Download chapter3_finalv

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 nuclei
What 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 ?
Rough visualization of an
electron "wave function". (Art by
Blake Stacey.)
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.
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!
These deviations occur at
short distances of approach.
1 the finite charge density
distributions overlap.
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.
2 strong nuclear forces between
the α particle and the target
nucleus comes into play.
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
(1)
To reveal more detail needs smaller de Broglie wavelength for the probing particle
To reveal more detail needs smaller de Broglie wavelength for the probing particle
h

p
(2)
We need to use proper incident particles with proper
kinetic energies for probing nuclear structures.
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.
mass of the neutron
M 3 Amn 3mn



3
V 4r0 A 4r03
neutrons and
protons have the same
mass to 0.14%
(3)
= about 1.82 x 1017 kg/m3
so nuclear density does not depend on the atomic mass number
The density distributions for some nuclei
2 Nuclear charge distribution ρc = total charge divided by its volume.
Q
3Ze
c  
V 4r03 A
(4)
Since N ~ Z ~ A/2
Q
3Ze
3e
3
c  

C/m
V 4r03 A 8r03
(5)
It is roughly a constant.
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.
R = r0A1/ 3
r0 = 1.4 F
for nuclear
matter distribution
= 1.2 F for charge
distribution
Nuclear matter distribution
Rough description Nuclear charge distribution
§ 3.2 The scattering of electrons by nuclei
When a gold ( 197
) foil is bombarded by energetic α particles:
79 Au
(1) R >
10-14
m
(2) R < 10-14 m
1
r
Coulomb scattering
Vcoul ~
Nuclear potential
e  ar
VN ~
r
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 scattering
Early 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
A typical electron
3
Small incident kinetic energy
Te ~ 1 MeV
de Broglie’s wave length
energy from beta
decay.
pe ~ 1.42 MeV/c
h 197.3  2


 873 F
pe
1.42
wave length
too long !!
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
 MeV 
197.3  2 
 F
 c
  123.96 MeV
pe 
10 F
c
And the kinetic energy of
an electron must be
2
2
T

m

p
e
e
e  me  124 MeV
around 124 MeV.
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.
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
dσ Z 2 2 2c 2

cosec 4 ( / 2)
2 2
d
4P v
(6)
If the incident particle is an electron which is unpolarized and can be relativistic (v→c),
the formula becomes
dσ Z 2 2 2c 2
v2
4
2

cosec
(

/
2
)[
1

sin
( / 2)]
d
4 P 2v 2
c2
This is the Mott formula.
(7)
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.
2
dσ Z 2 2 2c 2
v
4
2

cosec
(

/
2
)[
1

sin
( / 2)]
2 2
2
d
4P v
c
(7)
§ 3.3 The electric charge distribution
Nuclear 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.
Model I
Model II
t
 (r ) 
ρ
a
r
0
 r  a  (8)
1  exp 

d


Saxon-Woods form
ρ(r) = ρ0, r < a
Model I
ρ(r) = 0, r > a
Since the total charge in the nucleus is Ze.
Therefore the following equation must be
satisfied.
3 3
 a   0  Ze
4

(9)
Model II
t
 (r ) 
ρ
a
r
Ze  
Total Charge
In a nucleus
0dr 3
1  exp[( r  a) / d ]
 40 

0
0
 r  a  (8)
1  exp 

 d 
a  A1 / 3
d  0.52 F
t  4.39 d
t  2.28 F
dr
1  exp[( r  a) / d ]
For a spherical nucleus
(10)
“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 experiments
incident particle is a plane wave
Scattered wave fronts are spherical
intensity variations on wave-front
due to diffraction
diffraction pattern observed from
plane wave striking a target
analogy with optics: like optical
diffraction
but nucleus has blurred edges
The optical description of the Fraunhofer diffraction pattern
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.
  sin 1 (1.22 / D)
(11)
First minimum of the diffraction pattern
The Fraunhofer diffraction pattern
  sin 1 (1.22 / 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
h
momentum.

