Nuclear e 2012
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Transcript of Nuclear e 2012
Basic Concepts in Nuclear Physics
Paolo Finelli
Literature/Bibliography
Some useful texts are available at the Library:
Wong, Nuclear Physics
Krane, Introductory Nuclear Physics
Basdevant, Rich and Spiro, Fundamentals in Nuclear Physics
Bertulani, Nuclear Physics in a Nutshell
Introduction
Purpose of these introductory notes is recollecting few basic notions of Nuclear Physics. For more details, the reader is referred to the literature.
Binding energy and Liquid Drop Model
Nuclear dimensions
Saturation of nuclear forces
Fermi gas
Shell model
Isospin
Several arguments will not be covered but, of course, are extremely important: pairing, deformations, single and collective excitations,α decay, β decay, γ decay, fusion process, fission process,...
The Nuclear LandscapeThe scope of nuclear physics is
Improve the knowledge of all nuclei Understand the stellar nucleosynthesis
© Basdevant, Rich and Spiro
e−5
e−6
e−7
≥ e−4
Stellar Nucleosynthesis
Dynamical r-process calculation assuming an expansion with an initial density of 0.029e4 g/cm3, an initial temperature of 1.5 GK and an expansion timescale of 0.83 s.
The r-process is responsible for the origin of about half of the elements heavier than iron that are found in nature, including elements such as gold or uranium. Shown is the result of a model calculation for this process that might occur in a supernova explosion. Iron is bombarded with a huge flux of neutrons and a sequence of neutron captures and beta decays is then creating heavy elements.
The evolution of the nuclear abundances. Each square is a nucleus. The colors indicate the abundance of the nucleus:
©JINA
mNc2 = mAc2 − Zmec2 +
Z
i=1
Bi mAc2 − Zmec2
B = (Zmp + Nmn) c2 −mNc2 [Zmp + Nmn − (mA − Zme)] c2
B =Zm( 1
H) + Nmn −m( AX)
c2
Binding energy
Electrons Mass (~Z)
Atomic Mass Electrons Binding Energies (negligible)
© Basdevant, Rich and Spiro
E/A
(Bin
ding
Ene
rgy
per n
ucle
on)
A (Mass Number)
Average mass of fission fragments is 118
Fe Nuclear Fission Energy
Nuclear Fusion Energy
235U
© Gianluca Usai
The most bound isotopes
Binding energy
Binding energy and Liquid Drop Model
© Basdevant, Rich and Spiro
Volume term, proportional to R3 (or A): saturation
Surface term, proportional to R2 (or A2/3)
Coulomb term, proportional to Z2/A1/3
Pairing term, nucleon pairs coupled to JΠ=0+ are favored
Asymmetry term, neutron-rich nuclei are favored
Binding energy and Liquid Drop Model
© Gianluca Usai
Contributions to B/A as function of A
Comparison with empirical data
Nuclear Dimensions
Ground state
Excited States (~eV)
© Gianluca Usai
Ground state
Ground state
Excited States (~ MeV)
Excited States (~ GeV)
Nuclear Dimensions: energy scales
ρ(r) =ρ(0)
1 + e(r−R)/s
R : 1/2 density radiuss : skin thickness
Nuclear Dimensions
© Basdevant, Rich and Spiro
Fermi distribution
Nuclear forces saturation
An old (but still good) definition:
© E. Fermi, Nuclear Physics
Mean potential method: Fermi gas modelIn this model, nuclei are considered to be composed of two fermion gases, a neutron gas and a proton gas. The particles do not interact, but they are confined in a sphere which has the dimension of the nucleus. The interaction appear implicitly through the assumption that the nucleons are confined in the sphere. If the liquid drop model is based on the saturation of nuclear forces, on the other hand the Fermi model is based on the quantum statistics effects.
The Fermi model could provide a way to calculate the basic constants in the Bethe-Weizsäcker formula
H =A
i=1
Ti ≡A
i=1
− 22M
∇2i
Hψ(r1,r2, . . .) = Eψ(r1,r2, . . .)
ψ(r1,r2, . . .) = φ1(r1)φ2(r2) . . .
