUCTOR.

NucPhys_I_Syllabus

Click here for Syllabus

Introduction

The interaction of charged particles (electrons and positrons) through the exchange of photons is described by a fundamental theory known as Quantum ElectroDynamics(QED). QED has perturbative solutions which are limited in accuracy only by the order of the perturbation you have expanded to. As a result, the theory is quite useful in describing the interactions of electrons that are prevalent in Atomic physics.


Nuclear physics describes how Atomic nuclei interact via the strong forces as well as how the strong force binds the constituents of a nucleus (protons and neutrons, a.k.a. nucleons). Particle physics studies the interactions of fundamental particles, particles without substructure like quarks, photons, and electrons. Both Nuclear and Particle physics rely on the "Standard Model", a field theory description of the strong, weak and electromagnetic forces. Quantum ChromoDynamic (QCD) is one component to the Standard Model which represents the fundamental theory developed to describe the interactions of the quarks and gluons inside a nucleon, analogous to how QED describes the electromagnetic forces of electrons within the atom. The electroweak and Higgs field are the remaining components to the Standard model. Ideally, QCD is a field theory which could be used to describe how quarks interact to for nucleons and then describe how those nucleons interact to form a nucleus and eventually lead to a description of how the nucleus interacts with other nuclei. Unfortunately, QCD does not have a complete solution at this time. At very high energies, QCD can be solved perturbatively. This is an energy [math]E[/math] at which the strong coupling constant [math]\alpha_s[/math] is less than unity where

[math]\alpha_s \approx \frac{1}{\beta_o \ln{\frac{E^2}{\Lambda^2_{QCD}}}}[/math]
[math]\Lambda_{QCD} \approx 200 MeV[/math]


The objectives in this class will be to discuss the basic aspects of the nuclear phenomenological models used to describe the nucleus of an atom in the absence of a QCD solution.

Nomenclature

Variable Definition
Z Atomic Number = number of protons in an atom
A Atomic Mass
N number of neutrons in an atom = A-Z
Nuclide A specific nuclear species
Isotope Nuclides with same Z but different N
Isotones Nuclides with same N but different Z
Isobars Nuclides with same A
Nuclide A specific nuclear species
Nucelons Either a neutron or a proton
J Nuclear Angular Momentum
[math]\ell[/math] angular momentum quantum number
s instrinsic angular momentum (spin)
[math]\vec{j}[/math] total angular momentum = [math]\vec{\ell} + \vec{s}[/math]
[math]Y_{\ell,m_{\ell}}[/math] Spherical Harmonics, [math]\ell[/math] = angular momentum quantum number, [math]m_{\ell}[/math] = projection of [math]\ell[/math] on the axis of quantization
[math]\hbar[/math] Planks constant/2[math]\pi = 6.626 \times 10^{-34} J \cdot s / 2 \pi[/math]

Notation

[math]{A \atop Z} X_N[/math] = An atom identified by the Chemical symbol [math]X[/math] with [math]Z[/math] protons and [math]N[/math] neutrons.

Notice that [math]Z[/math] and [math]N[/math] are redundant since [math]Z[/math] can be identified by the chemical symbol [math]X[/math] and [math] N[/math] can be determined from both [math]A[/math] and the chemical symbol [math]X[/math](N=A-Z).

example
[math]{208 \atop\; }Pb ={208 \atop 82 }Pb_{126}[/math]

Historical Review

Rutherford Nuclear Atom (1911)

Rutherford interpreted the experiments done by his graduate students Hans Geiger and Ernest Marsden involving scattering of alpha particles by the thin gold-leaf. By focusing on the rare occasion (1/20000) in which the alpha particle was scattered backward, Rutherford argued that most of the atom's mass was contained in a central core we now call the nucleus.

Chadwick discovers neutron (1932)

Prior to 1932, it was believed that a nucleus of Atomic mass [math]A[/math] was composed of [math]A[/math] protons and [math](A-Z)[/math] electrons giving the nucleus a net positive charge [math]Z[/math]. There were a few problems with this description of the nucleus

  1. A very strong force would need to exist which allowed the electrons to overcome the coulomb force such that a bound state could be achieved.
  2. Electrons spatially confined to the size of the nucleus ([math]\Delta x \sim 10^{-14}m = 10 \;\mbox{fermi})[/math] would have a momentum distribution of [math]\Delta p \sim \frac{\hbar}{\Delta x} = 20 \frac{\mbox {MeV}}{\mbox {c}}[/math]. Electrons ejected from the nucleus by radioactive decay ([math]\beta[/math] decay) have energies on the order of 1 MeV and not 20.
  3. Deuteron spin: The total instrinsic angular momentum (spin) of the Deuteron (A=2, Z=1) would be the result of combining two spin 1/2 protons with a spin 1/2 electron. This would predict that the Deuteron was a spin 3/2 or 1/2 nucleus in contradiction with the observed value of 1.

