nuclear physics - · pdf filenuclear physics j.pearson april 28, 2008 abstract these are a set...

Nuclear Physics

J.Pearson

April 28, 2008

Abstract

These are a set of notes I have made, based on lectures given by D.Cullen at the University ofManchester Jan-June ’08. Please e-mail me with any comments/corrections: [email protected].

CONTENTS 1

Contents

0.1 Basic Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1 Nuclear Properties 1

1.1 Nuclear Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Experimental Methods to Measure Nuclear Distribution . . . . . . . . . . . . . . . . 2

1.2.1 Nuclear Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.2 Nuclear Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.3 Total Reaction Cross-Section . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Differential Cross-Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.4 Distribution of Nuclear Charge ρc(r) . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4.1 Electron Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4.2 Electron Scattering Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4.4 Conclusions From e Scattering on Stable Nuclei . . . . . . . . . . . . . . . . . 10

1.5 ρc(r) From Perturbation of Atomic Energy Levels . . . . . . . . . . . . . . . . . . . . 11

1.5.1 X-ray Isotope Shifts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.5.2 Isotope Shifts for Optical Transitions . . . . . . . . . . . . . . . . . . . . . . . 13

1.5.3 Muonic Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6 Nuclear Matter Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6.1 Elastic Scattering of p, α, π . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6.2 Pionic Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.6.3 α-barrier Penetration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Nuclear Models 17

2.1 The Liquid Drop Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.1.1 The Semi-empirical Mass Formula . . . . . . . . . . . . . . . . . . . . . . . . 17

2.1.2 SEMF Predictions & Explanations . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1.3 Microscopic Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

CONTENTS 2

2.2 The Shell Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.2.1 Solving Schrodinger Equation for Simple Central Potential . . . . . . . . . . 23

2.2.2 Spin-Orbit Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.2.3 Filling Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.2.4 Predictions of Ground State Spins & Parities Iπ . . . . . . . . . . . . . . . . 26

2.2.5 Excited States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Collective Excitations 29

3.1 Nuclear Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.1.1 Two Phonon States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Nuclear Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.2.1 Types of Deformed Shape . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2.2 Energies of Deformed Shapes . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2.3 Predictions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2.4 Excited Rotational Bands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2.5 Population of Excited States . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2.6 Information from γ-ray Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . 35

4 α-decay 36

4.1 Theory of α-emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.2 Fine Structure of α-decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5 β-decay 41

5.1 Q-values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.2 Fermi Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

5.2.1 β-decay Half-lifes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

5.3 Angular Momentum Rules for β-decay . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5.3.1 Allowed Decays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

5.3.2 First Forbidden Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5.4 Parity Violation in β-decay Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6 γ-ray Emission 46

CONTENTS 3

6.1 Angular Momentum Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6.2 Parity Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

6.2.1 Angular Momentum Considerations . . . . . . . . . . . . . . . . . . . . . . . 48

6.3 Parity Rule for E1 & Transition Rates . . . . . . . . . . . . . . . . . . . . . . . . . . 48

6.4 Single Particle Weiskopf Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

6.5 Internal Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

1 NUCLEAR PROPERTIES 1

0.1 Basic Terminology

Let us just start with basic terminology.

We write:AZXN

To denote an atom/nucleus. We say that Z is the atomic number, which is the number of protons,which is also the same as the number of electrons. A is the mass number of the nucleus, which isthe integer ratio of nuclear mass and fundamental mass unit u. Where u is defined such that 12Chas atomic mass 12.0000u. Note, we generally neglect the mass of the electrons. So, as examples,we write 1

1H0, 23892 U146; however, we usually just write 238U , as that we wrote U implies Z = 92.

1u = 1.66× 10−27kg

Nucleon Refers to either neutron or proton, when we dont want to specify which;

Isotope Two nuclei which have the same Z, but different N . That is, the same proton number,but different number of neutrons.

Isotone Converse: different Z, same N . For example, 21H1 and 3

2He1.

Isobar Nuclei with the same mass number A. For example, 3He and 3H. Note, we generallywork in MeV, as opposed to J, when we quote energies as opposed to masses.

1u = 931.502MeV/c2

1 Nuclear Properties

There is no simple theory which encompasses all nuclei: many models work in different situations.However, we can characterize nuclei by a relatively small number or parameters; such as radius,electric charge, mass, binding energy and energy of excited states.

1.1 Nuclear Size

Rutherford’s α−scattering experiment at Manchester. He found that almost all of the mass ofan atom (the nuclear mass) is concentrated in a very small volume at its centre, and had highelectric charge, which was bound by the short range strong interaction. He curiously found someα−particles bouncing back off the nucleus: it is like ‘an artillery shell bouncing off tissue paper’.Typical atomic radius is 10−10m, and proton radius 10−15m = 1fm.We shall see later that all nucleu have a similar density. Hence, nuclear volume:

43πR3 ∝ A


Where A is the atomic mass; experimentally, found to be:

R = r0A1/3 r0 ≡ 1.2fm

So, for example, for 238U , r = 7.4fm.

1.2 Experimental Methods to Measure Nuclear Distribution

1.2.1 Nuclear Charge

This will all be due to protons. We use the Coulomb interaction between the nucleus and a particleparticle/probe (such as e−, µ−). The EM interaction is exactly understood. There will be nocontribution from the Strong force.

1.2.2 Nuclear Matter

This will look at both protons & neutrons. We use probes which interact with the strong (nuclear)force (e.g. p, α, π).If the probe has sufficiently high energy, we say that it has overcome the Coulomb force, and wecan hence neglect it.

To go further, we need to discuss the idea of cross-section; which is the same as reaction probability.

1.2.3 Total Reaction Cross-Section

Early methods from the 1960’s are now used to very exotic nuclei; i.e. ones that are very shortlived, have small t 1

2. An example of such an exotic nucleus, is that of 11Li. It has t 1

2≈ 8.7ms. The

2 extra neutrons in going from the isotope 9Li to 11Li has the effect of creating a ‘halo’ - a so-calledhalo nucleus - which increases the size of the nucleus to something comparable with 208Pb.

Consider the classical collision between two spheres of radius R1, R2. We shall ignore any Coulombinteraction. Let their centres be a distance b apart. We call b the impact parameter. So, whathappens depends on b:

• If b > R1 +R2, then no interaction;

• If b < R1 +R2, then some interaction occurs.

Possible interactions are elastic scattering, exchange of nucleons (a transfer interaction), or fusionof nuclei.All of these processes remove particles from the beam/path.Let us define a geometric limit as being the effective area that will remove a particle from the beam:

σT = π(R1 +R2)2


Let us have a flux of N0 particles per second in a beam. Let us have a slab of target, thickness x,exposed surface area A. Each target nucleus has the effective area σT . The target has n nuclei perunit volume. Let the number of nuclei coming out of the target be N0 −∆N , that is, the numberthat interact and do not get through is ∆N . The interaction rate is thus ∆N .So, the total effective area across target is σT× no.nuclei in target:

= σTnxA

So, the probability of beam particle not getting through the thin target (i.e. that of interacting) isthe total effective area/total area

total effective areatotal area

=σTnxA

A= σTnx

Hence, the reaction rate ∆N is N0× probability of interaction:

∆N = N0σTnx (1.1)

Experiment measures N0,∆N,n, x; hence, we deduce the cross-section σT . σT has the units of area.Let us do a quick order of magnitude calculation.Let us use 11Li, so we have R = r0A

1/3 ≈ 3fm. Hence, σT is about:

σT = π(R1 +R2)2

≈ π(6× 10−15m)2

≈ 100× 10−30m2

= 10−28m2

So, we use a unit of nuclear reactions for cross-section:

1barn = 10−28m ≡ 1b

Now, the nuclear surface wont be well defined. There are a few ways to parameterize what we mean

Figure 1: A density profile for a general nucleus. The dotted line is the ‘ideal’ case, which wouldclassically be expected, but this is not the case,

by the nuclear surface:


• The point where ρ(r) ≈ ρ(0)2 ;

• By skin thickness: where the density drops from 90% to 10%;

• Use the mean square radius:

〈r2〉 =∫r2ρ(r) dτ∫ρ(r) dτ

This is equal to 35R

2 for a billiard ball.

For 11Li, the profile is even more convoluted. The 2n in the ‘halo’ completely mess things up!

Figure 2: A charge density profile for 9Li and 11Li. Notice the difference. The extra 2n sit very farfrom the core: neutron-skin. This is an example of ‘neutron rich material’.

1.3 Differential Cross-Sections

These are denoted by dσdΩ , and are used to describe the distribution of scattered particles. So, if dσ

is the effective area of the target nucleus that scatters a particle into a cone, which has solid angledΩ in a direction (θ, φ); where θ is measured from the direction of incidence of the beam, and φ theangle of rotation around the beam.Normally, there is no φ-dependance, unless the beam is polarised. So, dσ

dΩ is a function of θ only,So, the total scattering cross section is just:

σ =∫dσ

=∫ 2π

φ=0

∫ π

θ=0

dσ

dΩdΩ

=∫ 2π

φ=0

∫ π

θ=0

dσ

dΩsin θdθdφ

Using spherical polars.


