Introduction to Nuclear and Particle Physics Flashcards
ħ and c size approximation
ħc ≈ 200 MeV fm
DeBroglie relation
λ = h/p
What units should we be using
fm, MeV, fine structure constant α, factors of ħ and c
Heisenberg uncertainty principle
∆x∆pₓ ≥ ħ/2
Getting from momentum to energy
p = h/λ = 2πħc/λc = 2π 200 MeV fm / λc
Basically try to get ħc factor
Momentum in MeV/c corresponds to energy in MeV
Schrödinger equation
– ħ²/2M d²ψ(x)/dx² + V(x)ψ(x) = Eψ(x)
Time dependance of a wavefunction
Ψ(x,t) = ψ(x)exp(iEt/ħ)
Probability density of a wavefunction
P(x) = Ψ∗(x, t)Ψ(x, t)
which respects unitarity
∫±∞ P(x)dx = ∫±∞ Ψ∗(x, t)Ψ(x, t) dx = 1
Total energy of a particle
E² = p²c² + m₀²c⁴
Angular momentum operators for each component
L = r × p
L̂ₓ = –iħ (y ∂/∂z – z ∂/∂y)
L̂ᵧ = –iħ (z ∂/∂x – x ∂/∂z)
L̂₂ = –iħ (x ∂/∂y – y ∂/∂x)
Angular momentum commutations
[L̂ₓ, L̂ᵧ] = iħL̂₂
[L̂ᵧ, L̂₂] = iħL̂ₓ
[L̂₂, L̂ₓ] = iħL̂ᵧ
[L̂², L̂ᵢ] = 0
Atom notation
ᴬzX
Z = atomic number = number of protons
A = Z + N = mass number = number of protons + neutrons
X = Chemical symbol
What is binding energy
The energy that holds nucleons together in a nucleus, E_B
Equation for binding energy
Difference in mass from constituents
M(Z,A) = ZM(¹₁H) + NMₙ – E_B/c²
E_B/c² = ZM(¹₁H) + NMₙ – M(Z,A)
where M(Z,A) is atomic mass of isotope ᴬzX and M(¹₁H) is the mass of a hydrogen-1 atom
Relation between atomic and nuclear mass
Mₐₜ(Z,A) = Mₙᵤ꜀(Z,A) + Zmₑ
Mass excess
NOT actually related to any physics
∆(ᴬzX) ≡ M(Z,A) – A u
where M(Z,A) is atomic mass of isotope ᴬzX in atomic mass units, u
Direct method of measuring nuclear mass
Only able to be used for long-lived, stable particles
Ion moving in EM field, forces will be described by the Lorentz force
F = q(E + v × B)
If no E-field and velocity is perpendicular to B-field, ion exhibits circular motion,
F = mv²/r r̂
Equating forces
m = q|B|r/|v|
To select velocity, can include E field to only take ions passing through a slit. Could also just measure the speed.
Could also use a Penning trap, which uses an electric quadropole field. Motion of nuclei confined here can be measured and related to charge-mass ratio.
Transmutation equation notation
Inelastic collision of projectile a onto target nucleus, A, to produce B and outgoing projectile b
A + a → B + b
Can write as
A(a, b)B
For example:
¹⁴₇N(α, p)¹⁷₈O
Q-value of a reaction
Difference between nuclear rest masses of initial and final states
Q = Σᵢₙᵢₜᵢₐₗ mᵢc² – Σբᵢₙₐₗ mբc²
+ve Q values, energy is released in reaction, seen as kinetic energy of products. In general, larger Q, faster reaction, more energetically favourable it is.
-ve Q values, implies energy must be supplied for reaction to occur.
Indirect method of measuring mass
Use Q value, if we know masses of three constituents and look at final velocities, we can work out the fourth constituents mass.
Graph of binding energy per nucleon
Sharp rise, peak at Fe, slow decrease.
See fig 1.4, pg 14
Rate of nuclear decay
Activity
A(t) = dN/dt = –λN
where N is number of nuclei present and λ is decay constant.
A(t) has units of becquerels (Bq, decays/second)
Can solve differential to find number of nuclei present at a certain time, N(t), given how many there were at the start, N₀:
N(t) = N₀exp(–λt)
A(t) = λN₀exp(–λt)
Half life
t₁ᵢ₂
average time needed for number of nuclei of that isotope to halve.