The discovery of the neutron as an electrically neutral particle with a mass 0.1% larger than the proton led to the concept that the nucleus of an atom of atomic mass [math]A[/math] was composed of [math]Z[/math] protons and [math](A-Z)[/math] neutrons.

Powell discovers pion (1947)

Although Cecil Powell is given credit for the discovery of the pion, Cesar Lattes is perhaps more responsible for its discovery. Powell was the research group head at the time and the tradition of the Nobel committe was to award the prize to the group leader. Cesar Lattes asked Kodak to include more boron in their emulsion plates making them more sensitive to mesons. Lattes also worked with Eugene Gardner to calcualte the pions mass.

Lattes exposed the plates on Mount Chacaltaya in the Bolivian Andes, near the capital La Paz and found ten two-meson decay events in which the secondary particle came to rest in the emulsion. The constant range of around 600 microns of the secondary meson in all cases led Lattes, Occhialini and Powell, in their October 1947 paper in 'Nature ', to postulate a two-body decay of the primary meson, which they called p or pion, to a secondary meson, m or muon, and one neutral particle. Subsequent mass measurements on twenty events gave the pion and muon masses as 260 and 205 times that of the electron respectively, while the lifetime of the pion was estimated to be some 10-8 s. Present-day values are 273.31 and 206.76 electron masses respectively and 2.6 x 10-8 s. The number of mesons coming to rest in the emulsion and causing a disintegration was found to be approximately equal to the number of pions decaying to muons. It was, therefore, postulated that the latter represented the decay of positively-charged pions and the former the nuclear capture of negatively-charged pions. Clearly the pions were the particles postulated by Yukawa.

In the cosmic ray emulsions they saw a negative pion (cosmic ray) get captured by a nucleus and a positive pion (cosmic ray) decay. The two pion types had similar tracks because of their similar masses.

Nuclear Properties

NuclearProperties_Forest_NucPhys_I

The nucleus of an atom has such properties as spin, mangetic dipole and electric quadrupole moments. Nuclides also have stable and unstable states. Unstable nuclides are characterized by their decay mode and half lives.


Decay Modes

Mode Description
Alpha decay An alpha particle (A=4, Z=2) emitted from nucleus
Proton emission A proton ejected from nucleus
Neutron emission A neutron ejected from nucleus
Double proton emission Two protons ejected from nucleus simultaneously
Spontaneous fission Nucleus disintegrates into two or more smaller nuclei and other particles
Cluster decay Nucleus emits a specific type of smaller nucleus (A1, Z1) smaller than, or larger than, an alpha particle
Beta-Negative decay A nucleus emits an electron and an antineutrino
Positron emission(a.k.a. Beta-Positive decay) A nucleus emits a positron and a neutrino
Electron capture A nucleus captures an orbiting electron and emits a neutrino - The daughter nucleus is left in an excited and unstable state
Double beta decay A nucleus emits two electrons and two antineutrinos
Double electron capture A nucleus absorbs two orbital electrons and emits two neutrinos - The daughter nucleus is left in an excited and unstable state
Electron capture with positron emission A nucleus absorbs one orbital electron, emits one positron and two neutrinos
Double positron emission A nucleus emits two positrons and two neutrinos
Gamma decay Excited nucleus releases a high-energy photon (gamma ray)
Internal conversion Excited nucleus transfers energy to an orbital electron and it is ejected from the atom

Time

Time scales for nuclear related processes range from years to [math]10^{-20}[/math] seconds. In the case of radioactive decay the excited nucleus can take many years ([math]10^6[/math]) to decay (Half Life). Nuclear transitions which result in the emission of a gamma ray can take anywhere from [math]10^{-9}[/math] to [math]10^{-12}[/math] seconds.

Units and Dimensions

Variable Definition
1 fermi [math]10^{-15}[/math] m
1 MeV =[math]10^6[/math] eV = [math]1.602 \times 10^{-13}[/math] J
1 a.m.u. Atomic Mass Unit = 931.502 MeV

Resources

The following are resources available on the internet which may be useful for this class.


Lund Nuclear Data Service

in particular

The Lund Nuclear Data Search Engine

Several Table of Nuclides

BNL
LANL
Korean Atomic Energy Research Institute
National Physical Lab (UK)

Table of Isotopes at Lawrence Berkeley National Laboratory

French Nuclear data Java GUI

Quantum Mechanics Review

Quantum_Mechanics_Review_Forest_NucPhys_I

Nuclear Properties

NuclearProperties_Forest_NucPhys_I

The Nuclear Force

NuclearForce_Forest_NucPhys_I

Nuclear Models

Given the basic elements of the nuclear potential from the last chapter, one may be tempted to construct the hamiltonian for a group of interacting nucleons in the form

[math]H = \sum_i^A T_i + \sum_i\lt j^A V_{ij}[/math]

where

[math]T_i[/math] represent the kinetic energy of the ith nucleon
[math]V_{ij}[/math] represents the potential energy between two nucleons.