Figure 3: The radial charge distribution of various nuclei, found from electron scattering. The skinthickness is shown. Notice how the central density is approx constant.

1.4 Distribution of Nuclear Charge ρc(r)

The usual method to determine the size/shape of an object is to examine the radiation that scattersfrom it. We can only observe details that are of the same order of the de Broglie wavelength used.e.g. to see nuclear distribution using e−, we need λ ≤ 10fm. Which, via:

λ =h

p=hc

E

Is an energy E ≥ 100MeV. Typically however we use E ≈ 1GeV.We analyse with a precise spectrometer the electrons that are elastically scattered from the nucleus.

1.4.1 Electron Scattering

This is a direct analogy to light scattering from a 2D black disc. We would get an interferencepattern out, with first minimum at θ = 1.22 λ

D , where D is the diameter of the disc.

As is shown in Fig (5) (b), we see that the electron-scattering intensity dosent go to zero. This isbecause the nucleus is not an opaque disc, and that the nuclear surface is not well defined. However,we can still extract nuclear radii in the same way.

From these types of experiment, we get 2.6fm and 2.3fm for 16O and 12C, respectively. Comparewith 3fm and 2.7fm as predicted by the r0A

1/3 method.


Figure 4: The differential cross-section. A beam of particles is incident upon a nucleus, with effectivecross-section dσ. Beam particles are scattered into a cone of solid angle dΩ, which is inclined at anangle θ to the incident direction. Note how various angles are defined.

1.4.2 Electron Scattering Theory

We shall neglect the spin of the electron, and only look at Coulomb effects.Consider:

Mif =∫ψ∗fV (r)ψi dτ

So that e− scattering probability is governed by the overlap integral, containing initial and finalstates of the electron, and scattering potential V (r).It can also be thought of in terms of V (r) acting on the initial wavefunction ψi and converts it toa superposition of outgoing scattering waves. Mif then selects amplitude of the component ψf .The scattered intensity has the relation:

dσ(θ)dΩ

∝M2if

So, suppose that the incoming momentum of electrons is given by pi = hki, and outgoing bypf = hkf , so that the plane waves are:

ψi = e−iki·r ψf = e−ikf ·r

Using the concept of momentum transfer, and having q = ki − kf . and that for elastic scattering|ki| = |kf | = k, say. It can be shown that:

q = 2k sinθ

2

We derive it by considering the cosine rule. Consider:

q2 = (ki − kf )2

= 2k2 − 2k2 cos θ

Where we have used that ki = kf = k. Then, from the trig identity cos 2θ = 1− 2 sin2 θ, the abovebecomes:

q2 = 4k2 sin2 θ

2


Figure 5: Intensity against scattering angle. (a) is for light off a disc diameter D, and (b) forelectron scattering off a nucleus. The minima for electron scattering dont go to zero as the nucleusis not an opaque disc, as was the case for light scattering.

Square-rooting gives the desired result:

q = 2k sinθ

2

So, combining all this, we have:

Mif =∫V (r)eiq·r dτ

Which is just the 3D Fourier transform of the potential. So:

dσ(θ)dΩ

∝ |Mif |2 =∣∣∣∣∫ V (r)eiq·r dτ

∣∣∣∣2Where we generally use spherical polars, so that dτ = r2 sin θdrdθdφ.

1.4.3 Solutions

We consider various solutions to the above Fourier transform, and forms of the potential.

Special Case: Point Nucleus Here we consider the Coulomb potential energy between theprobing electron and the nucleus; that is:

V (r) =Z1Z2e

2

4πε0r


Figure 6: Electron scattering. Notice, the shape is similar to that obtained from diffraction experi-ments.

Where the Zi are the charges of the probe and target. So, for an electron we have Z1 = 1 and fora nucleus Z2 = Z. Hence, inserting this into our expression for the matrix element:

dσ(θ)dΩ

∝M2if =

[Ze2

4πε0

∫eiq·r

rdτ

]2

We quote the result to be:dσ(θ)dΩ

∝(Ze2

q2

)2

∝ 1sin4 θ

2

As we have that q = 2k sin θ/2.

So, let us consider back-scattering: i.e. when θ = 180. Intuitively, we expect this to have minimumprobability of happening:

dσ(180)dΩ

= 1

Which is a miniumu.


Real - Extended - Charge Distribution We need to consider the potential that the electronfeels, at position r due to an element of charge dQ at r′. That is:

V (r) =Ze2

4πε0

∫ρc(r′)|r − r′|

dτ ′

We have that the charge density is nomalised; that is:∫ρc(r′)dτ ′ = 1

Hence, our matrix element:

Mif =∫V (r)eiq·r dτ

=Ze2

4πε0

∫ ∫ρc(r′)|r − r′|

dτ ′ eiq·r dτ

Which, after a load of algebra (which we dont do), we get to:

Mif =Ze2

ε0

1q2

∫eiq·rρc(r′)dτ ′

So, we see that we get the product of what we had for our previous point nucleus, and the Fouriertransform of the charge density (the integral on the right). We call the Fourier transform the electricform factor of the nucleus F (q).We recall that dσ(θ)

dΩ ∝ M2if . Hence, we have that the probability of scattering an electron at an

angle θ is just given by:dσ(θ)dΩ

∣∣∣∣real

=dσ(θ)dΩ

∣∣∣∣point

F (q)2

That is, the actual cross-section is the product of the Rutherford (point nucleus) cross section withthe square of the form-factor.

Note:(1) We can measure dσ(θ)

dΩ , which can fall below the calculated Rutherford cross-section (i.e. thatfor a point-nucleus); which tells us about the Fourier transform of the charge distribution.(2) F (q) depends on the electron beam momentum, and scattering angle θ, only through momentumtransfer, via q = 2k sin θ/2. So, we can combine data from various energies (and labs) by plottingdσ(θ)dΩ vs q.

(3) We can get the charge distribution ρc(r′) by finding the inverse Fourier transform of the form-factor. That is, we measure:

F (q) = Fρc(r′)

Hence, we can calculate the inverse:

ρc(r′) = F−1F (q)

Note however, in principle, to invert properly, we need to know q up to infinity; which is obviouslyimpossible. Hence, we get experimental errors due to this, in ρc(r′) at smal r.


Figure 7: The plot on the left is the measured differential cross-section (for 450MeV electronsscattered by 58Ni), plotted as a function of q, where q = 2k sin θ/2; which is proportional to theFourier transform of the charge density, which is on the right, after finding the inverse Fouriertransform of the left.

1.4.4 Conclusions From e Scattering on Stable Nuclei

ρc has a similar central density in all nuclei; That is, the number of nucleons per unit volume isapproximately constant. Thus:

A

V=

A43πR

3= const

From which we get R = r0A1/3; and measurements give us r0 = 1.2fm.

We are also able to see that the nuclear surface is quite diffuse (FIG3.4); that is, the surfacethickness (going from 90% to 10% of ρ(r = 0) is about 2.3fm in all stable nuclei.

We are able to find the root-mean-square (rms) radius 〈r2〉1/2 directly from scattered electrons.Plotting R against A1/3 gives us a slope 1.2fm.We see:

〈r2〉 =∫r2ρc(τ)dτ

That is:〈r2〉 = 4π

∫ ∞0

r4ρc(r)dr


1.5 ρc(r) From Perturbation of Atomic Energy Levels

The nucleus is not a point source. Thus, this modifies the Coulomb potential in which e moves,but only within the small volume of the nucleus. An e spends some of its time inside the nucleus,where it ‘feels’ a very different charge distribution.

Figure 8: The deviation from ideal point charge, for inside the nucleus. Shown on the figure is thepotential: outside (i.e. r > R), we see that it is just equal to the point charge. Inside, we see thatit deviates off, with energy shift ∆E. The potential inside is denoted V ′(r).

