Nuclear Physics Flashcards
What is the gamow radius relation?
R = 1.23 fm A^(1/3)
How is mean-square radius defined?
What is the mean-square radius for a solid sphere of constant density?
integra{r^2 * density} / integral{density}
3/5 R^2
REMEMBER TO CONVERT TO MEAN SQUARE RADIUS
What is the classical (geometric) cross-section?
What kind of experiment could measure this cross-section?
Classically, corresponds to the effective area over which a beam particle will be absorbed/interact if there is a target particle present in that area.
= pi * (R1 + R2)^2
i.e: fire beam nuclei at some thickness x of target nuclei
The chance of a beam nucleus being attenuated = the cross-section * number density of target * thickness of target
What is meant by the differential cross-section?
The effective area of target nucleus which scatters beam nuclei into some specific region of solid angle.
d rho / d omega
What energy electrons do we need to probe a nucleus?
De broglie wavelength «_space;10 fm
–> energy»_space; 120 MeV
In a classical description of electron scattering, where would we expect a minima in the differential cross section?
What are the problems with this description?
Minima at [1.22 lambda / D] due to diffraction.
-D is the nuclear diameter.
The edge of a nucleus is not well defined, there is some skin thickness.
The coulomb force and nuclear charge distribution are not considered.
What is the matrix element from electron scattering theory?
The amplitude of some component of the scattered electron.
The cross section/differential cross section is proportional to the matrix element squared.
Considering the electron to be a plane wave makes the matrix element effectively a 3D inverse fourier transform of the scattering potential.
What is the result of electron scattering theory with a point nucleus scattering potential?
We recover the rutherford scattering formula:
diff. cross-section ~ (Z e^2 / q^2)^2
where q is the difference in k-vectors between the incident and scattered electron
(momentum difference / hbar)
q = 2k sin(theta/2)
CHARGE SIGN INVARIANT
What is the result of electron scattering theory with an extended charge distribution?
The potential is dependent on r (involves integral of the charge density).
Using the convolution theorem allows the matrix element to be written in terms of the Electric Form Factor:
m-element = ( Ze^2 / e_0 q^2 ) * electric form factor
i.e: rutherford result * electric form factor
The electric form factor is effectively the inverse fourier transform of the nuclear charge distribution.
This treatment only differs from the rutherford result if the electron penetrates the nucleus.
What is the electric form factor?
What about approximately, in the case of small deflection?
Multiplying the point-like nucleus matrix element by the electric form factor gives the matrix element for an extended charge distribution.
So the diff. cross-section involves the EFF squared.
The EFF is effectively the inverse fourier transform of the nuclear charge distribution.
For small q:
EFF ~ 1 - 1/6 * mean square charge radius * q^2
(given in formula sheet)
How can we use the electric form factor to find the nuclear charge distribution?
What is the limitation of this method?
The differential cross section is proportional to the EFF squared.
So we can perform an inverse fourier transform to determine the charge density:
- divide by rutherford scattering cross-section
- square root
- take inverse fourier transform
However, to perform the inverse fourier transform completely we would need knowledge of the diff. cross-section with momentum differences up to infinity.
(i. e. penetration into the centre of the nucleus)
- ——> there are large uncertainties on the result for the charge distribution for small radius
What is the idea behind investigating the nuclear charge radius using atomic energy shifts?
What is the problem with this?
Some electron orbitals (mainly ns) have finite density inside the nuclear region (~the origin).
Therefore, the binding energy for these electrons is less than it would be for a point nucleus.
We can use 1st-order perturbation theory to find the difference in energy as a function of the nuclear radius.
However, point-like nuclei do not exist, and the theoretical calculations result in large uncertainties compared to the energy difference. Therefore we can only really find the difference between the energies, not absolute energies.
How do X-ray isotope shifts allow us to compare the mean-square charge radii of isotopes?
The differences in energies of K X-rays (to the 1s orbital) for different isotopes reflect the difference in energy of the 1s electrons (as higher electron orbitals are ~ unaffected by nuclear size).
Different isotopes have varying mean-square charge radii due to the charge-independence of the nuclear force, but have the same charge, so are ideal to compare.
What is the equation for the approximate energy shift of an electron due to finite nuclear size? IMPORTANT TO MEMORISE
What assumption are we making in doing this?
|e-wavefunction at the origin|^2 * ( Ze^2 / 6 e_0 ) * mean square charge radius.
Assuming that the electron wavefunction is approximately constant across the nuclear volume. (appropriate for ns electrons
What is the scale of isotope shifts? (as compared to absolute energy of the X-ray)
delta E / E ~ 10^-6
What are the advantages/disadvantages of using optical isotope shifts as opposed to X-ray isotope shifts?
Optical energy range - investigating valence transitions (e.g: 6p -> 6s).
- therefore the changes in energy are far smaller (~ 10^-5 smaller)
- however, spectroscopy is far more precise at these photon energies
- also, we can take much more data, especially for large nuclei as we are investigating valence electron transitions
- the energy shift of the non-s state also becomes non-negligible!
- the reduced mass of the electron-muon system becomes non-negligible, and differs for different isotopes.
How is a muonic orbital different from an electronic one?
Muon mass is ~207 times larger than electron mass.
This can be substituted into the bohr radius and energies (adjusting rydberg energy).
The orbital is much closer to the nucleus, and the energy is much greater.
What are the advantages/disadvantages of using muonic energy shifts to investigate nuclear charge radius?
Muonic wavefunction significant in the nuclear region.
- > much larger energy shifts
- > allow us to determine ABSOLUTE mean square charge radii
However, this procedure requires muonic atoms -> very complicated procedure.