If you assume that the nuclear force is a two body force such that the force between any two nucleons doesn't change with the addition of more nucleons, Then you can solve the Schrodinger equation corresponding to the above Hamiltonian for A<5.

For A< 8 there is a technique called Green's function monte carlo which reportedly finds solution that are nearly exact. J. Carlson, Phys. Rev. C 36, 2026 - 2033 (1987), B. Pudliner, et. al., Phys. Rev. Lett. 74, 4396 - 4399 (1995)

Shell Model

Independent particle model

This part of the Shell model suggests that the properties of a nucleus with only one unpaired nucleon are determined by that one unpaired nucleon. The unpaired nucleon usually, though no necessarily, occupies the outer most shell as a valence nucleon.

SN-130 Example

The low lying excited energy states for Sn-130 taken from the LBL website are given below.

File:Sn-130 LowLyingE Levels.tiff


The listing indicates that the ground state of Sn-130 is a spin 0 positive parity [math](J^{\pi} = 0^+)[/math] state. The first excited state of this nucleus is 1.22 MeV above the ground state and has [math](J^{\pi} = 2^+)[/math]. The next excited state is 1.95 MeV above the ground state and has [math](J^{\pi} = 7^-)[/math].

Let's see how well the shell model does at predicting these [math](J^{\pi} )[/math] states

Liquid Drop Model

Bohr and Mottelson considered the nucleon in terms of its collective motion with vibrations and rotations that resembled a suspended drop of liquid.

Electric Quadrupole Moment

Electric_QuadrupoleMoment_Forest_NuclPhys_I


Nuclear Decay

Nuclear_Decay_Forest_NucPhys_I

Nuclear Reactions

Forest_NucPhys_I_Nuclear_Reactions

Electro Magnetic Interactions

TF_DerivationOfCoulombForce

Weak Interactions

Neutrino

[math]\nu_e + {40 \atop 18 }Ar_{22} \rightarrow {40 \atop 19 }K^*_{21} + e^- [/math]

The min neutrino energy needed for this reaction assuming the electron energy is ignorable:

[math]\Delta M = \left [ m\left({40 \atop 19 }K\right)- m\left({40 \atop 18 }Ar\right) \right] c^2 [/math]
[math]= \left [ 39.96399848 - 39.9623831225 \right] 931.502 \frac{\mbox{MeV}}{\mbox {u}} [/math]
[math]= 0.00162 931.502 \frac{\mbox{MeV}}{\mbox {u}} = 1.505 MeV[/math]

A reaction to the ground state of potassium is forbidden in this charged current electron neutrino reaction resulting in one or more gammas being emitted as the Potassium nucleus de-excited to the ground state. There are about 98 observed excited states of the K-40 nucleus.

Detecting this signal in DUNE means you need to see and electron and at least one gamma in coincidence.

Potassium 40 decay

89% of the time

[math]{40 \atop 19 }K_{21} \rightarrow {40 \atop 20 }Ca_{20} + \beta (E_{max} = 1.3 MeV) + \bar{\nu}[/math]

10.72% of the time

[math]{40 \atop 19 }K_{21} \rightarrow {40 \atop 18 }Ar_{22} + \gamma (E = 1.46 MeV) + \bar{\nu}[/math]

0.001% of the time

[math]{40 \atop 19 }K_{21} \rightarrow {40 \atop 20 }Ca_{20} + e^+ + \nu[/math]

Potassium has half life of [math]1.251×10^9[/math] years.

Potassium-Argon dating measures the amount of argon trapped in rock. Originally molten rock has no argon. After molten rock solidifies, the argon from decaying Potassium is trapped. In primordial material, Argon-39 is dominant. Argon-40 dominates in the earths atmosphere.


Argon-36,38,40 are stable.

Argon-39 is the longest lived isotope with a half life of 269 years

Argon-39 decay:

[math]{39 \atop 18 }Ar_{21} \rightarrow {39 \atop 19 }K_{20} + e^- + \bar{\nu}[/math]

Strong Interaction

Applications

Homework problems

NucPhys_I_HomeworkProblems


Midterm Exam Topics list

Basically everything before section 5.3 (The Nuclear Force). Section 5.3 and below is not included on the midterm.