This effect reduces the binding energy of an atomic electron, by ∆E. Let us make a rough approx-imation:

Assume the nucleus is a uniformly charged sphere with radius R. Thus:

V ′(r) = − Ze2

4πε0R

(32− 1

2r2

R2

)r ≤ R

And outside:

V0(r) = − Ze2

4πε0rr > R


The energy E of an electron in a state ψe depends partly on the expectation value of the potential:

〈V 〉 =∫ψ∗eV (r)ψe dτ

Now, we use first order perturbation theory, as the electron wavefunction varies slowly over thenuclear volume. Hence:

∆E =∫ψ∗e(r)V ′(r)− V0(r)ψe(t) dτ

The effect will be largest for the wavfunction which penetrates most into the nucleus: the 1swavefunction:

ψ(r) = 2(Z

e2

)3/2

e−Zr/a0

Hence:

∆E =Ze2

4πε04Z3

a30

∫e−2Zr/a0

1r− 3

2R+

12r2

R3

dτ

Now, the exponential argument is very small: so, we take its term to be 1. Also, dτ = 4πr2dr.Hence:

∆E =Ze2

4πε0

4Z3

a30

4π∫ R

0r − 3r2

2R+

12r4

R3dr

Which easily gives:

∆E =25Z4e2

ε0

R2

a30

However, just to make things more ‘correct’, R is actually just an expectation value. Hence:

∆E =25Z4e2

ε0

〈R2〉a3

0

(1.2)

Thus, we have that ∆E is the difference in energy of a 1s state in an atom with a point nucleus,and that for a 1s state in an atom with a uniformly charged nucleus with radius R.

However, we dont have an atoms with point nuclei, so we cannt directly measure ∆E, and hencededuce R. So, the next best thing would be to calculate the point value E, from the 1s atomicwavefunction.However, again, we have a problem: atomic wavefunctions are not known to enough accuracy to beable to do this.

∆E is of the order 10−4E, which is very small!

Thus, we cannot use this method to find ρc(r). However, there are 3 methods that can be used:

Consider an electron transition from the L-shell to K-shell (that is, 2p to 1s). This will liberate aphoton: an X-ray, whose energy we can measure.If we set up a Muonic atom (muons orbiting instead of electrons), we can measure the energy dropsfor muonic X-rays.We can excite an atom using lasers: laser flouresence spectroscopy.


1.5.1 X-ray Isotope Shifts

Here, we make a comparison of the rms 〈r2〉 between different isotopes A and A′ (remember, sameZ, different N). So, we can measure:

δE = h(νA − νA′)

That is, the measured shift of X-ray energy (K X-rays are produced by electron transitions fromthe 2p down to 1s state) between two isotopes; such as 200Hg and 202Hg.Now, we previously showed, from first order perturbation theory, that:

∆E =25Z4e2

ε0

〈R2〉a3

0

So, if we have ∆EA being the difference in the 1s state in atom A, with point nucleus, and uniformlyextended charge distribution. Similarly for ∆EA′ . Hence:

δE = h(νA − νA′)= ∆EA −∆EA′

=25Z4e2

ε0

1a3

0

R20(A2/3

A −A2/3A′ )

Where we have used that 〈R2i 〉 = (R0A

1/3i )2. So, we have subtracted out the point charge distribu-

tion. The shift is, however, very small: ≈ 10−6.

Typlically, δE = 0.15eV; and the transition energy (energy of the X-ray) is of the order 100keV

We also see that 〈R2〉 increases with N , that is, the proton distribution spreads out, as neutronsare added, because of the strong interaction between p, n in the nucleus.

So, X-ray isotope shifts work, but their shift is very small.

1.5.2 Isotope Shifts for Optical Transitions

Here we consider transitions which result in optical wavelengths, in the valence electrons. We seethat ∆E is smaller, as:

|ψ6s(r = 0)|2 << |ψ1s(r = 0)|2

That is, less overlap of wavefunction with the nucleus. Here, δE is typically 10−6eV, and E about3eV. Thus, greater shift detection.

The shift can be measured using laser spectroscopy, to great precision. Tune a laser, to excitefrom 6s to 6p (say), then as the electron relaxes, it emits a photon, which can be picked up byphotomultiplier tubes. As the frequency of the laser is scanned through, a resonance spectrum isseen; and two different isotopes will have slightly different excitation energies, and will hence beseparated slightly (for the same transition) on such a spectral plot. Hence the shift can be measured.

The laser method has its advantages, as it is the only method that can be used on short livednuclei; such as 11Li, which has t1/2 = 8.7ms.The method almost has single atom sensitivity.


1.5.3 Muonic Atoms

The problems with the previous two methods are the the effects are small. This is not the case ifwe use a muonic atom:

Get an atom, strip it of all its electrons, and fire a µ at it. The muon will spiral down, throughBohr-like orbits, emmiting photons as it goes.

Now, mµ ≈ 207me; hence, as the Bohr radius (for an electron) is given by:

a0 =1α

h

mec

Then we have that the equivalent, for a muon, is:

rµ =a0

207

So, for example, 82Pb, the µ 1s orbit is inside the nuclear surface. That is:

|ψµ1s(r = 0)|2 >> |ψe1s(r = 0)|2

Here, the X-ray shifts are of the order 130 , which is huge compared with previous methods.

Futhermore, wavefunctions are easier to calculate accurately enough, to be able to derive an abso-lute 〈r2〉 for the nucleus.

Muonic X-rays are 1MeV in Fe, and 5.5MeV in Pb; which are comparable to gamma-rays, whosedetectors are very accurate.

Infact, when spin-orbit coupling is taken into account, split states are visible.

1.6 Nuclear Matter Distributions

To do this, we need a probe that will interact, via the strong nuclear force, to be able to ‘see’ bothp, n. Such as proton, alpha-particles and pions.

1.6.1 Elastic Scattering of p, α, π

Here, we have a scattering potential, which is ‘Coulomb’ outside, and a deep attractive well inside.

So, the Coulomb potential, for target nucleus Z1, probe Z2 is just:

Vc =Z1Z2e

2

4πε0r= 1.44

Z1Z2

rMeV

If r in fm. So, if r > R (i.e. probe energy is less than Coulomb repulsion), then we get the standardGauss’ law; and the scattering particle behaves as if it scattered off a point like nucleus - Rutherfordscattering.


Figure 9: Scattering potential for a probe incident upon a target nucleus. Notice, outside, it is aCoulomb potential; and inside a (complicated!) attractive nuclear potential.

If the probes energy is greater than the Coulomb repulsion, then the strongly attractive force comesinto play. At this point, we expect deviation from the Rutherford formula, as other outcomes arepossible - such as adsorption - which are likely to reduce the probability of elactic scatter (σR).

Note, the analysis of data is far more complicated than for electron scattering, due to the complexnature of the strong nuclear force (the strong interaction).

What we ‘see’ depends on the wavelength λ of the probe:

E =hc

λ

If we use a large wavelegth, small energy; then we see the matter radius;At small λ, large E, start to see more detail.

The patterns we get on detection screens is akin to Fraunhoffer patterns in Optics: which providesinformation on the scattering surface, the internal density distribution.

Some notation: 40Ca(α, α) at 115MeV.Reading from left to right: target nucleus, incoming particle (probe), outcoming partcile, done atthe energy.

1.6.2 Pionic Atoms

This is another example of X-ray analysis.

Basicaly, let pions get captured by a nucleus, and as they spiral in towards the nucleus, they emitphotons. This time however, pions (as they are mesons) will interact with both the Coulomb &


Figure 10: Possibilities for nuclear scattering. The dotted lines show direction of increasing probeenergy E, and impact parameter b (i.e. the distance between centre of nucleus & probe). Thedifferent trajectories correspond to: 1) Fusion; 2) Scattering, with direct & indirect interaction;3) Coulomb interaction, elastic & inelastic scattering; 4) Mainly Coulomb interaction, Rutherfordscattering.

strong forces (c.f. just Coulomb for muons). The energy shifts are larger. The π-mesons emitπ-mesic X-rays as they cascade through the electron-like oribits.

When the π wavefunction begins to overlap with the nuclear w.f., then the energy levels are shiftedfrom pure Coulomb values (calculated using pure Coulomb interaction), and the π-mesons becomeabsorbed directly into the nucleus, from the inner orbits.

Hence, we get a dissapearence rate of π-mesons; hence another way to determine the nuclear radius.

1.6.3 α-barrier Penetration

We merely state that this method exists, but we will discuss later.

1.7 Summary

So, considering stable nuclei (i.e. sit along the line of stability):

• Charge & matter radii are very similar;

• The n− p attraction (via strong interaction) keeps nucleons in the same place;

2 NUCLEAR MODELS 17

• Heavy nucleu have about 50% more neutrons than protons, so, we may expect their neutronradius to be larger than proton radius. However, p repulsion pushes p out, dragging n withit; hence, neutrons and protons are mixed up; resulting in both the charge and matter radiibeing nearly equal.

Nuclei with large n excess appear to form n-rich skin; even a ‘halo’, as in 11Li.

The central density in all stable nucleu is similar;

All have a similar diffuse surface.ρ(r) =

ρ(0)

1 + er−R

a

Where a = 0.6fm.

We have the relation R = 1.2A1/3fm.

2 Nuclear Models

2.1 The Liquid Drop Description

This is a classical, macroscopic description of the nucleus, however, the SEMF will use some elementsof quantum theory in its formulation.

