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