january 23, 2001physics 8411 elastic scattering of electrons by nuclei we want to consider the...
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January 23, 2001 Physics 841 1
Elastic Scattering of Electrons by Nuclei
•We want to consider the elastic scattering of electrons by nuclei to see (i) how finite size can be accounted for, and (ii) how the electromagnetic potential (1/r) is converted to a quantum mechanical matrix element.
•The presentation here follows that of Perkins, Introduction of High Energy Physics, Third Edition, chapter 6.
•We will start by considering electrons and nuclei as spinless isolated particles with the nucleus at rest. We will later consider the nucleus as part of an atom, and we will review the effects of spin, etc.
January 23, 2001 Physics 841 2
Elastic Scattering of Spinless Electronsby Nuclei - 1
• In first order perturbation theory, the transition rate is
where
•Recall,
with
•That is, and are plane waves.
€
W =2πh
| Mif | 2 ρf
€
Mif = φ f*
∫ (r r )V(
r r )φi(
r r )d
r r
€
ψ i = φie−iωit ; ψ f = φ f e
−iω f t
€
φi = eir k i⋅
r r ; φ f = e
ir k f ⋅
r r
€
ψ i (r r , t)
€
ψ i (r r , t)
January 23, 2001 Physics 841 3
Elastic Scattering of Spinless Electronsby Nuclei - 2
•Begin the calculation formally:
where
•Given a charge density
then
€
Mif = e−i
r k f ⋅
r r V(
r r )∫ ei
r k i ⋅
r r d
r r
€
= e(
r k i −
r k f )⋅
r r
∫ V(r r )d
r r
€
= er q ⋅
r r ∫ V(
r r )d
r r
€
rq ≡
r k i −
r k f
€
ρ(r R ):
€
ρ(r R )d
r R = 1∫
€
V(r r ) =
Ze2
4πρ(
r R )
|r r −
r R |
∫ dr R
January 23, 2001 Physics 841 4
Elastic Scattering of Spinless Electronsby Nuclei - 3
• Just a bit of algebra gives:
where we have defined the elastic scattering form factor
€
Mif =Ze2
4πei
r q ⋅
r r
∫∫ρ(
r R )
|r r −
r R |
dr R d
r r
€
=Ze2
4πei
r q ⋅
r R ρ(
r R )d
r R ∫
eir q ⋅(
r r −
r R )
|r r −
r R |
∫ dr r
€
=Ze2
4πF(q2 )
eir q ⋅(
r r −
r R )
|r r −
r R |
∫ dr r
€
F(q2 ) = eir q ⋅
r R ρ(
r R )d
r R ∫
January 23, 2001 Physics 841 5
Elastic Scattering of Spinless Electronsby Nuclei - 4
•With , and the polar angle between and
€
rs ≡
r r −
v R
€
rs
€
rq
€
Mif =Ze2
4πF(q2 )
eiqscosα ⋅2π ⋅s2dsd cosαs
∫
€
=Ze2
2F(q2 ) sds∫ eiqscosα ⋅d cosα∫
€
=Ze2
2F(q2 )
sds(seiqs − e−iqs )iqs
∫
€
=Ze2
2F(q2 )
2 sin qsq
∫ ds
January 23, 2001 Physics 841 6
Elastic Scattering of Spinless Electronsby Nuclei - Add Atomic Screening
•The term which depends on the nature of the potential:
diverges as so we say that the range of the electromagnetic potential is infinite.
•Atoms have clouds of electrons as well as nuclei, and the effect is to modify the electric potential:
and the exponential factor is referred to as screening.
€
2sinqsq
∫ ds
€
q2 → 0
€
′ V (r r ) = e−r /aV(
r r )
January 23, 2001 Physics 841 7
Elastic Scattering of Spinless Electronsby Neutral Atoms -1
•With the screening potential, the matrix element becomes:
The charge density is significantly greater than zero only for .