N(t₁ᵢ₂) = N₀/2
t₀ = ln(2/λ)
Gamow’s saturation hypothesis and nuclear radius
As A increases, volume per nucleon remains constant.
Spherical nuclei,
4/3 πR³ = V₁A
where V₁ is related to volume of a single nucleon.
Implies
R ∝ A¹’³
where R is nuclear radius
Using scattering for charge dist. of a nucleus
Electron scattering. Process is identical to diffraction; beam incident on a small obstacle produces same diffraction pattern as a slit of equal area.
Rough estimate, assuming nucleus is a 2D disk, thus relating diameter to the first minima by
D = 1.22 λ/sinθ
where λ is de Broglie wavelength for the incoming electron.
See fig 1.7, pg 17
Cross section of a reaction
General scattering experiment, beam incident onto a target, production rates of the final states are counted.
Rate of specific reaction, Wᵣ, is proportional to flux of the beam, J, (# of particles passing through surface at the target perp. to beam direction) and number of target nuclei illuminated by the beam, N
Wᵣ ∝ JN
Constant of proportionality is the cross-section, σᵣ, such that
σᵣ ≡ Wᵣ/JN
Barn unit
1 barn = 100 fm²
Rate of particles passing through an area of solid angle dΩ (= d(cosθ)dϕ)
dWᵣ ≡ JN dσᵣ(θ,ϕ)/dΩ dΩ
Calculating differential cross section for electron scattering
Differential cross-section is a probability, so must be related to quantum-mechanical amplitude of the reaction being studied, Aᵣ
σᵣ/dΩ ∝ |Aᵣ|²
Aᵣ ∝ ∫ ψբ∗ V(r) ψᵢ d³r
Assuming point-like nucleus, can use Coulomb potential
V꜀(r) = –Zαħc/r
where α is fine-structure constant.
End up finding
A꜀ ∝ Ze² 4π/k²sin²(θ/2)
which gives differential cross section
dσ/dΩ ∝ Z²e⁴/k⁴sin⁴(θ/2)
where k is magnitude of the momentum of the incoming electron and θ is angle from initial trajectory to final (fig 1.8)
Actual differential cross section
Need to account for nucleus having finite size, so include charge density for potential
Charge density
∫ ρ(r) d³r = Z
produces potential
V(r) = –αħc ∫ ρ(r’)/|r–r’| d³r’
Eventually end up with
d/dΩ ∝ Z²e⁴/q⁴ × F(q²)² × (rel. factor)
OG Rutherford term x finite size modifier x relativity factor
F(q²) is the form factor, depends on momentum transfer, q, between incoming and outgoing electrons.
Skin thickness
Skin thickness of a nucleus is linked to the fact that nuclei dont have sharp edges.
Quantified by density falling from 90% to 10% of its original value.
What can we learn from a graph of radial charge distributions
See fig. 1.9, page 20
- Density within core appears more or less the same regardless of A. Suggests nuclear force is saturated and nucleons experience repulsion at very short distances.
- Skin thickness, t, roughly the same for all nuclei. Attractive nuclear force must be very short ranged.
- Combining these, nucleon close to
the centre of ¹²₆C is as tightly bound as a nucleon close to the centre of ²⁰⁸₈₂Pb, and similarly for nucleons close to their respective surfaces. - Densities resemble Fermi-Dirac distribution, expected as we are considering fermions. Can approximate charge distribution as
ρ(r) = ρ₀ / 1+exp[(r–R)/a]
where ρ₀ can be read off graph and
a = t/(4 ln(3))
Nuclear radius from charge radius
Mean square charge radius
⟨r꜀²⟩ ≡ 1/Z ∫ r²ρ(r) d³r
Fitting this to data for A ≥ 40 gives
⟨r꜀²⟩¹’² = 0.95 A¹’³ fm
Approximating nucleus as a homogeneously charged sphere
R꜀ₕₐᵣ₉ₑ = √5/3’ ⟨r꜀²⟩¹’² = 1.23 A¹’³ fm
Nuclear density, distribution, radius from charge density, distribution, radius
Differ by a factor of A/Z, assuming protons and neutrons are distributed evenly throughout nuclei.
