Download PHY492: Nuclear & Particle Physics Lecture 4 Nature of the nuclear force

PHY492: Nuclear & Particle Physics
Lecture 4
Nature of the nuclear force
Reminder: Investigate
www.nndc.bnl.gov
Physics of nuclei
Topics to be covered
•size and shape
•mass and binding energy
•charge distribution
•angular momentum (spin)
•symmetries (parity)
•magnetic moments
Nuclear size
Binding energy per nucleon is < 1% of
the nucleon mass. The protons and
neutrons in the nucleus retain their
particle properties. Assume nuclei
are spherical and have a constant
density (not compressed).
1
V  r3 ⎫
⎬ r  A3
VA⎭
•radioactivity
•energy levels
•reactions
Nuclear radius
(
r = 1.2 × 10
January 22, 2007
−15
)
m A
1
3
Carl Bromberg - Prof. of Physics
2
Nuclear mass and binding energy
The mass of a bound system is always less than the mass of its
component parts. For example, the mass of the hydrogen atom is
13.5 eV/c2 less than proton mass plus electron mass. When the
hydrogen atom is formed, 13.5 eV is released in photons.
mH c 2 − m p + me c 2 = −13.5 eV electron binding energy in hydrogen
(
)
What is the sign of the binding energy?
D&F: the binding energy of a bound system is negative. So be it.
Nuclear binding energy
(
)
B.E. = M ( A, Z ) c 2 − Zm p + Nmn c 2 where N = A − Z
B.E.
A
binding energy per nucleon,
~ the energy to remove one nucleon
Sometimes “mass excess” Δ is given in the tables.
where u = 931.5 MeV, the average
Δ = M ( A, Z ) c 2 − A ⋅ u
nucleon “mass” in 12C.
(
)
B.E. = Δ + A ⋅ u − Zm p + Nmn c 2
January 22, 2007
Carl Bromberg - Prof. of Physics
3
Nuclear charge distribution via electron scattering
Electron scattering
Relativistic invariants: t,s
Momentum transfer squared : t
(difference of final & initial momentum)2
2

 2
t = ( Eb′ − Eb ) − ( pb′ − pb )
Center of Mass Energy squared : s
ee2
4
(effective m c of beam + target)


⎧
p′b , Eb′ )
(
pb , Eb )
Feynman
(
⎪
2

 2⎪
s = ( E b + E t ) − ( pb + pt ) ⎨
diagram

⎪

( pt , Et )
⎪⎩
( pt′, Et′)
A
For target in the Lab frame
2

t = Et′ − mt c 2 − pt′2 c 2
2 2
2
= Et′ − pt′ c − 2 E ′mt c 2 + mt 2 c 4
(
(
)
(
)
)
= −2mt c 2 E ′ − mt c 2 = −2mt c 2Tt ′ = − pt′2 c 2
January 22, 2007
N.R.
A
2
  2
t = ( Et′ − Et ) − ( p′t − pt )
In lab, target starts at
rest, picks up the
transferred momentum
− momentum2 of target
Carl Bromberg - Prof. of Physics
4
Electron scattering to probe nucleus
Rutherford scattering (θ)
Electron scattering
Momentum
transfer2
2
⎛ ZZ ′e2 ⎞
dσ
1
(θ ) = ⎜
dΩ
⎝ 4E ⎟⎠ sin 4 θ
2
Assume only the direction of beam
changes so that E,|p| = E’,|p’|

