Chapter 3

The Size and Shape of Nuclei

◎ The size of nuclei

● The scattering of electron by nuclei

◎ The nuclear electric charge distribution

● The nuclear electric form-factor

◎ Nuclear scattering and nuclear size

◎ The shape of nuclei

“7And I will establish My covenant between Me and you and your seed after you throughout their generations for an everlasting covenant, to be God to you and to your seed after you. 8And I will give you and to your seed after you the land of your sojournings, all the land of Canaan, for an everlasting possession; and I will be their God.” (Genesis 17:7 ~ 8)

Borders of the land of Canaan

§ 3.1 The size of nucleiWhat we are interested here is

1 the size of nuclei

2 how the nuclear matter is distributed inside a sphere, if the nucleus is spherical.

In fact

Nuclei are not always spherical!

A prolate deformed nucleus

What we mean by the size of a nucleus ?

The electron probability density for the first few hydrogen atom electron orbitals shown as cross-sections. These orbitals form an orthonormal basis for the wave function of the electron.

Rough visualization of an electron "wave function". (Art by

Blake Stacey.)

In atomic physics, the boundary of an atom is not

sharp since the wave function of the outer electrons

decreases monotonically.

We have a similar situation in the nuclear physics!

Rutherford’s formula breaks down when the kinetic energy of the incoming α particle is too high and it comes too close to the target nucleus.

1 the finite charge density distributions overlap.

2 strong nuclear forces between the α particle and the target nucleus comes into play.

These deviations occur at short distances of approach.

Size - Rutherford’s result

measurements by Rutherford et al nuclei size of about 10-14 m

Nuclear radii, R, are seen to increase with the total number of nucleons, A,

R = r0A1/ 3

To reveal more detail needs smaller de Broglie wavelength for the probing particle

(1)

To reveal more detail needs smaller de Broglie wavelength for the probing particle

We need to use proper incident particles with proper kinetic energies for probing nuclear structures.

ph

(2)

Charge(1) The charge of the nucleus is just the sum of the proton charges(2) The hydrogen atom is precisely neutral(3) The neutron is precisely neutral - but has a magnetic moment - an internal charge structure - and so a composite object

net hydrogen charge is less than 1 x 10-21 e

Hydrogen n=4 level

Neutron charge is less than 2 x 10-21 e

However the neutron does have a magnetic moment

moving charges

a composite object - not fundamental

There are two different kinds of distributions which we are considering.

1 Nuclear matter distribution

2 Nuclear charge distribution

1 The nucleus has mass density, ρ = mass divided by its volume.

30

30 4

343

rm

ArAm

VM nn

mass of the neutron

neutrons andprotons have the samemass to 0.14%

= about 1.82 x 1017 kg/m3

so nuclear density does not depend on the atomic mass number

(3)

The density distributions for some nuclei

ArZe

VQ

c 304

3

2 Nuclear charge distribution ρc = total charge divided by its volume.

Since N ~ Z ~ A/2

33

03

0

C/m 83

43

re

ArZe

VQ

c

It is roughly a constant.

(4)

(5)

We may use electrons to probe the charge distribution in a nucleus because electrons do not experience the nuclear force.

Accelerated neutrons can be used as a probe for nuclear matter distribution since neutrons are electrically neutral and do not experience Coulomb force.

Nuclear matter distribution Nuclear charge distribution

R = r0A1/ 3

r0 = 1.4 F for nuclear matter distribution

= 1.2 F for charge distribution

Rough description

§ 3.2 The scattering of electrons by nuclei

When a gold ( Au19779 ) foil is bombarded by energetic α particles:

(1) R > 10-14 m Coulomb scattering r1 ~Vcoul

(2) R < 10-14 m Nuclear potential r

e ar

~VN

a-1 is an estimation of the short nuclear interaction range

a-1 ≈ 1 ~ 2 F

We may use electron beams as probes to study nuclear charge distribution. ― electron scatteringEarly problems with electron beams:

1 Beam energy is not mono-energetic.

2 Electron’s linear momentum too small and easily experiences large angle scattering as well as multiple scatterings

3 Small incident kinetic energy

MeV 1~ eT MeV/c 42.1~ep

F 87342.1

23.197

ep

hde Broglie’s wave length wave length too long !!