ρₙᵤ꜀ₗₑₐᵣ ≈ A/Z × ρ꜀ₕₐᵣ₉ₑ
= A/Z × Z / ⁴⁄₃πR³꜀ₕₐᵣ₉ₑ
= A / ⁴⁄₃πR³꜀ₕₐᵣ₉ₑ
Find nuclear radius
rₙᵤ꜀ₗₑₐᵣ ≈ 1.2 A¹’³ fm
Charge symmetry and independence of the nuclear force
Charge here means different nucleons.
Consider elastic nucleon scattering at low energies.
Like-nucleon scattering:
Two particles are identical, cannot distinguish between small and large angle scattering. After taking Coulomb interaction into account for p-p, experimental results show proton-proton and neutron-neutron interactions are identical and hence charge symmetric.
Vₚₚ ≈ Vₙₙ
Proton-neutron interactions: See peaks of similar magnitude at both low and high angles. Low peak is from glancing collisions, not surprising. High angles would be from almost head-on collisions, which aren’t as common as glanging blows.
dσ/dΩ ∝ 1/sin⁴(θ/2)
Nuclear force not strong enough to produce backscatters.
Instead can be explained by an exchange force causing the proton to turn into a neutron, and vice versa.
Virtual exchange particles:
Different pions have similar mass, so pn scenario is approx. equivalent to pp and nn. Nuclear force is charge independent.
Find exchange particle of the nuclear force
-
Spin
Both proton and neutron are spin-1/2 particles, quanta must have integer spin. -
Charge
Protons and neutrons differ in electric charge, must have charge. For consistency with like-nucleon, must also be a neutral case. - Energy and mass
for nn scattering
n₁ → n₁ + x
x + n₂ → n₂
Energy is ‘borrowed’ using the uncertainty principle
∆x∆pₓ ≥ ħ/2
∆E τ ≥ ħ
Energy must at least be rest mass
∆E ≥ mₓc²
and it can only exist for time τ
τ ~ ħ/∆E ≤ ħ/mₓc²
Cannot travel faster than c, range is limited to
R ≲ cτ = ħc/mₓc² = 200 MeV fm / mₓc²
Nuclear force saturated, can argue range is of nucleon size order ~1fm
Rest mass of x is of order 200 MeV
Meson!
Pion factfile
Lightest mesons, come in three varieties based on electric charge: π±, π₀ with electric charges ±e and 0 respectively.
Masses vary slightly depending on if they carry electric charge.
All have 0 spin.
Lowest order Feynman diagrams for pion exchange between a proton and neutron.
See figure 1.14, pg 27
Basic form of the SEMF
M(Z,A) = ZM(¹₁H) + (A–Z)Mₙ – E_B/c²
Binding energy is composed of 5 terms.
The five binding energy terms in SEMF
- Volume
- Surface
- Coulomb
- Asymmetry
- Pairing
Volume term
Nuclear force is short range, so a nucleon only experiences the presence of its nearest neighbours, independent of the size of the nucleus.
Nuclear radius is proportional to A¹’³, so volume term is proportional to A.
Term is attractive, so positive, increasing total binding energy.
aᵥ A
Surface term
Corrects for over-estimation of the volume term. Nucleons at the surface are not surrounded by other nucleons and so are less bound.
Negative term, decreasing the binding energy, proportional to nuclear radius squared.
–aₛ A²’³
Coulomb term
Accounts for electromagnetic repulsion between protons within the nucleus.
Electric potential experienced by each proton re-written in variables common to the rest of the terms:
V = 3/5 1/4πε₀ q²/r = a꜀ Z(Z–1)/A¹’³
where all constants have been absorbed into a꜀.
For large values of Z, can approximate this to
a꜀ Z(Z–1)/A¹’³ ≈ a꜀ Z²/A¹’³
–a꜀ Z(Z–1)/A¹’³
Asymmetry term
Purely quantum mechanical effect from nuclei tending to have equal numbers of protons and neutrons.
Originates from Pauli exclusion principle:
a large excess of either type of nucleon would have to occupy increasingly higher energy states. Would then be energetically favourable for excess nucleons to decay to the other type to balance the numbers of each.