Also, it can only be performed for stable isotopes.
How could we directly investigate the matter radius of nuclei?
Must use strongly interacting probes, e.g: pion, alpha particle. (these also will interact with the coulomb force)
At probe energy equal to the coulomb barrier, we see a change in the diff. cross. section of the deflected probes.
-> will be below that of rutherford scattering, due to inelastic scattering/absorption
How can we derive an expression for the momentum transfer q in terms of the initial momentum and the angle of deflection? (small angle)
Simply write out the vector equation for q, and square it, giving a term in cos(theta).
Then assume that the final momentum = initial momentum. (due to small deflection angle)
Remember that HBAR q is the actual momentum transfer.
What are the key features of nuclear charge distribution for medium-mass nuclei?
Central density roughly constant (and constant among nuclei).
Some “skin thickness” at the edge, also roughly constant among nuclei around 2.3fm
Sketch the shape of a coulomb potential for a finite size nucleus.
(check notes)
Coulomb potential (remember negative) up to the nuclear radius, at which point the potential is curtailed.
What is the classical motivation for the quantum idea of magnetic moment due to a valence nucleon?
Classical magnetic dipole moment = current * area of loop
-> this can be shown to be proportional to angular momentum
remember to use quantum angular momentum
What is a gyromagnetic ratio?
The ratio between magnetic moment and angular momentum.
(be careful with factors of hbar and nuclear magneton)
[ magnetic moment in units of nuclear magneton / angular momentum in units of hbar]
i.e: magnetic moment = gyromagnetic ratio * nuclear magneton / hbar
What is the magnetic moment of a nucleus (general vector)?
mu = ( g_l * l + g_s * s) mu_N / hbar
g_l is the orbital gyromagnetic ratio for proton or neutron depending on the valence nucleon.
g_s is intrinsic gyromagnetic ratio for proton/neutron.
mu_N is the nuclear magneton.
What are the values for orbital gyromagnetic ratio for protons and neutrons?
1 for protons (charge e)
0 for neutrons (zero charge)
What is the difference between the general nuclear magnetic moment vector and the schmidt limits?
In practice, we measure the magnitude of a magnetic moment COMPONENT. Thus, we must construct our theoretical moment in some basis: choose maximum projection along the z-direction. i.e: m_j = j
What are the approximations made by the schmidt limits?
What are they appropriate for?
What assumptions do we make when using the schmidt estimates?
Nuclear magnetic moment due to a single valence nucleon.
Assumes the nucleons to be free. (they are not in reality) –> so the schmidt limits give upper and lower bounds on nuclear magnetic moment
(upper for j = l + 1/2, lower for j = l - 1/2)
We are assuming a very simplistic shell model - levels may change order due to deformation or nucleon imbalance. No collective effects included (usually).
How can we estimate the nuclear magnetic moment due to multiple valence nucleons in the framework of schmidt estimates?
We must also consider the magnetic moment due to the TOTAL nuclear angular momentum I (collective motion).
Add a factor of g_R I to the general nuclear magnetic moment vector.
g_R ~ Z/A as only protons contribute to magnetic moment, but neutrons contribute to the total angular momentum.
What is some evidence for nuclear deformation?
Charge/matter mean square charge radii increases significantly more than expected for unstable nuclei. This is particularly seen using isotope shifts.
(we expect “picket fence” of X-ray energies for different isotopes, but this is not what we see)
Also, for deformed nuclei, there are rotational excited states, characterised by enhanced E2 transitions and energy ratio 4+/2+ ~ 3.33.
We could measure the spectroscopic quadrupole moment, but this only works if I is not 0 or 1/2.
What is the character of a deformed nucleus?
Why is this?
Nuclear potential can be expanded as a multipole (far away from the nucleus).
Monopole term is simply the point charge result.
Dipole term = 0 as the nucleus has good even parity.
Quadrupole term introduces deformation (axially symmetric) (prolate nucleus). It is dependent on Z.
Higher-order deformation would significantly increase energy.
What is the difference between intrinsic quadrupole moment and spectroscopic quadrupole moment?
Qo is the moment in the frame of the nucleus (no angular momentum).
-> we can only measure Qs in the lab frame, and it is dependent on the orientation of the nucleus. We measure the component aligned with maximal angular momentum.
However, there is a relation, valid for all I except 0, 1/2.
What does a positive value of quadrupole moment correspond to?
Prolate nucleus (extended in the angular momentum axis).
What is the difference between the hyperfine structure and fine structure of electron energy levels?
Fine structure - due to interaction of electron spin mag. mom. and electron orbital mag. mom.
Hyperfine structure - due to interaction of electron orbital mag. mom. and nuclear mag. mom.
(~1000 times smaller effect)
(allows us to investigate nuclear magnetic moment)
(same energy scale as isotope shift)
What is the hyperfine energy splitting related to?
- the nuclear magnetic moment (dot) the electron magnetic field at the nucleus.
(aligned -> lower energy)
This means that the energy splitting is proportional to the expectation value of I dot J, where I is the nuclear ang. momentum and J is the electron total ang. momentum. The constant of proportionality is different for each multiplet (electron state) (hyperfine A parameter).
We define F = I + J to characterise the multiplets.
Why might g-factors often be “quenched” before comparing to experimental data?
Schmidt g-factor estimates assume free nucleons. This is not the case in practice, for example, due to virtual mesons in the nucleus. These quantum fluctuations decrease the magnetic moments of nuclei.
A correction of g’ = 0.6g is often made.
What factor is it important to remember when using the schmidt limit formulae?
They give the nuclear magnetic moment IN TERMS OF the nuclear magneton mu_N