A typical electron energy from beta decay.

In early years there were not comprehensible results through electron scattering data due to insufficient electron’s kinetic energy.

If we are to acquire sensible results out of electron scatterings the de Broglie wavelength must be of the order of 10 F.

The corresponding linear momentum (pe) for an electron is

cMeV 96.123

F 10

Fc

MeV23.197

ep

And the kinetic energy of an electron must be around 124 MeV. MeV 12422 eeee mpmT

The scattering of energetic electrons (Te > 100 MeV) is a very important tool in the investigation of the nuclear size.

The formula for the differential cross-section for the elastic scattering of relativistic electrons was derived by Mott using relativistic quantum mechanics. Mott formula is analogous to Rutherford’s for α particle scattering.

)2/(cosec4d

dσ 422

2222

vP

cZ

The Rutherford formula for the differential angular cross-section for the elastic scattering of a non-relativistic, spin-less particles of unit charge, momentum P, and velocity v, at a fixed (no recoil) target nucleus of atomic number Z, is

If the incident particle is an electron which is unpolarized and can be relativistic (v→c), the formula becomes

(6)

)]2/(sin1)[2/(cosec4d

dσ 22

24

22

2222

cv

vPcZ

(7)

This is the Mott formula.

)]2/(sin1)[2/(cosec4d

dσ 22

24

22

2222

cv

vPcZ

Assumption in the Mott formula:

1. Relativistic quantum mechanical treatment for the electron;2. First-order perturbation theory is adequate to calculate the scattering

cross section;3. There is no nuclear recoil;4. The nuclear electric charge is point-like;5. The nuclear spin is zero.

(7)

§ 3.3 The electric charge distribution

There are two models which are to be used in the study of nuclear charge distributions.

Model I is a sharp-edged charge distribution which is very unlikely but can be tested.

Model II softens the hard edges by assuming a charge distribution with a mathematical form normally associated with the Fermi-Dirac statistics but which, applied to nuclei, is called the Saxon-Woods form.

Nuclear charge distribution

dar

rexp1

)( 0

Saxon-Woods form

(8)

Model I Model II

a

t

ρ

r

Model Iρ(r) = ρ0, r < a

ρ(r) = 0, r > a

Since the total charge in the nucleus is Ze. Therefore the following equation must be satisfied.

Zea

03

43 (9)

Model II

dar

rexp1

)( 0 (8)

3/1Aa F 52.0d

F 28.2t

00

30

]/)exp[(14

]/)exp[(1 dardr

dardrZe (10)

Total ChargeIn a nucleus For a spherical nucleus

a

t

ρ

rdt 39.4

“19But God said, No, but Sarah your wife will bear you a son, and you shall call his name Isaac; and I will establish My covenant with him for an everlasting covenant for his seed after him. 20And as for Ishmael, I have heard you; indeed, I have blessed him and will make him fruitful and will multiply him exceedingly. Twelve princes will he beget, and I will make him a great nation.” (Genesis 17:19 ~ 20)

Scattering experimentsincident particle is a plane wave

Scattered wave fronts are spherical

intensity variations on wave-frontdue to diffraction

diffraction pattern observed fromplane wave striking a target

analogy with optics: like opticaldiffraction but nucleus has blurred edges

In optics if the obstacle (lens) size D is much larger than the wavelength λ of an

incoming wave (D >>λ) there would appear diffraction pattern. The observed pattern on

a screen which is located behind the obstacle is called the

“Fraunhofer diffraction pattern”.

Mathematically we perform a Fourier transform on the obstacle and is able to describe diffraction patterns seen on the screen. In other words diffraction pattern is no less than the Fourier transformed image of the obstacle.

For a disk lens of diameter D the first minimum appearing on the screen satisfies the following relation.

(11)

The optical description of the Fraunhofer diffraction pattern

)/22.1(sin 1 D First minimum of the diffraction pattern

The Fraunhofer diffraction pattern

)/22.1(sin 1 D First minimum of the diffraction pattern

In the electron scattering experiment we may treat the incoming electrons as quantum mechanical waves. The wavelength of incident electrons is inversely proportional to the electron’s linear momentum.