2.1.1 The Semi-empirical Mass Formula

- SEMF

The binding energy BE is defined as the energy needed to break an atom into its constituents.

So, an atom, plus BE will give Zp, Nn and Ze; which we will simplify to just ZH + Nn (i.e. neglectBE of electrons, which is 13.6eV, compared with MeV for nucleons; and where Zp = ZH). The BEis dominated by the nuclear contribution. Usually, we write:

M(Z,A) = ZMH+NMn − 1

c2BE(Z,A) (2.1)

Since the BE increases with A, we generally plot BE/A as a function of A. See plot.

The SEMF attempts to parameterise this shape by a series of 6 (including mass) terms:

BE(Z,A) = aVA− aSA2/3 − aCZ(Z − 1)A−1/3 − asym(A− 2Z)2

A+ δ (2.2)

We have, in order:

• The volume term: The nucleons only interact with near-neighbours. If each nucleon has thesame number of nearest- neighbours, then the BE has such a dependancs;

• The surface term: This takes into account that there are nucleons on the surface, which areless tightly bound. Surface area ∝ R2; and we have seen that R ∝ A1/3.

2 NUCLEAR MODELS 18

• The coulomb term: We have that the coulomb repulsion between protons, at a distance R(∝ A1/3) is just given by the ‘test’ charge, and whatever is left. That is, Z and (Z−1) protons.

• The symmetry term: This acts to minimise the difference in the number of protons andneutrons; i.e. Z = N preffered.

• The pairing term: We will come to this later. However, δ = 34A−3/4MeV; and we use +δ foreven (both Z,N even) nuclei; 0 for odd (odd-even or even-odd) A nuclei; or −δ for odd-oddnuclei. This is due to nucleons ‘liking’ to couple into pairs with zero spin.

2 NUCLEAR MODELS 19

Figure 11: The breakdown of Rutherford scattering. When incident α particles get close enoughto the target (here, Pb), they interact through the nuclear force, hence the Rutherford formula nolonger holds. The point at which breakdown occurs gives a measure of the size of the nucleus.

2 NUCLEAR MODELS 20

Figure 12: Binding energy per nucleon. Notice that helium has a disproportionally high BE; thatiron is near the peak; that most have BE/A around 8MeV

2 NUCLEAR MODELS 21

2.1.2 SEMF Predictions & Explanations

Odd Mass For a given odd mass A, there is only one stable isobar (same neutron, differentproton). We can show this by computing:

∂M

∂Z

)A

= 0

That is, at fixed A. Which gives:

Zmin =A

21

1 + 14A

2/3 aCasym

Thus, for a constant A, M(A,Z) is a quadratic function of Z.Suppose the position of the minimum is Zmin, then, for Zmin + 1, whatever nuclei exists there willwant to change a proton into a neutron; and will do so via β+ decay, which is just e-capture. For anucleus at Zmin − 1, a neutron will change into a proton, via β− decay; that is, electron emission.

Even Mass For a given even A, there can be several even-even stable isobars, due to the pairingterm δ in the SEMF.

Figure 13: Even mass nuclei decay chains. Decays to the ‘right’ go via β+, and the ‘left’ by β−.The two parabolas are sparated by 2δ.

Energy From Fusion Consider:2H +2 H →4 He

The binding energy is:

BE(2H) = 2× 1.1MeV/A = 2.2MeV

BE(4He) = 4× 7MeV/A = 28MeV

2 NUCLEAR MODELS 22

Where these numbers have been found from data. So:

2H +2 H →4 He+ 23.6MeV

Where the 23.6MeV comes from mf −mi = 28− (2× 2.2) = 23.6MeV. Thus, in the above reaction,23,6MeV is released.

Energy From Fission If we look at the binding energy curve, we see that nuclei with A = 60 aremost tightly bound - they are at the top of the plot. So, when 238U fissions (splits up), it produces2 more strongly-bound nuclei. Hence, a releas of energy. Consider:

238U →100 A1 +138 A2

Then, we see that, from BE data:

BE(238U) = 238× 7.5MeV/A

BE(100A1) = 100× 8.2MeV/A ≈ BE(138A2)

So, this will release (100× 8.2 + 138× 8.2)− (238× 7.5) = 166MeV.

α-decay In heavy nuclei, the disruptive Coulomb energy increases at a faster rate (≈ Z2) thannuclear BE with A. Hence, emission of 4He nucleus of a favoured way to reduce charge.

The BE for an alpha particle is about 7MeV/A; which is relatively high, for something with so fewnucleons.

A Q-value is the difference in masses before and after:

Q =∑i

minitali −

∑i

mafteri (2.3)

2.1.3 Microscopic Effects

These are effects which are not explained by the liquid drop model. These show that we needquantum/shell structure of nucleus:

Systematic Deviation in the smooth BE curve, from the SEMF: there is extra binding energyfor nuclei with Z or N being 28, 50, 82, 126.

Proton & Neutrons Separation Energies denoted Sn, Sp, S2nS2p - equivalent to the ionisationenergy of atomic electrons. Where S2n is the energy needed to pull a neutron pair from the nucleus- hence, a measure of the BE of last pair of n. We see that S2p greatest where N,Z are 8, 20, 28,50, 82, 126.

2 NUCLEAR MODELS 23

Energy Required to Excite Nucleus to first excited state. We see that significantly moreenergy is required when N is 8, 20, 28, 50, 82, 126

So, we see a pattern of the same ‘magic numbers’.

2.2 The Shell Model

Available evidence suggests that protons & neutrons fill orbits & shells similar to atomic electrons.We may initally think that a nucleon could not orbit independantly in dense nuclear interior.Internal scatterings are forbidden by the Pauli exclusion principle, as the nucleons are all belowtheir ‘fermi surface’.

Indeed, it has been possible to map out the density distribution, |ψ|2, for the last (i.e. 82nd) protonin 205

82 Pb, by comparison of charge radius ρc(r) for 20582 T l and 206

82 Pb by electron scattering.It turns out that |ψ|2 is very close to that of predicted 3s orbit which actually peaks in nuclearmaterial at r = 0.

2.2.1 Solving Schrodinger Equation for Simple Central Potential

Let us assume that each nucleon move independantly in a poential V (r); which we will take to bethe average interaction with all other nucleons.

Figure 14: The Woods-Saxon potential model.

The potetial is called a Woods-Saxon shape; and is similar to the ρ(r) distribution. We canparameterise the shape thus:

V (r) = − V0

1 + er−R

a

2 NUCLEAR MODELS 24

Where a is the surface diffusiveness parameter - which we have looked at previously.

The single particle levels are calculated by solving the 3D Schrodinger equation:− h2

2m∇2 + V (r)

ψ(r) = Eψ(r)

We apply the usual solving methods: separation of variables. If V (r) has no θ, φ dependance, thenwe have a central potential exactly like the H atom:

ψn`m(r, θ, φ) = Rn,`(r)Y`m(θ, φ)

Where we have the usual spherical harmonics, and rules for the various quantum numbers:

n = 1, 2, 3, . . .` = 0, 1, . . .m = −`, . . . , 0, . . . , `

For the nuclear potential, we have solutions which are similar to the 3D harmonic oscillator HO.We see that if we count up all the nucleons in a particular state, up to that state, we start to get

Figure 15: Occupancies of the energy levels in the HO problem. Notice the ‘magic numbers’.

the ‘magic numbers’ fall out: 2, 8, 20. However, the next predicted is 40; however, experimentally,we get 28. So, this model is only of limited use in explaining whats going on.

So, we have made an approximation to the Woods-Saxon potential, using the HO potential; butwe have seen that the prediction breaks down for high ` states. The higher ` states feel a deeperpotential; thus, nucleons are bound more. We chose V0 to reflect the density disribution; so,V0 ≈ −35MeV.We should add in Coulomb energy term for protons, but this is a small effect, especially in lightnuclei; so, we ignore it.

2 NUCLEAR MODELS 25

Figure 16: The difference between the Woods-Saxon potential, and the approximate harmonicoscillator potential.

2.2.2 Spin-Orbit Interaction

Early on, it was apparent that the a spin-orbit term was present. There is a direct analogy to atomicphysics, but the direction is reversed. We consider the interaction that a nucleon has with it self,when it is orbiting, with orbital angular momentum, and ‘rotating’, with spin angular momentum.We see that s parallel to ` is favored. So, we must add a term (−VSO` · s) to the Hamiltonian.The result of which is that each nucleon orbit ` (except the s-state, where ` = 0) is split into 2components, labelled by the total spin j = `+ s.