Topics of emphasis:

  1. 1-D Schrodinger Equation based problems involving discrete potentials ( wells, steps) and continuous potentials (simple Harmonic, coulomb).
  2. Calculating form factors given the density of a nucleus
  3. Determining binding and nucleon mass separation energies
  4. [math]\vec{I}[/math], [math]\vec{\ell}[/math], and [math]\vec{s}[/math] angular momentum operations
  5. Calculating scattering rates given the cross-section and a description of the experimental apparatus

Formulas given on test

Schrodinger Time independent 1-D equation

[math]\left ( \frac{- \hbar^2}{2m} \frac{ \part^2}{\part x^2} + V \right ) \psi = E \psi[/math]

Particle Current Density

[math]j = \frac{ \hbar}{2im} \left ( \Psi^* \frac{\part \Psi}{\part x} - \Psi \frac{\part \Psi^*}{\part x}\right )[/math]

Form Factor

If the density has no [math]\theta[/math] or [math]\phi[/math] dependence

[math]F(q) = \frac{4 \pi}{q} \int \sin(qr) \rho(r) rdr[/math]

Coulomb energy difference between point nucleus and one with uniform charge distribution

[math]\Delta E = \frac{2}{5} \frac{Z^4e^2}{4 \pi \epsilon_0} \frac{R^2}{a_0^3}[/math]

Nucleus Binding Energy

[math]B(^A_ZX_N) = \left [ Z m(^1H) + Nm_n -m(^AX) \right ]c^2[/math]

Neutron Separation energy

[math]S_n = B(^A_ZX_N) - B(^{A-1}_ZX_{N-1})[/math]

Proton Separation energy

[math]S_p = B(^A_ZX_N) - B(^{A-1}_{Z-1}X_N)[/math]

Semiempirical Mass Formula

[math]M(Z,A) = Z m(^1H) + Nm_n - B(^A_ZX_N)/c^2[/math]

where

[math]B(^A_ZX_N) = = \alpha_V A - \alpha_S A^{2/3} - \alpha_C \frac{Z(Z-1)}{A^{1/3}} - \alpha_{sym} \frac{(A-2Z)^2}{A} - \delta[/math]
Parameter Krane
[math]\alpha_V[/math] 15.5
[math]\alpha_S[/math] 16.8
[math]\alpha_C[/math] 0.72
[math]\alpha_{sym}[/math] 23
[math]\alpha_p[/math] [math]\pm[/math]34
[math]g_{\ell} =\left \{ {1 \;\;\;\; proton \atop 0 \;\;\;\; neutron} \right .[/math]
[math]g_{s} =\left \{ {5.5856912 \pm 0.0000022 \;\;\;\; proton \atop -3.8260837 \pm 0.0000018 \;\;\;\; neutron} \right .[/math]


Final

1.) Calculate the magnetic moment of a proton assuming that it may be described as a neutron with a positive pion [math](\pi^+)[/math] in an [math]\ell =1[/math] state.


2.) Show that the phase shift ([math]\delta_0[/math]) for the scattering of a neutron by a proton can be given by the equation

[math]\delta_0 = \tan^{-1} \left( \frac{\tan(k_1R) - \frac{k_1}{k_2} \tan(k_2R)}{ \tan(k_1R) \tan(k_2R)+\frac{k_1}{k_2}}\right )[/math]

where

[math]k_1^2 = \frac{2m(V+E)}{\hbar^2}[/math]
[math]k_2^2 = \frac{2mE)}{\hbar^2}[/math]

V = 36.7 MeV R = 2.1 fm

3.)

a.) Write the reaction equations for the following processes. Show all reaction products.

i.)[math]{226 \atop\; }Ra \; \; \; \alpha \mbox{- decays}[/math]

ii.)[math]{110 \atop\; }In \; \; \; \beta^+ \mbox {-decays}[/math]

iii.)[math]{36 \atop\; }Ar (2^{nd} 0^{+}) \mbox{internal conversion}[/math]

iv.)[math]{12 \atop\; }C (2^+) \; \; \; \gamma \mbox{- decays}[/math]

b.) Determine the Q-values for the first two reactions above.


4.) Find the Quadrupole moment of [math]{209 \atop\; }Bi(9/2^-)[/math] using the shell model and compare to the experimental value of -0.37 barns.


5.) Find [math]\frac{\mu}{\mu_{NM}}[/math], using the shell model, for the following nuclei

a.)[math] {75 \atop\; }Ge[/math]

b.) [math]{87 \atop\; }Sr[/math]

c.) [math]{91 \atop\; }Zr[/math]

d.) [math]{47 \atop\; }Sc[/math]

6.) Use the shell model to predict the ground state spin and parity of the following nuclei:

a.) [math]{7 \atop\; }Li[/math]

b.)[math] {11 \atop\; }B[/math]

c.) [math]{15 \atop\; }C[/math]

d.)[math] {17 \atop\; }F[/math]

7.) Tabulate the possible [math]m[/math] states for a nucleus three quadrupole phonon state ([math]\lambda = 3[/math] ). Show that the permitted resultant states are [math]0^+[/math], [math]2^+[/math], [math]3^+[/math], [math]4^+[/math], and [math]6^+[/math].

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