− 22M
∇2iφ(ri) = Eφ(ri)
E = E1 + E2 + E3 + . . . =A
i=1
Ei
k2i ≡ (k2ix + k2iy + k2iz) =2MEi
2 > 0
φi(r) ≡ φi(x, y, z) = N sin(kixx) sin(kiyy) sin(kizz)
d2φi(x)
dx2= −k2ixφi(x)
Fermi gas model (I)
Hamiltonian
Wavefunction factorization
Boundary conditions
Separable equations
Gasiorowicz, p.58
φi(x) = B sin(kixx)
1 =
L
0dx|φi(x)|2 = B2
L
0dx sin2(kixx) = B2L
2B =
2
L
φi(r) =
2
L
3/2
sin(kixx) sin(kiyy) sin(kizz)
kix =π
Ln1i , kiy =
π
Ln2i , kiz =
π
Ln3i
(n1i, n2i, n3i = positive integers)
Ei =2k2i2M
=22M
(k2ix + k2iy + k2iz)
Ei(n1i, n2i, n3i) =2π2
2ML2(n2
1i + n22i + n2
3i)
Fermi gas model (II)
Solution
Normalization
∆kx,y,z =π
L(n1,2,3 + 1− n1,2,3) =
π
L
dn(k) =1
84πk2dk
1
(π/L)3≡ Ω
(2π)3dk
n(k) =
k
0dn(k) =
Ω
(2π)34π
3k3
A = 4
kF
0dn(k) =
Ω
(2π)344π
3k3 = Ω
2k3F3π2
ρ0 =2k3F3π2
Ω ≡ L3
ρ0 = A/Ω
Fermi gas model (III)
Density of states
Numberof particles
Densityof particles
spin-isospin
Fermi momentum
θ(kF − k)
Fermi gas model (IV)
Fermi gas distribution:N(k)
kkF
1
0
Step function
filled empty
ρ0 = 0.17 fm−3 kF = 1.36 fm−1
F =2k2F2M
= 38.35 MeV
T = 23 MeV
4dn(k)N(k) = 4Ω
(2π)3θ(kF − k)dk
T = Ω2
π2
2k22M
k2dkθ(kF − k) = Ω2k3F3π2
3
5
2k2F2M
= A3
5F
(BE)vol = −bvolA (bvol = 15.56 MeV)
< U >= −15.56− < T > −39 MeV
Fermi gas model (V)
The fermi level is the last level occupied
Evidences of Shell Structure in Nuclei
© Basdevant, Rich and Spiro
En = (n + 3/2)ω
H = Vls(r)l · s/2
l·s2 = j(j+1)−l(l+1)−s(s+1)
2= l/2 j = l + 1/2= −(l + 1)/2 j = l − 1/2
Mean potential method: Shell modelThe shell model, in its most simple version, is composed of a mean field potential (maybe a harmonic oscillator) plus a spin-orbit potential in order to reproduce the empirical evidences of shell structure in nuclei
© Basdevant, Rich and Spiro
Mean potential method: Shell model
Hi =1
2mp2i +
1
2Mω
20r
2i − V0
p2
2M+
1
2Mω2
0r2
ψ(r) = (E + V0)ψ(r)
ψ(r) = Rnl(r)Ylm(θ,φ)
Rnl(r) = (−1)n
2
(l + 1/2)!
l + n+ 1/2
n
rle−λr2/21F1
−n, l +
3
2,λr2
Shell model (I)
H =A
i=1
Hi
1F1 (−n, µ+ 1, z) =Γ(n+ 1)Γ(µ+ 1)
Γ(n+ µ+ 1)Lµn(z)
EN =
N +
3
2
ω0 N = 2n+ l
d = 2N
l=0
(2l + 1) = 2
[N/2]
n=0
(2(N − 2n) + 1) =
= 2(2N + 1)
N
2+ 1
− 8
[N/2]
n=0
d = (N + 1)(N + 2)
Shell model (II)
Degeneracy
Shell model (III)
Shell model (IV)
Shell model (V)
Shell model (V)
Isospin
In 1932, Heisenberg suggested that the proton and the neutron could be seen as two charge states of a single particle.