So, we have a state being split; with total energy splitting ∆E, and each level being split by ε′j(j = `− s, i.e. up) and εj (j = `+ s i.e. down) from the centroid. Note that s = 1

2 . So, we requirethe energy shifts εj = −VSO〈` · s〉. So:

j = ` + s⇒ j2 = `2 + s2 + 2` · s⇒ ` · s = 1

2(j2 − `2 − s2)

⇒ −VSO〈` · s〉 = −VSOh2

2[j(j + 1)− `(`+ 1)− s(s+ 1)]

Hence, for the ε′j shift (i.e. that for j′ = `− 12), we have:

ε′j =VSO

2[(`− 1

2)(`+ 12)− `(`+ 1)− s(s+ 1)]

Similarly:

εj =VSO

2[(`+ 1

2)(`+ 32)− `(`+ 1)− s(s+ 1)]

Thus:

∆E = ε′j − εj =VSOh

2

2(`+ 1

2)[(`+ 32)− (`− 1

2)]

2 NUCLEAR MODELS 26

After we have used that s = 12 . Hence, this is just:

∆E =VSOh

2

2(2`+ 1) (2.4)

The introduction of the S-O interaction is able to account for the experimental shell closures at 2,8, 20, 28, 50, 82, 126.

NOTE: all magic numbers (i.e. shell gaps) above N,Z = 20 are produced by S-O splitting.

2.2.3 Filling Orbits

The shell model uses the standard quantum-numbers system, except that n is the number of timesa particular ` has occurred in the level sequence.

As examples, we have:1s1/2: ` = 0, j = `+ s = 1

2 .1p3/2: ` = 1, j = 1 + 1/2 = 3/2.

We also have parity π = (−1)`. We also use the standard ` = 0, 1, 2, 3, . . . or s, p, d, f, . . . system.Shells are filled according to the Pauli principle.

2.2.4 Predictions of Ground State Spins & Parities Iπ

We assume the pairing interaction couples pairs of protons and pairs of neutrons into Iπ = 0+ pairs(this is not always a good approximation). All even nuclei have Iπ = 0+ ground state.

For odd mass nuclei, spin and parity is determined by orbital of the last unpaired nucleon. So, forexamples, see the figure

For odd-odd nuclei, the low lying states are made by vector coupling of the spins of the unpairednucleons, j

πpp and jπn

n . Then, the possible spins are found from:

I = |jp − jn|, . . . , |jp + jn|

All will exist as states, but difficult to predict which is lowest in energy. That is, it is hard todetermine which is the ground state. The overall-partity is multiplicative πpπn. That is: (+)(+)→+ and (+)(−)→ −.

So, as an example, in 169 F7, we have that the unpaired proton is a 1d5/2, and the neutron is 1p1/2.

Hence, jπpp = 5

2

+ and jπnn = 1

2

−. Hence, we expect either 2− or 3− states.Suppose we promoted the proton to a higher energy state. That is, to a 2s1/2 state. Then, we mayexpect 0− or 1−.

There is a one-to-one correspondance with all low-lying excited states.

2.2.5 Excited States

Such states can be made in all nuclei by either promoting nucleons up to higher levels, and/orbreaking an 0+ pair, and recoupling their spins.

2 NUCLEAR MODELS 27

Figure 17: Nuclear shell filling. The nuclei shown as ‘doubly magic’, as both the protons andneutrons occupy filled shells; and are hence very stable nuclei.

PPT

This happens at low excitation energy in f7/2 shell. This is all to do with the size of the shell-gap.

This illustrates the difference between the complete shell model & extreme independant particlelimit. 3 particles in 43Ca can affect each other, and allow many more possibilities.

2 NUCLEAR MODELS 28

Figure 18: Nuclear shell filling, for various nuclei, for even-odd nuclei (odd mass). The nucleonwhich contributes to the spin & partity is the unpaired nucleon, which has been circled in all cases.The parity is found from π = (−1)`, where ` = 0, 1, 2, 3 according to the usual s, p, d, f labelling.

3 COLLECTIVE EXCITATIONS 29

3 Collective Excitations

Here we consider the nucleus as a collective, as opposed to being composed of nucleons.

3.1 Nuclear Vibrations

Consider a liquid drop vibrating at high frequency, about a spherical equilibrium shape. We candescribe instantaneous shape by radius vector R(θ, φ) of the surface, at time t.

Figure 19: Vibrations of a ‘liquid drop’. The radius R(t) at some θ, φ.

So, we have:

R(t) = Rav[1 + β2 cosωtY`m(θ, φ)] (3.1)

Where we have that Rav = r0A1/3, as usual. β2 is the deformation parameter - see later.

SEE PPT - MODES OF VIBRATIONS

We see the modes of vibration for ` = 1, 2, 3. A quantum vibration is called a phonon. Adding Y`mterm into nuclear wavefunction introduces `h units of angular momentum, with parity (−1)`. Thevibrations carry angular momentum `h, and are bosons. Consider the following possibilities:

Adding an ` = 1, dipole phonon This gives the term Iπ = 1−; and is equivalent to a netdisplacement of the centre of mass of an isolated system. Hence, this is not observed.

Adding an ` = 2, quadrupole phonon Here, Iπ = 2+. That is, it creates an excited 2+ state.Usually the lowest excited state of even spherical nucleus.

Adding an ` = 3, octupole phonon So, we create a 3− state. This state is commonly seen athigher excitation energies than 2+ quadrupole vibrations.

3.1.1 Two Phonon States

Consider two phonons, so an energy of 2hω. There are 25 combinations of m = m1 +m2. However,by symmetry & indistinguishability arguments, this is reduced to 15. Leaving us with ` = 4, 2, 0


states. That is, we excite a triplet of states 0+, 2+, 4+ ; each with twice the energy of the first 2+

state. By similar considerations, 3 quadrupole phonon states can be constructed. That is, energy

Figure 20: Ordering of the first few quadrupole-phonon-states. The subscript refers to how manytimes that particular state has occured. The energy is the 2+

1 state is E(2+1 )500− 1200keV.

of 3hω. Giving the states 6+, 4+, 3+, 2+, 0+.

The best examples of vibrational states are in nuclei within the region 50 ≤ N ≤ 82 closed shels,with A = 120. Such as 120Te.

So, the predictions from the vibrational model:

• 0+, 2+2 , 4

+ at the same energy 2hω, and at 2× E(2+1 ). And is roughly ok!

• 0+, 2+3 , 3

+, 4+2 , 6

+ states, energy 3hω; should be at 3 × E(2+1 ). But, this is not always true,

because 3hω states start to overlap with shell model states.

• The number of possible states increases rapidly with energy (or phonon number).

3.2 Nuclear Rotations

Nuclei which lie close to the magic numbers are roughly spherical, and therefore cannot rotate inQM.

However, in certain regions of the nuclear chart: 150 ≤ A ≤ 190 or A > 220; that is, regions nearmiddle of shell gaps; nuclei have substantial distortions from spherical shape. e.g. 152

66 Dy86.

Many nucleons participate in the motion: hence, a collective effect collective rotation.

SEE PPT FOR EVIDENCE


3.2.1 Types of Deformed Shape

The most common deformed shape of rugby-ball shaped: axially symmetric, prolate shape (note:‘prolate’ is when the equator is squashed in; and ‘oblate’ when the poles are squashed in. Thinkthat the earth is oblate).

We describe the surface by:

R(θ, φ) = Rav[1 + β2Y2,0(θ, φ)] (3.2)

Notice that we use an axially symmetric spherical harmonic; which is (up to factors) a Legendrepolynomial.We say that β2 is the quadrupole deformation parameter ; so, if β2 = +0.3, then prolate; and ifβ2 < 0, then oblate.

3.2.2 Energies of Deformed Shapes

A deformed nucleus has a rotational degree of freedom. The classical expression for rotationalkinetic energy is just:

T =12Jω2

Where J is the moment of inertia. If we have angular momentum Ih, then it is related to J by:

Ih = Jω

Hence, we have:

T =(Jω)2

2J=

(Ih)2

2JHence, we have the expression for our QM rotor:

T =h2I(I + 1)

2J

For even nuclei, the 0+ state is always the ground state. Then, the next states are Iπ = 2+, 4+, 6+, 8+, . . .;on symmetry grounds.

3.2.3 Predictions

If we take the moment of inertia for a rigid body:

Jrigid =25MR2

av(1 + 0.31β2)

Then, we end up with:h2

2Jrigid= 6keV

After taking A = 170, β2 = +0.3. So, from this, we expect:


Figure 21: The energy level scheme for the rotational band formed, on top of the ground state.Shown are experimental values obtained for 164Er.

E(2+) = 36keV. However, experimentally, we find that this is more like 90-100keV; as per thefigure. However, if we take the ratio E(4+)

E(2+)= 20

6 = 3.33, then this is an extremely good prediction;as it divides out the moment of inertia.

Thus, by using Irigid, we have overestimated the actual moment of inertia.