Effect is less dominant for heavy nuclei (proton Coulomb potential raises all proton energy levels), accounted for by the factor of A in the denominator.
–aₐ (A–2Z)²/A
Pairing term
Quantum mechanical effect that is justified based on observations.
Nuclei with even numbers of both protons and neutrons are the most favourable, and nuclei with odd numbers of both are least favourable.
It is energetically favourable for nucleon spins to be paired.
Form is based on fitting to data.
+aₚA⁻¹’² for even-even
0 for odd-even or even-odd
–aₚA⁻¹’² for odd-odd
Sizes of SEMF contributions
Constants must be extracted from data. Approach doesnt work for light nuclei, results will depend on where we choose to start fitting from (e.g. A ≥ 20)
Terms are ordered from greatest to least contribution.
Rough plot of contributions to E_B/A in SEMF
See fig 2.4, pg 32
Valley of stability from SEMF
Minimising SEMF w.r.t. Z gives max binding energy, most stable isotope for a given mass number.
Zₘᵢₙ = A (a꜀A⁻¹’³ + 4aₐ + Mₙ – M(¹₁H)) / 8aₐ+2a꜀A²’³
Plotting this follows trend of line of stability, including trend for light nuclei, N=Z.
What is beta decay
Nuclear reaction whereby a nucleon can change type and is accompanied by the emission of an electron (or positron) and an electron anti-neutrino (or neutrino)
β decay reaction equations
n → p + e⁻ + ν̄ₑ
p → n + e⁺ + νₑ
Generic β⁻ decay
ᴬzX → ᴬz₊₁Y + e⁻ + ν̄ₑ
Q value of β⁻ decay
Q = (Mₙᵤ꜀(ᴬzX) – Mₙᵤ꜀(ᴬz₊₁Y) – mₑ – mᵥ)c²
where Mₙᵤ꜀ is nuclear mass.
Neutrinos are thought to have masses of a few eV at most, we approximate their mass to zero;
mᵥ ≈ 0
In terms of atomic masses,
Mₐₜ(Z,A) = Mₙᵤ꜀(Z,A) + Zmₑ
Q = (Mₐₜ(ᴬzX) – Zmₑ – (Mₐₜ(ᴬz₊₁Y) – (Z+1)mₑ) – mₑ)c²
= (Mₐₜ(ᴬzX) – Mₐₜ(ᴬz₊₁Y))c²
Electron capture and its equation
Proton rich nuclei, orbiting electron is absorbed by a proton, which then turns into a neutron emitting a neutrino.
p + e⁻ → n + νₑ
Isobar
Same A, differing Z
Beta decay and SEMF
Beta decay allows nuclei to balance asymmetry between proton and neutron fermi levels.
Odd nuclear mass isobars have min. mass excess accessible through beta decays. (fig 3.1 pg 36)
Even nuclear mass isobar graphs appear as two parabola, with two local minima accessible through beta decays. Double beta decay is possible to go from the local to true minimum, but this is highly unlikely.
(fig 3.2 pg 37)
Fusion vs fission using binding energy
For A < 56, binding energy per nucleon increases with A. Energy is released when light nuclei fuse.
Opposite happens for A > 56, energy is released when nuclei split.
What are s & r processes
Paths to reach higher A elements.
Slow and rapid neutron-capture processes.
Path taken depends on intensity of neutron flux.
If neutron flux is such that the nucleus is more likely to β-decay before it absorbs another neutron, s-process will be followed.
Example,
⁵⁶₂₆Fe → ⁵⁷₂₆Fe → ⁵⁸₂₆Fe all stable.
⁵⁹₂₆Fe has half-life of ~45 days.
If likelihood of absorbing another neutron is less than once per 45 days, this unstable isotope will β-decay to ⁵⁹₂₇Co.
Process could repeat to produce ⁶⁰₂₈Ni and beyond.
Exact path followed will vary, but in general will zigzag along the line of stability until ²⁰⁹₈₃Bi, where there are no more isotopes stable enough for the s-process to continue.
Rapid neutron-capture process, r-process, is where nuclei are bombarded with neutrons faster than they can β-decay. Such environments are thought to only occur in violent astrophysical processes such as supernovae and neutron star mergers.
