Chapters 1-5 (Nuclei) Flashcards
Nuclide
Nucleus w/ specific A and Z values
Isotone
Nuclei w/ same no of neutrons (i.e. fixed (A-Z) value) but diff no of protons
Isobar
Nuclei w/ same A but diff Z
Mirror nuclei
2 nuclei w/ odd A where no of p+s in one nuclei = no of neutrons in other and vice-versa
Binding energy and why is it -ve
energy required to free nucleons -> -ve as need to put energy IN
What force is binding energy due to?
Strong force
Why is the binding energy term (over c^2) known as the “mass defect”?
As the strong force that causes the binding energy causes reduction in total mass of nucleus from just sum of mass of constituent nucleons.
Describe relationship b/w A and binding energy
Binding energies per nucleon increases sharply as A increases
Peaks at iron then slowly decreases for more massive nuclei.
The Mass of a nucleus contains the binding energy, which can be given by a semi-empirical formula. This formula assumes that the nucleus can be thought of as what? Why was this eventually changed and what was it changed to?
Assumes nucleus is like a liquid drop, where the nucleons are all on the surface of this drop. (might be wrong)
However, this model underestimated binding energies of “magic nuclei” , whose number of p+s or ns equal to one of the numbers in the list: 2,8,20,28,50,82,126. (Doubly magic if both p+s and ns equal numbers in list)
This however was explained by “Shell Model”, where each nucleon is moving in some potential and their energy levels are classified in terms of the quantum numbers n, l and j. [like multi-layered liquid drop I think?]
Note 5 features of the Shell model
- For spherically symmetric potential, the wavefunction of any nucleon is: psi_nlm = R_nl (r) * Y^m _l (theta, phi) [same as wavefunction of electron in H]
- The energy eigenvalues will depend on n and l but are degenerate in m.
- Energy levels come in “bunches” called “shells” w/ large energy gaps b/w each shell.
- In ground state, nucleons fill available energy levels from bottom up, w/ 2 in each energy level.
- Potential follows Saxon-Woods model.
Why would using a Simple Harmonic Potential to model the potential of nucleons be unsuccessful?
Would yield equally spaced energy levels and would not see the shell structure and hence the magic no.s
What quantum numbers does the energy level depend on?
s, l and j
One of the magic numbers was predicted to be 40 but turned out to be 50. What caused this increase? Write a general equation for its contribution.
Spin-Orbit coupling contribution to potential.
V(r) -> V(r) + W(r)L.S [L and S are vectors]
What values does total angular momentum “j” take?
j = l +/- 1/2
Equation for energy shift caused by spin-orbit coupling. [Hint: J_Lower(J_Lower + 1) - J_Higher(J_Higher + 1)]
delta E proportional to j(j+1) - l(l+1) - s(s+1)
Do states with higher j have higher or lower energy?
lower
The spin-orbit effect is v. large. What implications does this have?
Leads to crossing over of energy levels into diff shells. Hence, there are now more energy levels in shell with magic number 50 than previously thought, giving the reason for why magic no increased from 40 to 50.
Spin of a pair of nucleons
0
Parity of a pair of nucleons
1 (“even parity”)
If Z and A are even, what is the spin and parity?
0 spin
1 (even parity)
If A or Z is odd, there will be an unpaired nucleon. What is the spin and parity now?
spin = j value of unpaired nucleon
parity = (-1)^l where l is the orb ang mom of unpaired nucleon.
If Z and A are odd, there will be an unpaired p+ w/ j1 and unpaired neutron w/ j2.
spin: takes values b/w mod(j1-j2) and mod(j1+j2) in integer steps.
parity = (-1)^{l1 + l2}
Why do nuclei w/ and odd no. of p+s and/or ns have a magnetic dipole moment?
Because they have a non-zero intrinsic spin
5 Assumptions made in Rutherford experiment
- Nuclear charge of atom = Ze and that mass of proton and neutron are the same.
- Nucleus is seen by alpha particle as a point particle because alpha particle has low energy, therefore has large wavelength, so nucleus appears as a point to it.
- m_nucleus»_space; m_alpha, therefore can neglect nuclear recoil
- Laws of CM and Em apply and there are no other forces present
- ELASTIC collision