939.6 MeV938.3 MeV
EM ≠ 0 EM = 0n
pN
Protons and neutrons have almost identical mass
Low energy np scattering and pp scattering below E = 5 MeV, aftercorrecting for Coulomb effects, is equal within a few percent
Energy spectra of “mirror” nuclei, (N,Z) and (Z,N), are almost identical
ψN (r, σ, τ) =
ψp(r, σ, 12 ) proton
ψn(r, σ,− 12 ) neutron
η 12 , 1
2= |π =
10
η 1
2 ,− 12
= |ν =
01
Isospin is an internal variable that determines the nucleon state
One could introduce a (2d) vector space that is mathematical copy of the usual spin space
proton state neutron state
Isospin (II)
τ3|π = |πτ3|ν = −|π
ψN = a|π + b|ν =
ab
[ti, tj ] = iijktk
Pp = 1+τ32 = Q
ePn = 1−τ3
2
τ1, τ2, τ3
ti =12τi
t+|ν = |πt−|π = |νt+|π = 0t−|ν = 0
t± = t1 ± it2
Isospin
eigenstates of the third component of isospin
In general
The isospin generators
Projectors Raising and lowering operators
Pauli matrices
neutron to proton proton to
neutron
Fundamental representations
T = t1 + t2 T = 0, 1
T = 0 η0,0 = 1√2(π1ν2 − ν1π2)
T = 1
η1,1 = π1π2
η1,−1 = ν1ν2
η1,0 = 1√2(π1ν2 + π2ν1)
Isospin for 2 nucleons
|T = 1, Tz = 1 = |pp
|T = 1, Tz = −1 = |nn1√2
[|T = 1, Tz = 0+ |T = 0, Tz = 0] = |pn
Proton-proton state
Neutron-neutron state
Proton-neutron state
Isospin for 2 nucleons
ψ(1, 2) = ψpp(r1, σ1, r2, σ2)η1,1 + ψnn(r1, σ1, r2, σ2)η1,−1 + ψanp(r1, σ1, r2, σ2)η1,0 + ψs
np(r1, σ1, r2, σ2)η0,0
PT=0 =1− τ (1)τ2
4PT=1ν=1 =
1 + τ (1)3
21 + τ (2)
3
2
PT=1ν=0 =
14(1 + τ (1)τ (2) − 2τ (1)
3 τ (2)3 )
η0,0η1,1
PT=1ν=−1 =
1− τ (1)3
21− τ (2)
3
2 η1,−1 η1,0
antisymmetric symmetric
Wavefunction
Ψ(r,s1,s2,t1,t2) = φ(r)fσ(s1,s2)fτ (t1,t2)
(−)L+S+T = (−)
Symmetry for two nucleon states
the overall wavefunction must be antisymmetric
L=0, S=1 T=0 3S1isospin singlet
Sistema di 2nucleoni identici
(pp,nn)
Sistema di 2nucleoni distinti
(pn)
ISOSPIN SPAZIO SPIN
Tz = ±1
Tz = 0
Funzione simmetrica(tripletto T=1)
Funzione antisimmetrica(singoletto T=0)
L dispari
L dispari
L pari
L dispari
ψ(x)
ψ(x)
ψ(x)
ψ(x)
antisimmetrica
antisimmetrica
simmetrica
simmetrica
S=1
simmetrica(no onda S)
S=0
ψ(σ)
ψ(σ) antisimmetrica
1S0
L pari
L pari
ψ(x) simmetrica
ψ(x) antisimmetrica
S=1
S=1
simmetricaψ(σ)
simmetricaψ(σ)
S=0
S=0
ψ(σ) antisimmetrica
ψ(σ) antisimmetrica
(no onda S)
1S0
(no onda S)
3S1Tz = 0
Funzione simmetrica(tripletto T=1)
pp np nn
0.0
60 eV
-2.23 MeV3S1 (T=0)
1S0 (T=1)1S0 (T=1)1S0 (T=1)
Coulomb
Additional slides
...many open questions
v(r − r) = −v0δ(r − r)
V (r) =
dr v(r − r)ρ(r)
dr v(r) ∼ 200 MeV fm3
V (r) =V0
1 + e(r−R)/R
Mean potential methodThe concept of mean potential (or mean field) strongly relies on the basic assumptionof independent particle motion, i.e. even if we know that the “real” nuclear potentialis complicated and nucleons are strongly correlated, some basic properties can be adequately described assuming individual nucleons moving in an average potential (it means that all the nucleons experience the same field).
a rough approximation could be
where v0 can be phenomenologically estimated to be
Then one can use a simple guess for V: harmonic oscillator, square well, Woods-Saxon shape...