The nuclear charge distribution is regarded as the “optical obstacle” which stays in the path of incoming “electronic wave”.

The angular distribution of differential cross sections measured on various scattering angles can be regarded as the diffraction patterns after the “electronic wave” passing

through the nuclear charge distribution.

eph

0, ,0

arar

D = 2a = 2 4.1 = 8.2 F

Te = 450 MeV → = 2.76 F

The first maximum occurs at

242.876.2

22.1sin 22.1sin 1-1

D

θ ≈ 24°

Electron scattering Proton

scattering

Tp = 1000 MeVλ= 0.73 F

Te = 502 MeVλ=2.47 F

λ=2.47 F

Angular differential cross section

Optical diffraction pattern

First minimum

§ 3.4 The nuclear electric form-factor

For electron scattering we take the nucleus to have charge Ze where e is the charge on the proton.

If the nucleus is point-like the measured differential cross section (dσ/dΩ) is dependent on the scattering amplitude Zef(θ) at large distance at an angle θ, so that

222

Mott

)( feZdd

(12)For a point-like nucleus

2222

Mott

)()( AfeZdd

)()( ZefA where

2)(A is dependent on various scattering details such as the wavelength of the incident wave, energy, or momentum etc.

(12)

(13)

In general A(θ) is a complex function which is called the scattering amplitude.

In the case of point Coulomb scattering

)2/(sin1

42

2

EZze16

)( dd 2

Coulomb

)2/(sin41)( 2

E

f

and the scattering amplitude for point Coulomb scattering is

)2/(sin4)( 2

2

EZzeA (14)

In the case of point Mott scattering

)]2/(sin1)[2/(cosec4d

d 22

24

22

2222

Mott

cv

vPcZ

2/1

Mottdd)(

A

and the scattering amplitude for the point Mott scattering is

(15)

While the nuclear charge is no longer a point we need a form factor F(θ) to modify our scattering formula. Namely,

(16)2

Mott

2

)(

)(dd)F()(

dd

A

Mott

2 )(dd/)(

dd)F(

(17)

Form factor F(θ)

Mott

2 )(dd/)(

dd)F(

(17)

The form factor F(θ) is connected to the internal charge distribution of a nucleus. It is actually the Fourier transform of the nuclear charge density ρ(r).

Form factor F(θ) Scattering amplitude A(θ)

Differential cross section [dσ/dΩ(θ)]

If that charge is spread out then an element of charge d(Ze) at a point r will give rise to a contribution to the amplitude of eiδf(θ)d(Ze) where δ is the extra ‘optical’ phase introduced by wave scattering by the element of charge at the point r compared to zero phase for scattering at r = 0.

The incident and scattered electron have momentum

p and p’ with p =∣p =∣ ∣p’∣.

The momentum transfer q [q = 2psin(θ/2)] is along OZ, O being the nuclear center.

p

p’

The ‘optical ray’ P1OP1’ (path a) is taken to have zero relative path length.

The ray P2SP2’ (path b) has equal angles of incidence and reflection with the ray P1OP1’ at the plane AXA’ which is perpendicular to OZ.

Due to different lengths of two paths there will be a phase difference δ for waves coming from two paths.

ab ll

2

(18)

la

lb

(path a)

(path b)

r

The contribution to the scattering amplitude for a ‘point charge’ with volume element dV and located at the position r is

ii edVfefdQ )()()()( r (19)

r

And the scattering amplitude is the volume integral of the eq. (19).

(20)dVefA i )()()( r

The line OZ bisect the angle (∠ P1OP1’) and intersect with the line AA’ at the point X.

Every path that is reflected from the plane AX’A ( line ⊥ OZ) is of the same ‘optical path length’ . Therefore all paths reflected from the plane AX’A have the same ‘optical path length’ .