Example,
Fe isotopes would continue until the isotope is so unstable that it does decay faster than it can absorb an additional neutron.
r-process path follows more or less straight horizontal and vertical lines along the neutron drip lines.
Alpha decay and its reaction equation
For A > 56, E_B/A decreases with A. Eventually energetically favourable for the nucleus to split into two lighter nuclei with larger total binding energy (smaller total mass).
Fission is an example of cluster decay. When the cluster is an α-particle, ⁴₂He, this is α decay.
⁴₂He is good because it is light, has a small number of protons, but is strongly bound. Probability of it happening involves quantum tunneling out of the nucleus.
α decay reactions have form
ᴬzX → ᴬ⁻⁴z₋₂Y + ⁴₂He
Energy released through α decay
Q = (Mₙᵤ꜀(ᴬzX) – Mₙᵤ꜀(ᴬ⁻⁴z₋₂Y) – Mₙᵤ꜀(⁴₂He))c²
In terms of atomic masses
Q = (Mₐₜ(ᴬzX) – Zmₑ – (Mₐₜ(ᴬ⁻⁴z₋₂Y) – (Z–2)mₑ) – (Mₐₜ(⁴₂He) – 2mₑ))c²
= (Mₐₜ(ᴬzX) – Mₐₜ(ᴬ⁻⁴z₋₂Y) – Mₐₜ(⁴₂He))c²
In terms of binding energies
Q/c² = ZMₐₜ(¹₁H) + (A–Z)Mₙ – E_B(Z,A)/c²
– [(Z–2)Mₐₜ(¹₁H) + (A – 4 – (Z–2))Mₙ – E_B(Z–2, A–4)/c²]
– [2Mₐₜ(¹₁H) + (4–2)Mₙ – E_B(2,4)/c²]
Mass terms cancel
Q = E_B(Z–2, A–4) + EB(2,4) – EB(Z,A)
Decay only allowed if Q greater than zero, only favourable if
E_B(2,4) > EB(Z, A) – EB(Z–2, A–4)
Alpha decay constant
λ = PfT
where P is the preformation factor (probability of finding four nucleons as an α particle in a given nucleus), f is the frequency the α hits the barrier (v_α/2R), and T is the transmission
factor (Gamow)
Spontaneous fission
Nuclear process where a heavy parent nucleus breaks apart into two daughter nuclei of similar mass without external action. Process is random and the products can vary. As heavy nuclei are neutron rich, the decay is often accompanied by release of free neutrons and neutron rich daughter nuclei that may β decay towards stability.
Variation of potential for spontaneous vs induced fission
Fig 3.6, pg 44
Nucleus shape and fission
If nucleus surface becomes prolate, surface term will increase whilst Coulomb term decreases. Physically the protons repel one another, which leads to a deformation, stretching the nucleus until ultimately it falls apart. SEMF estimates this process to be likely for A ≥ 270.
Activation energy in fission
Naturally occurring isotopes rarely undergo spontaneous fission. Instead, fission can be induced, typically by adding a neutron to an isotope such that it overcomes the activation energy to become fissile.
In reactors, a fuel susceptible to induced fission is chosen such that it emits more neutrons than needed to be absorbed. This leads to a chain reaction and is a very efficient source of energy.
SEMF failings and magic numbers
Difference in SEMF predicted and measured binding energes shows unexpected stability (higher binding energies) for certain Z and N numbers.
These magic numbers are experimentally found to be
N = 2, 8, 20, 28, 50, 82, 126
Z = 2, 8, 20, 28, 50, 82
3D harmonic oscillator equations
–ħ²/2M ∇²ψ(r) + 1/2 Mω²r²ψ(r) = Eψ(r)
Splitting into 1D eqs, get e-vals
Eₙₓ,ₙᵧ,ₙ₂ = (nₓ + nᵧ + n₂ + 3/2) ħω
Splitting instead into radial and angular, ψ(r) = Rₙₗ(r)Yₗₘ(θ,ϕ), get e-vals
Eₙₗ = (2n + l + 3/2) ħω
where n = 0, 1, 2, … and l is the angular momentum of the state