•Therefore .
recall that .
€
Mif =Ze2
4πF(q2 )
eir q ⋅(r
r −r R ) ⋅e−r /a
|r r −
r R |
∫ dr r
€
ρ(r R )
€
|r R |~ O(10−13 cm)
€
e−|r r −
r R |/a = e−s/a → e−r /a
€
a ~ O(10−8 cm)
January 23, 2001 Physics 841 8
Elastic Scattering of Spinless Electronsby Neutral Atoms -2
€
Mif →Ze2
4πF(q2 )
eiqscosα ⋅e−s/a ⋅2π ⋅s2dsd cosαs
∫
€
=Ze2
2F(q2 ) s ⋅e−s/a ⋅ds∫ eiqscosα ⋅d cosα∫
€
=Ze2
2F(q2 ) se−s/a ⋅
(eiqs − e−iqs )iqs
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
∫ ⋅ds
€
=Ze2
2F(q2 )
iqe
−s1
a−iq ⎛
⎝ ⎜ ⎞ ⎠ ⎟− e
−s1
a+iq ⎛
⎝ ⎜ ⎞ ⎠ ⎟
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
∫ ⋅ds
January 23, 2001 Physics 841 9
Elastic Scattering of Spinless Electronsby Neutral Atoms -2
€
=Ze2
2F(q2 )
iq1
1a
− iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟
−1
1a
+ iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥
€
Mif =Ze2
2F(q2 )
iqe
−s1
a−iq ⎛
⎝ ⎜ ⎞ ⎠ ⎟− e
−s1
a+ iq ⎛
⎝ ⎜ ⎞ ⎠ ⎟
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
∫ ⋅ds
€
=Ze2
2F(q2 )
iq
1a
+ iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟ −
1a
− iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟
1a
− iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟ ⋅
1a
+ iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥
January 23, 2001 Physics 841 10
Elastic Scattering of Spinless Electronsby Neutral Atoms -3
€
Mif =Ze2
2F(q2 )
iq
1a
+ iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟ −
1a
− iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟
1a
− iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟ ⋅
1a
+ iq ⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥
€
=Ze2
2F(q2 )
iq2iq
1
a2+ q2
⎡
⎣
⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥
€
=Ze2 ⋅F(q2 )
q2 + 1 / a2
January 23, 2001 Physics 841 11
•For ,
•When is ?
Elastic Scattering of Spinless Electronsby Neutral Atoms -4
€
q2 >> 1 / a2
€
Mif →Ze2 ⋅F(q2 )
q2
€
q2 >> 1 / a2
€
k = hc / a; hc = 197MeV ⋅fm = 197MeV × 10-13 cm
€
a = 10−8 cm⇒ k =197MeV × 10-13 cm
10-8 cm
€
~ 2 × 10−3 MeV = 2keV
January 23, 2001 Physics 841 12
Form Factors and Central Potentials
•We can use the same formalism for and central potential that we used for the 1/r potential, and the form factor will emerge again as a factorizable term.
