Forest NucPhys I

Advanced Nuclear Physics

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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

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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.

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

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

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

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)

Quantum Mechanics Review

• Debroglie - wave particle duality

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]\lt f\gt = \int \Psi^* f \Psi dx =\lt \Psi^*| f| \Psi\gt [/math] = average (expectation) value of observable [math]f[/math] after many measurements of [math]f[/math]

example: [math]\lt p_x\gt =\int \Psi^* \left ( -i\hbar \frac{\partial}{\partial x}\right ) \Psi dx[/math]

• Constraints on Quantum solutions

1. [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)
2. the solution is normalized:[math]\int \psi^* \psi dx =\lt \psi^* | \psi \gt =1[/math]

• Current conservation: the particle current density associated with the wave function \Psi is given by

[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:

[math]-\frac{\hbar^2}{2m} \nabla^2 \Psi(x,t) + 0 = i\hbar\frac{\partial \Psi(x,t)}{\partial t}[/math]

Using separation of variables Let:

[math]\Psi(x,t) = \psi(x) f(t)[/math]

Substituting we have

[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]

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]

Solving the temporal (t) part:

[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.

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 = \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]\lt \Psi(x,t) | \Psi(x,t)\gt = \lt \psi(x)| \psi(x)\gt \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]\lt \Psi(x,t) | A | \Psi(x,t)\gt = \lt \psi(x)| A| \psi(x)\gt \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 \gt 0[/math]

[math]V(x) =\left \{ {0 \;\;\;\; x \lt 0 \atop V_o \;\;\;\; x\gt 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 =\frac{2mE}{\hbar^2}[/math] and [math]\omega =\frac{E}{\hbar}[/math]

x>0

[math]-\frac{\hbar^2}{2m} \nabla^2 \Psi(x,t) + V_o = i\hbar\frac{\partial \Psi(x,t)}{\partial t}[/math]

[math]-\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V_o = i\hbar\frac{\partial \Psi(x,t)}{\partial t}[/math]

separation of variables: [math]\Psi(x,t) = \psi(x)f(t)[/math]

[math]\frac{\hbar^2}{2m} f(t)\frac{\partial^2 \psi(x)}{\partial x^2} - V_o\psi(x)f(t) = -i\hbar \psi(x)\frac{\partial f(t)}{\partial t}[/math]

Homework problems

NucPhys_I_HomeworkProblems