Infact, the completely filled (inner) shells cannot contribute to this angular momentum, as theyare paired up. So, we should really just regard the rotational motion as due to the valence nucleonsonly. So, we imagine that we have an 0+ core, with the rest of the nucleus rotating around it. Thisleads to other extremes of motion, where we consider the nucleus as a fluid inside an ellipsoidalvessel. In this case, we calculate:

h2

2Jfluid= 90keV

But, this is too high, in comparison with experiment.


Hence, we conclude:Jfluid < Jnuclear < Jrigid

That is, the actual value of the moment of inertia is somewhere between that predicted by thenucleus being fluid, and that being a rigid body.

3.2.4 Excited Rotational Bands

So far, we have only discussed collective rotational bands built upon the ground state of even nuclei,such as 164Er. We do actually have more than one rotational band however; each of which is builton different states. There is one ground state rotational band, and two bands built on differentvibrational states.

β-Vibration Here, β2 oscillates about an equilibrium value. So, the picture to have in mind hereis that of a rugby ball, on its tip (i.e. prolate), rotating about its centre (not its axis of symmetry),and vibrating along its axis of symmetry.

γ-Vibration Here, β2 remains constant, but the axial symmetry is now lost.

Figure 22: (a) β-vibrations: notice that the axial symmetry is kept. (b) γ-vibrations: notice thataxial symmetry will be lost. The bottom figures show how the shape will vibrate. NB: it is veryhard to draw these things in 3D, so I wont even bother!

3.2.5 Population of Excited States

We use the technique of heavy ion fusion evaporation


Figure 23: The process of heavy ion fusion evaporation. Shown is for the process 48Ca+130Te→178

Hf∗ →174 Hf + 4n. As the beam particles will not hit the target ‘dead-on’, the compound nucleuswill rotate, in states of anything in excess of 60h. This whole process takes place about 10−12secsafter collision.

Heavy ion fusion evaporation reactions populate high spin states (i.e. high ` states); as long as theenergy of the beam is above the Coulomb barrier; which is given by:

E >Z1Z2

4πε0r

So, we see many excited states and sequences of states, as the products undergo gamma-decay.We use an accelerator to provide the beam, and have large arrays of high resolution germianiumsemi-conductor detectors.ASIDE: the accelerator (such as the Oak Ridge National Lab, Tennessee) ‘shoots’ beams of particlesalong tunnels, funneling them towards a van-der-Graff generator. The generator has a pd of MV -when it sparks, and hits the cavity in which it is housed, sparks hit with a force equivalent to thatof a car going at 40mph, hitting a brick wall! And makes a massive cracking sound!


3.2.6 Information from γ-ray Spectroscopy

Spins of the States Since ` = r × p, the beam imparts a large angular momentum ` to thecompound nucleus. The angular momentum (with reference to the previous figure), will be intoor outof the page, depending on which side the beam hits. That is, the compound nucleus will bealigned (perpendicular to the beam direction), in a particular direction, relative to the lab beamline. So, we can measure the angular distribution of γ-rays in the final nucleus, by moving a detectoraround the nucleus (or, as how it is actually done, have arrays of hundreds of detectors around thenucleus).

Figure 24: The diagram above shows the classical radiation pattern for both a dipole and quadrupoleγ-ray transition; the beam direction θ = 0 is shown. (a) Dipole transition. This is for ` = 1. Noticethat most intensity will be picked up at θ = 90, and virtually none along the beam direction. (b)Quadrupole transition. Here, ` = 2. Notice that the main difference here, is that there is a moreeven distribution of intensity along the beam direction.

We measure γ-ray intensity, as a function of detector angle, and use to determine what type ofradiation (i.e. angular momentum) is carried by the γ-ray.

An important technique is Directional Correlation from Orientated states DCO. For example, ifeven-even nucleus (e.g. 180Os) has ground state Iπ = 0+; we measure the lowest γ-ray transitionto first excited state (about 90keV). If we find (roughly) equal intensities at 90, 0, then it mustbe a quadrupole transition. Hence, the spin of the first excited state must be 2.

Rotational Bands Rotational bands in even-even nuclei de-excite by cascades of γ-ray transi-tions, between adjacent states.

4 α-DECAY 36

Figure 25: Here, we see the energies of the transitions between different spin states; and the spectrumwe will see. The spacing of the states (i.e. γ-ray energy increases linearly) will be the same, unlessthe moment of inertia changes J , which is seen in experiment.

So, with reference to the figure, the energy of a state with angular momentum I is given by:

E(I) =h2

2JI(I + 1)

Thus, γ-ray energies:

Eγ(I) = E(I)− E(I − 2) =h2

2J[I(I + 1)− (I − 2)(I − 1)]

Giving:

Eγ(I) =h2

2J(4I − 2)

And, as the caption on the figure says, the energy of γ-rays should increase linearly with angularmomentum; but this is only the case if the moment of inertia is constant.

4 α-decay

This is a process where a heavy nucleus reduces its charge, and therefore reduces the disruptiveCoulomb force. α-particle emission is particularly favours, as it is very tightly bound. It has a largeQ-value.

Q = mbefore −mafter

For example:23492 U →230

90 Th+42 He

4 α-DECAY 37

Where, looking up the masses:

23492 U = 234.040947u

23090 Th = 230.033128u

42He = 4.002603u

Giving a Q-value for the reaction of Q = 5.216× 10−3u = 4.86MeV. By way of terminology, the Uis the parent nucleus, and Th the daughter nucleus.

4.1 Theory of α-emission

Assume that an α-particle is preformed, moving in the spherical potential of what will become thedaughter nucleus. There are two parts to the potential: the attractive nuclear (Woods Saxon) andCoulomb repulsion.

Figure 26: The Woods-Saxon and Coulomb potential. The Coulomb potential is given by the chargeof the daughter nucleus, and α-particle: V = (Z−2)2e2

4πε0r. These two potentials combine, to form the

nuclear potential, below.

The velocity of the α particle is (non-relativistic):

v =√

2Eαmα

=

√2(Q+ V0)

mα≈ 105ms−1

Classically, the alpha particle has KE = Q+ V0 inside the nucleus, and can never escape.The region between a and b form a potential barrier: the Coulomb barrier. It has height B andwidth (b− a); and it is these parameter which determine the probability of tunnelling.

For example, in 238U , the α particle hits the barrier on average 1038 times before it escapes!To calculate this number, consider that the α particle travelling at 106ms−1 (as we just calculated),traverses a distance of 10−15m. Thus, the time that it takes to do this is just the distance over thespeed, giving 10−21s. Then, given that the half-life is t1/2 = 109yrs ≈ 1016secs. Hence, the half-lifedivided by the time taken for the α particle to traverse the nucleus, gives the number of ‘goes’ theα particle has before it escapes the nucleus.

4 α-DECAY 38

Figure 27: An α particle inside the nucleus will experience a potential shaped like this. Notice howthings are defined. The KE the α has, when it leaves the nucleus, is the Q-value. The bottom ofthe well is at V0 ≈ −35MeV

We can calculate this via quantum mechanics, but, very crudely, the barrier penetration probabilityP is:

P ≈ e−k(b−a) k ≡√

2mh2

12

(B −Q) B ≡ (Z − 2)2e2

4πε0a

Where B −Q gives the barrier height; and a the nuclear radius.

So that, from the standard:N = N0e

−λt

We have:λ = fP

Where the assault frequency f is the velocity of the α-particle divided by a f = v/a.

This is able to account for the very rapid Q-energy dependance of half-lifes: SEE GEIGER-NUTTALL PLOT.

For example, in 232Th, we have Q = 4.08MeV, and t1/2 = 1.4 × 1010yrs. However, for 218Th, wehave Q = 9.85MeV, and t1/2 = 1 × 10−7s. Hence, for a factor of 2.4 in Q, we get 4 × 1027 in thehalf life.

• Almost all comercial α-sources have Eα in the range 5-8MeV.

• In deformed nuclei, the barrier height and width depend on the surface where the α assault istaking place: α emission is favoured from the pole of a prolate nucleus (i.e. where the radiusis smallest). As barrier is smallest at pole.

• The alpha energies are discrete. 242Cm has decays to all possible daughter states in 238Pu.This gives fine structure. Most decays go to 0+ or 2+ states of daughter nucleus.

4 α-DECAY 39

Figure 28: The spectrum from α-decay. Notice that most have high energy, but a few have lower.This is because most alphas leave low exicted states of the daughter nucleus, but some do exictethe daughter nucleus, hence having lower energy themselves.

4.2 Fine Structure of α-decay

Let us use 242Cm as an example. We note:

In general, higher Q-values have higher decay rates.

Figure 29: The decay branches for 242Cm →238 Pu + α. Notice the branching ratios (i.e. therelative percentages) of a particular decay to that excited state of the daughter.