 2
2
t = ( Eb′ − Eb ) − ( pb′ − pb ) c 2
Momentum transfer q
q 2 c 2 = −t = 2 p 2 c 2 (1 − cosθ )
= −2 pb pb′ c 2 + 2 pb pb′ c 2 cosθ
= −2 p 2 c 2 (1 − cosθ )
q 2 = 2 p 2 (1 − cosθ )
Trigonometry
θ 1 − cosθ
=
2
2
dq 2 = −2 p 2 d ( cosθ )
sin 2
dΩ = −2π d ( cosθ )
January 22, 2007
v = velocity of electron ≈ c
Z =1
Rutherford electron scattering (q)
dσ
(θ ) =
2
dq
(
4π Z ′e2
v2
)
2
1
q4
Carl Bromberg - Prof. of Physics
5
Point-like scattering
Rutherford scattering
(
dσ 4π Z ′e
=
2
dq
v2q 4
)
2 2
Mott included effects of electron
spin (relativistic quantum mechanics)
⎡ dσ
⎤
⎤
2 θ ⎡ dσ
θ
=
4
cos
θ
(
)
(
)
⎢ dq 2
⎥
⎢ dq 2
⎥
2
⎣
⎦ Mott
⎣
⎦ Rutherford
What happens if the charge distribution is not point-like?
In other words, the electron approaches the nuclear surface?
dσ
2 ⎡ dσ ⎤
= F (q) ⎢ 2 ⎥
2
dq
⎣ dq ⎦ Mott
Nuclear form factor
January 22, 2007
Carl Bromberg - Prof. of Physics
6
Form factors -nuclear charge distribution
Nuclear charge distribution
Nuclear form factor
ρ ( r ) = Ze f ( r )
F ( q ) = ∫ d 3r f ( r ) eiqir /
Modified differential cross section
Fourier transform of distribution
dσ
2 ⎡ dσ ⎤
= F (q) ⎢ 2 ⎥
2
dq
⎣ dq ⎦ Mott
( )
f (r )
F q2
(1 / 4π )δ ( r )
(a
3
(a
2
8π ) exp ( −ar )
2π
)
32
⎛ r2 ⎞
exp ⎜ − 2 ⎟
⎝ 2a ⎠
3
(r < R)
4π R 3
January 22, 2007
2
1
⎛
q2 ⎞
⎜⎝ 1 + 2a 2  2 ⎟⎠
−2
⎛ −q 2 ⎞
exp ⎜ 2 2 ⎟
⎝ 2a  ⎠
3 3 ⎡ qR qR
qR ⎤
sin
−
cos
 ⎥⎦
( qR )3 ⎢⎣  
Carl Bromberg - Prof. of Physics
7
Obtain Rutherford scattering via quantum mechanics
Assume:
•particles are plane waves (known as the Born Approximation)
•target recoil energy is neglected
•particles have spin 0 (electrons are fermions - spin 1/2)
•point like scattering (charge distribution included later)
Fermi’s Golden Rule
Probability per unit time to make a transition from
2
2π
initial state ψ to any one of many final states ψ’ ,
P=
ψ ′ H ψ ρ ( E′)
with a density of final states ρ(Ε’ ) .

tricky




1 ipi r /
1 ip ′ i r /
e
,ψ′ =
e
Free particle wave functions: ψ =
V
V
1 −ip ′ i r /
∗
3
∗
ψ′ =
e
ψ ′ H ψ = d r ⎡⎣ψ ′ H ( r )ψ ⎤⎦
Matrix
V
element
Coulomb potential
V (r) =
Z ′e
r
January 22, 2007
∫
 
e
3
iqi r /
= ∫ d r V( r ) e
V
Z ′e2
ψ′ H ψ =
V
momentum transfer
  
q = p′ − p
eiqr cos θ / 2
∫ r r drd ( cosθ ) dφ
Carl Bromberg - Prof. of Physics
8
Evaluation of the matrix element
Z ′e2 eiqr cos θ / 2
ψ′ H ψ =
r drd ( cosθ ) dφ
∫
V
r
∞
1
2π Z ′e2
iqr cos θ /
=
rdr
e
d ( cosθ )
∫
∫
V 0
−1
∞
2π Z ′e2
iqr /
−iqr /
⎡
=
rdr
2
/
qr
e
−
e
/ 2i ⎤⎦
(
)
∫
⎣
V 0
(
)
∞
4π Z ′e2 
4π Z ′e2  2
∞
=
dr sin ( qr /  ) = −
⎡⎣ cos ( qr /  ) ⎤⎦ 0
2
∫
qV 0
qV
4π Z ′e2  2
=
q 2V
January 22, 2007
Scattering probability
per unit time
Carl Bromberg - Prof. of Physics
9
Other factors in golden rule
Density of final states
ρ(E)
Probability ---> cross sections
2
2π
P=
H fi ρ ( E ′ )

dpx dx = 2π  (each state)
dn ( p′ ) =
Vd 3 p′
( 2π  )
3
=
Vp′ 2 dp′dΩ
( 2π  )
(scatters/second/beam particle/target particle)
3
dE ′ = v′dp′
V ψ normalization
=
; v = v′
v′
beam velocity
dn ( p′ ) V p′ 2 dΩ
ρ ( E′) =
=
dE ′
v′ ( 2π  )3
2
V
V 2π
P=
H fi ρ ( E ′ )
v
v 
V
dσ = P
v
January 22, 2007
(scatters/second)
( beam/second ) ( targets/unit-area )
Carl Bromberg - Prof. of Physics
10
Fermi’s “Golden Rule” ---> Rutherford scattering
2
V 2π
dσ =
H fi ρ ( E ′ )
v 
2π V ⎡ 4π Z ′e  ⎤ ⎡ V p′ 2 ⎤
dσ =
⎢
⎢
⎥
3 ⎥ dΩ
2
v ⎣ q V ⎦ ⎢⎣ v ( 2π  ) ⎥⎦
2
2
2
2
2π ⎡⎣ 4π Z ′e ⎤⎦ p′ 2 ⎛ π dq 2 ⎞
=
( 2π )3 v 2 q 4 ⎜⎝ p′ 2 ⎟⎠
2
(
dσ 4π Z ′e
=
2
dq
v2q 4
January 22, 2007
)
2 2
Trigonometry
q 2 = 2 p 2 (1 − cosθ )
dq 2 = −2 p 2 d cosθ
dΩ = −2π d cosθ
Rutherford scattering
Carl Bromberg - Prof. of Physics
11