P0

P0’

The optical length difference between P2SP2’ and P1OP1’ = The optical length difference between P0XP0’ and P1OP1’

d

)2/sin()(2 OXd

P0

P0’

θP1

P1’

X

O

)(21

d/2

θ

(21)

It is evident from the figure that (OX) = (OS) cos α = r cos α

la

lb

)2/sin()cos(2)2/sin()(2 rOXdll ab (22)

)2/sin()cos(2)2/sin()(2 rOXdll ab

The phase difference δcan then be expressed as

)2/sin()cos(2222

rdll ab

Finally

)2/sin()cos(4 r (23)

Here λ = p/h, and p is the linear momentum of the incident electron.

)2/sin()cos(4 r (23)

With λ = p/h the equation (23) can also be written as

cos)2/sin(21 rp

(24)

rq

Since the magnitude of the momentum transfer (q) for the electron elastic scattering on a nucleus is [2p sin(θ/2)] it

can be shown geometrically that

(25)

dVefA i )()()( r (20)

Substitute the expression in the equation (25) into the equation (20) we get

dVefrA i /)()()( rq (26)

More specifically

rqierrfA

dβ dαdr αsin)()()(2

0 0 0

2 (27)

Now we have for the total charge

dαdr αsin)(20 0

2

rrZe

So we see that we can write

)()(αdrdαsin)(

αdrdαsin)()()(

2

2

FZef

rr

errZefA

rqi

(28)

Thus the Mott (or Rutherford) scattering amplitude Zef(θ) is changed by a factor F(θ) and the scattering formula becomes

Mott

2

dd)

dΩdσ

F(θ (29)

F(θ) is called a form factor.Somewhat formally it could be written

dVerZedVr

dVerF rqi

rqi

)(1)(

)()(

(30)

dVerZedVr

dVerF rqi

rqi

)(1)(

)()(

(30)

Since the form factor is more properly a function of q than of θ it has become useful to write F(q2) rather than F(θ).

It is immediately recognized that the form factor F(q2) [or F(θ)] is actually the Fourier transform of the nuclear charge distribution.

The physical meaning of the form factor F(q2).

1. When q → 0 (without momentum transfer), F(q2) → 1 and the scattering is no different from that for a point-like nucleus. As q increases the oscillatory nature of the exponential in equation (30) for an extended nucleus reduces from 1 and the scattering is reduced.

dVerZe

qF rqi)(1)( 2 (30)

An extended electric charge has greater difficulty in taking up the momentum transfer than does the point-like arrangement of the same total charge.

If the nuclear charge distribution is of spherical symmetry please show that

(a).

0

2

0

0

/2

02

sin)(

sin)()(

rrd

errdrdqF

rqi

0

2 )/sin()(4)( drqrrrZeq

qF (b).

(31)

(32)

If the nuclear charge distribution is like a step function shown

0, ,0

arar

please show thatρ0

a

3

2 cossin 3)(x

xxxqF

qax where

(33)

2. If a nucleus is seen as a point charge without any extensive distribution then we have the form factor simply as 1.

)()( 3 rZer The point charge is located at the origin.

1

)(

)(1

)(1)(

/3

/3

/2

dVer

dVerZeZe

dVerZe

qF

rqi

rqi

rqi

For the form factor value of other than 1 there is some sort of extended charge

distribution being detected.

3. High-q transfer measurement shows characteristics on the nuclear surface.

ppq ' )2/sin(2 pq

0

2 )/sin()(4)( drqrrrZeq

qF

(32)

1 )/( qr is a rapidly changing periodic function.)/sin( qr

The value of integration in the equation (32) would be very small if the charge distribution function ρ(r) is almost constant.

It is only when the distribution function ρ(r) itself changing rapidly can we expect a notable value from the integral.

The foregoing argument points out that data collected from high-q transfer measurement actually reveal

characteristics on the nuclear surface on which charge distribution changes rapidly with coordinate locations

22

2

0

32

611

61)(4)(

rq

drqrqrrrZeq

qF

3

61)/sin(

qrqrqr

0

222 )()(4 drrrrZe

r The mean square average of the nuclear charge extension

4. Low-q transfer measurement shows the mean square average of the nuclear charge extension.

In the mathematical description the form factor F(q2) is the Fourier transform of the nuclear charge distribution function.