Let us use the following data:

M(242Cm) = 242.05883uM(238Pu) = 238.04955u

M(α) = 4.0026033u

To compute Q: add up masses before, add up masses after, subtract after from before; convert intokg, convert into an energy. We then find Q = 6.216MeV.

4 α-DECAY 40

As a second thing to note, we see that decays to states of higher spin appear to be hindered.Experimentally, decays to 8+ have branching ratio 0.00002%, and to 1−, 0.00024%; but have roughlythe same Q-value.

Thirdly, decays to some states do not occur at all; such as to 2−, 3+, 4−. Such decays are forbiddendue to parity conservation.

To explain the last two points, we must consider the angular momentum carried away by the αparticle.Now, by considering the α particle to be a doubly magic nucleus, with spin/parity 0+, we see thatthe only angular momentum it can carry away is orbital.

So, considering the emission of an α particle, from some point r from the centre of the nucleus. Theα particle will then have energy Eα and momentum pα. Hence, angular momentum L = r × pα.And L = `h with ` = 0, 1, 2, . . .. Hence, we see that the maximum possible angular momentumarises from emission at the maximum possible radius (i.e r = r0A

1/3). Classically:

pα =√

2mαEα =√

2mαQ

Where by definition, the energy of the α particle is the Q-value.Thus, for our 242Cm decay, with Q = 6.216MeV = 9.95 × 10−13J, and mα = 4 × 1.673 × 10−27 =6.692× 10−27kg, we have that pα = 1.15× 10−19kgms−1. Hence, using the maximum possible, onecan (apparently) show that `max = 8h.Hence, 2 reasons why high ` is unfavoured:

First, α particle must originate from a more restricted region of the nucleus: the larger radius-bithas a higher Coulomb barrier. Secondly, there is an additional contribution to the barrier height-width: the angular momentum barrier, which is a QM effect.

Now, spin difference between parent (Ii) and daughter (If ) can change by `h, where:

Ii = If + `

Where parity change is (−1)`.So, if parent nucleus is Iπi = 0+, then If can be, for ` = 0 0+; for ` = 1 → 1−, ` = 2 → 2+ etc.Where ` is that carried by the α. Hence, all others are forbidden: 1+, 2−, 3+, 4−.

5 β-DECAY 41

5 β-decay

This allows nucleus to reach most stable p to n ratio.

On the mass parabola from the SEMF, for odd-mass (i.e. one parabola only), get β− decay to theright, and β+ and ε to the left. ε is electron capture, where the nucleus absorbs an electron fromits orbital.

The processes are:

β− : n → p+ e− + νe

β+ : p → n+ e+ + νe

ε : p+ e− → n+ νe

Note, all particles involved are fermions, having spin 1/2. Energy, momentum and angular momen-tum must be conserved in the decay.

5.1 Q-values

This is the KE released by a reaction.

Qβ−,ε = [mN (parent)−mN (daughter)] c2

Qβ+ = [mN (parent)−mN (daughter)] c2 − 2mec2

See the examples sheet for derivation.

In β-decays, the Q-value is shared between the decay products, in a way that considers momentumconservation. That is, if Pm,Pβ,Pν are the daughter nucleus, electron an neutrino recoil momenta,then:

Pm = Pβ + Pν = 0

The Q-value is given by:

Qβ =p2

2m+KE(β) + Pνc

Where the m in the first term makes its contribution very small; and the last two terms arerelativistic, assuming ν is massless.

We see from the Figure that the β spectrum is continuous, up to a maximum Qβ. The shape ofthe spectrum is determined by the density of final states: electron and neutrino Fermi theory; aswell as the Coulomb effect on the electron/positron as it leaves the nucleus.

From the plot, we see that we can gain information on mν ; which has important consequences forelectroweak decays. Experimentally, it gives the limit:

mνc2 < 60eV

For Qβ = 18.6keV .

5 β-DECAY 42

Figure 30: (Left) The spectrum from β-decay. Notice that it is continuous (unlike α-decay). Theshape is determined by the density of final states, from Fermi theory. (Right) Notice on the rightplot, the intercept gives information about the mass of the ν.

5.2 Fermi Theory

The Fermi theory of β-decay states that the decay rate λ is given by the Fermi Golden rule:

λ =2πh

∣∣∣∣∫ ψ∗fVβψi d3r

∣∣∣∣2 ρ(EF ) (5.1)

Where ρ(EF ) is the density of states, for electrons and neutrinos KE.

It is the overlap integral involving the β-decay operator Vβ, connecting initial and final states.Hence, it is particularly large when decaying nuceon keeps the same wavefunction; hence, shorthalf-life. These are called superallowed transitions.

From Krane, we write that the number of electrons with a specific kinetic energy Te is:

N(Te) ∝√T 2e + 2Temec2(Qβ − Te)2(Te +mec

2)

Where Qβ ≡ Te + Tν ; shared between two particles.

5.2.1 β-decay Half-lifes

From the Figure, we see that the different half-lifes are determined by Q; and infact, the decay rateis ∝ Q5

β:

λ ∝ Q5β (5.2)

5 β-DECAY 43

Figure 31: Two decay modes to Nitrogen. Notice the very different half-lifes, for slightly differentQ-values of the parents.

ft-Values If the energy dependance on half-life is taken out, a comparitive half-life, or ft-valuecan be deduced for a β-transition, which depends on the inverse of the matrix element. That is:

λ ∝∣∣∣∣∫ ψ∗fVβψi d

3r

∣∣∣∣2 τ ∝ 1λ

Since such values range from 10−3 → 103secs, a log10ft is usually quoted; having value 3-4 for thesuperallowed decays.

Below is a table of decays to 147 N7, and the parents which decay to it, and with their corresponding

data.

14C β− log ft = 9.03 0+ → 1+ Q = 0.156MeV14O(0.6%) β+ log ft = 7.3 0+ → 1− Q = 5.144MeV

14O(99.3%) β+ log ft = 3.4 0+ → 0+ Q = 2.83MeV

The table shows that the last entry is of the type ‘superallowed’.

It turns out that:

ft =2π3h ln(2(hc)6)

(mec2)5G2V 〈mν〉2

∝ 1G2V

Where GV is the weak interaction coupling constant; and is thus a fundamental test of the StandardModel.

5.3 Angular Momentum Rules for β-decay

Remember that for α-decay, involving `h, we had an α emitted a distance:

r =`h√

2mαEα

5 β-DECAY 44

From the centre. Since me,mν << mα, it is much more difficult for an electron to carry awayorbital angular momentum. If Eβ ≈ 5MeV, then for an electron of ` = 1, we compute that it wouldhave to be emitted from r = 100fm from the centre. This is obviously very hard to do! We thenclassify:

• ` = 0 as allowed decay ;

• ` = 1 as first forbidden decay ;

• ` = 2 as second forbidden decay ;

• etc

Forbidden decays have a much longer half-life, smaller decay rates.

So, for allowed decays, there must be no orbital angular momentum change between the daughterand parent nuclei. And, as ` = 0, we therefore have no parity change between daughter and nucleus.

5.3.1 Allowed Decays

Now, there are two types of allowed decays, and they depend on the intrinsic spin carried away bythe electron and neutrino (both being spin 1/2 particles).

• Fermi Type. This is for the electron and neutrino having opposite spins: ↑↓. Hence, s =1/2 + 1/2 = 0. Therefore, no angular momentum carried away. Therefore, considering:

Ii = If + S

We therefore conclude that ∆I = Ii − If = 0, as s is zero. Therefore, a decay of this typeretains both its partity & angular momentum. For example, the decay:

14O(0+)→14 N∗(0+)

• Gamov-Teller Type. This is for the electron and neutrino having aligned spins ↑↑. Then,s = 1/2 + 1/2 = 1. Hence, Ii = If + 1. And thus that ∆I = |Ii − If | = 0,±1.

We can then classify some decays as being one of the above. Consider:

n→ p+ e− + νe

This has t1/2 = 11mins. Notice, all particle are spin-1/2. Hence, the combinations of the electronand neutrino are s = 0, 1. Hence, as ∆I = 0, it satisfies both the Fermi & Gamov-Teller types.Hence, we call it a mixed Fermi/Gamov-Teller transition.

5 β-DECAY 45

5.3.2 First Forbidden Decay

Here, we have ` = 1, and hence a parity change of −1. We have:

Ii = If + S + `

The Fermi-first forbidden has s = 0, and hence ∆I = 0,±1.The GT-first forbidden has s = 1, and thus that δI = 0,±1,±2.

Considering an example:115In(9/2+)→115 SN(1/2+)

We see that ∆I = 9/2− 1/2 = 4. And that ∆π = 0. Hence, we deduce that ` must be even. Now,as ∆I = `+ s, we see that s can either be 0 or 1 (for ` = 4, 3). However, we also saw that ` must beeven . Therefore, we conclude that it must be of Fermi-type (i.e. s = 0), and the fourth forbiddendecay (as ` = 4). Infact, such a reaction has t1/2 ≈ 1014yrs!

