What is the nuclear spin of a deuteron?
I = 1
Total angular momentum of deuteron, I eqn
I = s_n + s_p + L
Expectation values for L^2 and L_z (and s^2 and s_z)
<L^2> = L(L+1) hbar^2
<L_z> - hbar m_L
</L_z>
Magnetic quantum numbers m_L (and m_s)
m_L = -L, … , 0 , … , L
Total angular momentum J
J = L + s
What are the allowed values of J
integer steps from |L-s| to L+s
What are allowed values of s for deuteron
s = 1 and s = 0
What are the allowed values of I,
what do we want for deuteron?
integer steps from |s-L| to s+L
we want I = 1 for deuteron
If s = 0 what are values of L that allow I = 1
L = 1
If s = 1 what are the allowed values of L for I = 1
L = 0, L = 1, L = 2
What are the 4 states that satisfy I = 1
s = 0 , L = 1
s = 1 , L = 0
s = 1 , L = 1
s = 1 , L = 2
Parity of deuteron, proton and neutron are all even so L must be
even
What are the 2 states that satisfy I = 1and parity
s = 1 , L = 0 (96%)
s = 1 , L = 2
What are the key properties of the strong nuclear force (5)?
Strong (~100× electromagnetic), short range (~2 fm), attractive with repulsive core, charge independent, spin dependent.
What does spin dependence of the nuclear force mean?
The strength of the force depends on the relative spin orientation of nucleons.
What are the 2 states that satisfy I = 1 and parity
s = 1 , L = 0 (96%)
s = 1 , L = 2
How is the deuteron potential modeled in the square well model?
As a 3D square well with radius R and depth V₀.
What state do we assume the deuteron is in?
L = 0
What does r represent in the deuteron square well model?
The separation distance between proton and neutron.
Why is there no Coulomb barrier in the deuteron square well model?
Because the deuteron consists of a proton and neutron, and the neutron has no charge.
How can a two-body quantum problem be simplified?
By using the reduced mass to convert it into an effective one-body problem.
What happens to the wavefunction inside the square well?
It oscillates.
What happens to the wavefunction outside the square well?
It decays exponentially.
Why must the wavefunction remain finite?
Because physical wavefunctions cannot diverge to infinity.