In principle we should be able to know all the details of nuclear charge distribution if we can acquire all necessary information out of

the momentum transfer measurement.

In reality it is impossible to make a thorough measurement for our needful purpose. We are in a situation of fragmental information as far as the nuclear charge distribution is

concerned.

With this limitation in mind we can only fit our data taken from the electron scattering to the

Wood-Saxon function at this stage.

a

t

ρ

r

By varying two parameters, a and d, in the function until data are reasonably fit to the shape of the presumed distribution can we

more or less tell how charges are distributed in a nucleus.

dar

rexp1

)( 0

The measured differential scattering cross-section dσ/dΩ for the scattering of 450 MeV electrons by Ni-58. The positions of the diffraction minima should be matched against those of the figure below.

0, ,0

arar

D = 2a = 2 4.1 = 8.2 F

Te = 450 MeV → = 2.76 F

The first maximum occurs at

242.876.2

22.1sin 22.1sin 1-1

D

θ ≈ 24°

Here shows results from some nuclei with momentum transfer q = 800 MeV/c. It is apparent that our Woods-Saxo

n model is only an approximation.

If we neglect the finer details, our Woods-Saxon model gives a modestly reasonable description of nuclear size.

An analysis of the data made by Hofstadter and Collard (1967) gave for the half-point radius.

F 07.055.0 d

Where the ± indicates the range of values found in nuclei with A > 40. Below A = 40 there are marked

changes in thickness with A.

F 48.018.1 3/1 Aa

Problems (I)

1. Show that, for a spherically symmetric charge distribution,

2

2

02

2

6d)(d

2

r

qqF

q

where〈 r2〉 is the mean square of the electric charge distribution.

2. Show that the form factor for the charge distribution of model I is

3

2

)/()/cos()/()/sin(3)(

qaqaqaqaqF

3. Find the form factor for a charge distribution

rr

ar /0e)(

4. An electron of momentum 330 MeV/c is scattered at an angle of 10° by a calcium nucleus. Assuming no recoil, find the m

omentum transfer and its reduced de Broglie wavelength. Also calculate the Mott differential cross-section (point-like nucleus), and by what factor it is reduced if the calcium nucleus (A = 40) can be assumed to be represented by model I with a = 1.2A1/3 F.

Problems (II)

§ 3.5 Nuclear scattering and nuclear size

Electron scattering the distribution of electric charge in a nucleus

Neutron scattering the distribution of matter in a nucleus

Other particles which can be employed in measurements

protons

and even the long-lived elementary particles….

α-particles He42 deuterons H2

1

tritons H31

helions He32

These particles are charged so that scattering is caused by the combined effect

of both the nuclear and Coulomb forces. Considerable task is required to take away

effect from Coulomb force.

protons

and even the long-lived elementary particles….

α-particles He42 deuterons H2

1

tritons H31

helions He32

Elastic scattering of 14 MeV neutrons by nickel

The observed diffraction pattern with peaks and valleys is the characteristic of scattered waves from an absorbing object with

a moderately well-defined boundary.

The diffraction pattern can be interpreted by the optical model.

The Optical Model Potential

SLr

frcm

V

riWfr Vf

rVrV

LSLS

c

dd1

)( )(

)()(

2

2

1

The Coulomb potential for proton only

The nuclear potential well

The imaginary potential representing absorption of the incident nucleon

The spin-orbit interaction

1. V, W and VLS are expected to and do vary with energy.

2. The functions f1(r), f2(r), fLS(r) are usually taken to have the familiar Woods-Saxon form

dar

rfexp1

1)(

dar

rfexp1

1)(For f1(r), f2(r), and fLS(r)

Roughly We have the following results

a = 1.2A1/3 F

d = 0.75 FNuclear matter distribution

a = 1.18A1/3 ± 0.48 F

d = 0.55 ± 0.07 FNuclear charge distribution

Possible shapes of nuclei apart from spherical

1. If the charge distribution is not spherically symmetric the nucleus

can have electric moments other than monopole. This will manifest by its effect on the optical spectroscopy of the atom.

2. A non-spherical nucleus will have rotational states of motion and are identifiable in the spectrum of excited states.

~ The End ~