Another example:9/2− → 1/2+

So, ∆I = 4, and ∆π = 1. Thus, we see that ` must be odd. s can be either 0 (Fermi) or 1 (GT).If 0, then ` must be 4 (which is even, which we have discounted); or 1, then ` must be 3, and weconclude that this must be the case.Hence, it is the third forbidden GT decay.

5.4 Parity Violation in β-decay Process

Most laws of nature are parity invariant; they dont change under r → −r, a reflection through theorigin. Examples are the strong force and EM: they all conserve parity.

If the Hamiltonian conserves parity, then the eigenstates are said to have good parity. Note:ψ(−r) = ±ψ(r); if the -, then even, and if +, then odd.

In 1957, Madame Wu (et al) discovered the β-decay (weak int) process violated parity:

Vnuclear = Vstrong + VEM + Vweak

Where Vweak is the parity violating part of nuclear potential, and is of the order 10−7.

6 γ-RAY EMISSION 46

6 γ-ray Emission

6.1 Angular Momentum Considerations

Photons carry an intrinsic spin of s = 1, with a component of either sz = ±1 (but not zero) alongthe direction of propogation. Like β-particles, γ-rays have difficulty in carrying away orbital angularmomentum `h.

If a γ-ray is emitted from postion r from the centre of a nucleus, carrying energy Eγ , and momentump = hk, where Eγ = pc = hkc; with the useful relation:

hc = 197MeV fm

Thus, we have that `h = r × p. Hence:

r =`h

p=`hc

Eγ= 100`fm

Where we have computed for Eγ = 2MeV. That is, for a photon to carry away angular momentum`, it must have been emitted from a distance 100`fm from the centre. This is way out in the tail ofthe nuclear wavefunction.Also, via h` = r × p and p = hk, we can see that:

`h = r × hk

That is:` = kr

But, kR << 1, from the maximum nuclear radius.

Hence, we say that γ-ray transitions with ` 6= 0 are inhibited, and are inhibited proportional to(rK)2`.

If a γ-ray carries away a total angular momentum L, it can be considered a combination of itsintrinsic spin & an orbital part:

L = ` + s

Now, as a rule, L = 0 (i.e. monopole) is forbidden. This is because ` is perpendicular to p, andthis combination just cannot be formed.

The character σL of γ-ray transtions come in two types: σ = E, electric transitions, and σ = B,magnetic transitions; and L = ` + s.


σL Transition Type Comments

E1 Electric Dipole Must be ` = 0 as s = 1. Therefore, not hindered,and thus fast.

M1 Magnetic Dipole Must be ` = 0 as s = 1. Therefore, not hindered,and thus fast, but slower than E1.

E2 Electric Quadrupole Must be ` = 1, as s = 1. It is hinderd via(kR)2` = (kR)2; and parity change w.r.t dipole.

M2 Magentic Quadrupole Same as E2.E3,M3 Electric/Magnetic Octupole ` = 2, s = 1. Hindered via (kR)2` = (kR)4,

compared to dipole.E4,M4 Electric/Magnetic Hexadecupole Hindered by (kR)6 (` = 3).

Experimentally, we see the lowest order most of the time, as it is least hindered.

We use DCO (directional correlation from orientated states) to distinguish between states.

6.2 Parity Rules

Now, we mentioned that parity changes for different transitions, but we did not say from what towhat.

Suppose we have a transition Iπii → I

πf

f . We have the following rules:

EL Parity Change Example, Iπii → I

πf

f .

E1 Yes 1− → 0+

E2 No 2+ → 0+

E3 Yes 3− → 0+

E4 No 4+ → 0+

And, for magnetic:

ML Parity Change Example, Iπii → I

πf

f .

M1 No 1+ → 0+

M2 Yes 2− → 0+

M3 No 3+ → 0+

M4 Yes 4− → 0+

So, notice that E and M are opposite. We will derive E1 later. Experimentally, we can distinguishbetween these EL & ML by measuring γ-ray polarisation.


6.2.1 Angular Momentum Considerations

Radiation carries away Lh of angular momentum:

Ii = If + L

Under the usual vector addition rules. That is, the possible L values:

|Ii − If | ≤ L ≤ |Ii + If |

Note, L = 0 is not possible. Transition proceed by least hindered process, usually the one withlowest L. However, there is a common exception: an “enhanced” E2 can often compete with M1,as they have the same parity change. When more than one L contribute, we call it a “mixedtransition”.

Figure 32: Some possible transitions. (a) We have that |2 − 0| ≤ L ≤ |2 + 0| as the possiblevalues. Hence, L = 2 only, and must be E2 as no parity change. (b) Same reasoning as for (a). (c)Again, the only possible value of L is 2, and there is no parity change, hence E2. (d) We have that|2− 2| ≤ L ≤ |2 + 2|, hence possible L = 0, 1, 2, 3, 4. We exclude 0. Hence possible are M1, E2, M3,E4 (by parity considerations: not changing). The higher L values are hindered; hence we get themixed transition of both M1 & E2. (e) Possible |4− 2| ≤ L ≤ |4 + 2|, i.e. L = 2, 3, 4, 5, 6. However,lower L favoured. Hence E2 or M3 mixed transition.

6.3 Parity Rule for E1 & Transition Rates

Now, the electric dipole moment is charge times the distance between the two charges: ez. Hence,the E1 transition rate goes like:

∝∣∣∣∣∫ ψ∗fezψidτ

∣∣∣∣2Via the Fermi Golden rule.


Since z is an odd function, then the integrand is zero if ψi, ψ∗f have the same parity.

The rule is that E1 transitions connect states of opposite parity. Transition rates are given by theFermi Golden rule:

λ =2πh

∣∣∣∣∫ ψ∗fM(σL)ψidτ∣∣∣∣2 dNdE

Where M(σL) is the operator for a σL-type transition, having the form:

M(σL) ≈ 1k

(kr)LPL(cos θ)

Notice the (kr)L dependance coming in. The factor of dN/dE is just the density of final states foremitted photon, and is ∝ E3

γ .

6.4 Single Particle Weiskopf Estimates

Now, we calculate: ∣∣∣∣∫ ψ∗fM(σL)ψidτ∣∣∣∣2

For a single proton making a simple transition between two shell-model states. We shall say thatψi, ψf are known wavestates. We are able to calculate (but not actually do) rates for varioustransitions:

λ(E1) = 1× 1014A2/3E3γ

λ(E2) = 7.3× 107A4/2E5γ

λ(E3) = 34A2E7γ

λ(E4) = 1.1× 10−5A8/3E9

With a set for magnetic transitions:

λ(M1) = 5.6× 1013E3γ

λ(M2) = 3.5× 107A2/3E5γ

λ(M3) = 16A4/3E7γ

λ(M4) = 4.5× 10−6A2E9

Where the transition rate λ is in s−1; and Eγ in MeV. So, for example, a 152keV transition in 142Tbhas:

τ(E1) =1

λ(E1)= 1.04× 10−13secs τ(M1) = 5.08× 10−12secs

We have used that the lifetime τ of a σL-decay is:

τ =1

λ(σL)

And also, its half-life:

t1/2 =ln 2λ


Now, a state may decay via different modes: 4+ → 2+, via E2, or 4+ → 3+ via M1. The total decayrate is:

λtotal = λ(E2) + λ(M1)

And hence the lifetime of the state is:

τ(4+) =1

λtotal

The experimental data can be compared with these single particle rates.; to determine wether“enhanced rate” or “hindered rate”.

For example, in rotational nuclei, where large number of nucleons N are coherently involved in themotion, then the E2 rate would be enhanced.∫

ψ∗fM(E2)ψidτ ≈ N × single particle value

Thus, the decay rate:

λ ∝∣∣∣∣∫ ∣∣∣∣2 ≈ N2λ(single particle)

This is why E2 transitions can compete with single particle M1 transitions (note, M1, as sameparity change). For example, superdeformed E2’s may have 2000 times single particle values.

6.5 Internal Conversion

This is a process which competes with γ-decay. The nucleus makes the transition by giving theenergy to one of the orbiting electrons. So, the total decay rate is the sum of γ-decay and internaldecay rate:

λtotal = λγ + λe

Then, τ is shortened. We define the internal conversion coefficient:

α ≡ λeλγ

So, if α = 1, then the decay rate is doubled by internal conversion, and thus the lifetime is halved.Also, the coefficient is the sum for contributions from different electron shells.

Also, L = 0 transitions are not allowed, via γ-decay. However, they can occur by internal conver-sion.

nuclear physics - · pdf filenuclear physics j.pearson april 28, 2008 abstract these are a set...

Documents