€
V(r r ) = ρ(
r R ∫ )V(
r r −
r R )d
r R
€
Mif = eir q ⋅r r ρ (
r R ∫∫ )V(
r r −
r R )d
r R d
r r
€
= eir q ⋅
r r ρ(
r R ∫ )d
r R ⋅ ei
r q ⋅(
r r −
r R )
∫ V(r r −
r R )d
r r
€
=F(q2 ) eiqscosα V(s) ⋅s2ds ⋅dφ ⋅d cosα∫
€
=F(q2 )2π sV(s)2 sin(qs)
q∫ ⋅ds
January 23, 2001 Physics 841 13
The Finite Rutherford Proton - 1
•When scattering spinless electrons from protons, we found that for
•Using the notation electron + nucleus electron + nucleus
•With the assumptions that (i) the recoiling nucleus is non-relativistic, and (ii) that the nuclear recoil momentum is small compared to the momentum of the incident electron:
€
q2 >> 1 / a2
€
Mif →Ze2 ⋅F(q2 )
q2
€
′ p N = q << pe
€
rp e
€
rp e
€
r′ p Nat rest
January 23, 2001 Physics 841 14
The Finite Rutherford Proton - 2
•One can approximate
so that
•With ,
€
q2 = 2pe2 − 2pe
2 cosθ = 4 pe2 sin2 θ
2
€
dσdΩ
=Z2(e2 / 4π )2 F(q2 )[ ]
2
4 pe2 sin4 θ / 2( )
€
dΩ = 2π ⋅d(cosθ ) =2π ⋅dq2
2pe2
€
dσ
dq2 =4πα 2Z2 F(q2 )[ ]
2
q4
January 23, 2001 Physics 841 15
Point-like Elastic Scattering - 1
•Rutherford Scattering - non-relativistic quantum mechanics, first Born approximation, no spin or magnetic moments.
•Mott Scattering - spin 1/2 electrons, spin 0 protons, single photon exchange
•Dirac Scattering - spin 1/2 electrons, spin 1/2 protons, point-like , single photon exchange.
€
→ μ p = eh / 2mpc
January 23, 2001 Physics 841 16
Point-like Elastic Scattering - 2
•Rutherford Scattering:
•Mott Scattering:
•Dirac Scattering:
€
dσdΩ ⎡
⎣ ⎢
⎤
⎦ ⎥R
=Z2(e2 / 4π )2
4 pe2 sin4 θ / 2( )
€
dσdΩ ⎡
⎣ ⎢
⎤
⎦ ⎥M
=dσdΩ ⎡
⎣ ⎢
⎤
⎦ ⎥R
×cos2 (θ / 2)
1 +2pe
mp⋅sin2 (θ / 2)
⎧
⎨
⎪ ⎪
⎩
⎪ ⎪
⎫
⎬
⎪ ⎪
⎭
⎪ ⎪
€
dσdΩ ⎡
⎣ ⎢
⎤
⎦ ⎥D
=dσdΩ ⎡
⎣ ⎢
⎤
⎦ ⎥M
× 1 +q2
2mp2 × tan2(θ / 2)
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪
January 23, 2001 Physics 841 17
Rosenbluth Scattering -1
•Rosenbluth extends the Dirac formula to a finite proton:
where and
€
dσdΩ ⎡
⎣ ⎢
⎤
⎦ ⎥RB
=dσdΩ ⎡
⎣ ⎢
⎤
⎦ ⎥M
×
GE2 +
q2
4mp2 GM
2
1+q2
4mp2
⎛
⎝
⎜ ⎜ ⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟ ⎟ ⎟
+q2
4mp2
× 2GM2 × tan2 (θ / 2)
⎧
⎨
⎪ ⎪ ⎪
⎩
⎪ ⎪ ⎪
⎫
⎬
⎪ ⎪ ⎪
⎭
⎪ ⎪ ⎪
€
GE = GE(q2 ); GM = GM (q2 )
€
GEP(0) = 1; GE
N (0) = 0; GMP (0) = 2.79; GM
N (0) = −1.91
January 23, 2001 Physics 841 18
Rosenbluth Scattering -2
•To extract the form factors, one may fix andvary the scattering angle.
€
q2
€
dσdΩ ⎛ ⎝ ⎜
⎞ ⎠ ⎟ = A(q2 ) + B(q2 ) × tan2 θ{ } ×
dσdΩ ⎛ ⎝ ⎜
⎞ ⎠ ⎟M
January 23, 2001 Physics 841 19
Rosenbluth Scattering -3
Experimentally, the form factors obey a simple scaling law:
€
GEp(q2 ) =
GMp (q2 )μ p
=GM
n (q2 )μn
; GEn (q2 ) = 0
€
GMp
μ p
€
GEp
€
GMn
μn