pe
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.
  0 , r  a
0,
ra
D = 2a = 2  4.1 = 8.2 F
θ ≈ 24°
Te = 450 MeV →  = 2.76 F
The first maximum occurs at


  sin 1 1.22

2.76 
-1 


sin
1
.
22


  24
D
8.2 

Electron
scattering
Te = 502 MeV
λ=2.47 F
Proton
scattering
Tp = 1000 MeV
λ= 0.73 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
For a point-like nucleus
d
2
2 2
 Z e f ( )
d Mott
(12)
d
2
2
2 2
 Z e f ( )  A( )
d Mott
where
A( )  Zef ( )
(12)
(13)
A( ) is dependent on various scattering details such as the
2
wavelength of the incident wave, energy, or momentum etc.
In general A(θ) is a complex function which is called the
scattering amplitude.
In the case of point Coulomb scattering
d
( Zze 2 )2
1

d Coulomb
16 E 2 sin 4 ( / 2)
f ( ) 
1
4 E sin 2 ( / 2)
and the scattering amplitude for point Coulomb scattering is
Zze 2
A( ) 
4 E sin 2 ( / 2)
(14)
In the case of point Mott scattering
d
Z 2 2 2c 2
v2
4
2

cosec
(

/
2
)[
1

sin
( / 2)]
2 2
2
d Mott
4P v
c
and the scattering amplitude for the point Mott scattering is
1/ 2
 d

A( )  

 d Mott 
(15)
Form factor F(θ)
While the nuclear charge is no longer a point we need a form factor F(θ)
to modify our scattering formula. Namely,
d
2  d

( )  F( ) 
( )
d
 d  Mott
 A( )
(16)
2
d
 d

F( ) 
( ) / 
( )
d
 d  Mott
2
(17)
d
 d

F( ) 
( ) / 
( )
d
 d  Mott
2
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)]
(17)
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.
p’
p
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.
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.
  2
(path a)
la
lb
(path b)
lb  la

(18)
r
The contribution to the scattering amplitude for a ‘point charge’
with volume element dV and located at the position r is
(dQ) f ( )e
 i
  (r )dVf ( )e
 i
(19)
r
And the scattering amplitude is the volume integral of the eq. (19).
A( )    (r ) f ( )e dV
 i
(20)
P0’
P0
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’ .
The optical length difference between P2SP2’ and P1OP1’
= The optical length difference between P0XP0’ and P1OP1’
d
P1’
1
(   )
2
P0’
θ
P1
O
θ
X
P0
d/2
d  2(OX ) sin(  / 2)
(21)
la
lb
It is evident from the figure that
(OX) = (OS) cos α = r cos α
lb  la  d  2(OX ) sin(  / 2)  2(r cos  ) sin(  / 2)
(22)
lb  la  d  2(OX ) sin(  / 2)  2(r cos  ) sin(  / 2)
The phase difference δcan then be expressed as
  2
lb  la


2

d
2

2(r cos  ) sin(  / 2)
Finally

4

(r cos  ) sin(  / 2)
(23)
Here λ = p/h, and p is the linear momentum of the incident electron.

4

(r cos  ) sin(  / 2)
(23)
With λ = p/h the equation (23) can also be written as
1
  2 p sin(  / 2)  r cos  

(24)
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
 
qr

(25)
A( )    (r ) f ( )e  i dV
(20)
Substitute the expression in the equation
(25) into the equation (20) we get
A( )    (r ) f ( )eiqr /  dV
(26)
More specifically
A( )  f ( )
2
0


0
0

 (r )r 2 sin α dr dα dβ ei qr
(27)
Now we have for the total charge
Ze  

0


0
2(r )r 2 sin α dr dα
So we see that we can write
A( )  Zef ( )
2
i qr

(
r
)
r
sin
αdrdα
e

  (r )r
2
sin αdrdα
 Zef ( ) F ( )
(28)
Thus the Mott (or Rutherford) scattering amplitude
Zef(θ) is changed by a factor F(θ) and the scattering
formula becomes
dσ
2 d
 F(θ )
dΩ
d Mott
(29)
F(θ) is called a form factor.
Somewhat formally it could be written
F ( ) 
i q r

(
r
)
e
dV

  (r )dV
1
i q r


(
r
)
e
dV

Ze
(30)
F ( ) 
i q r

(
r
)
e
dV

  (r )dV
1
i q r


(
r
)
e
dV

Ze
(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.
1
i qr
F (q ) 

(
r
)
e
dV

Ze
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

F (q ) 
2
(a).

0

d  dr (r )r 2 sin  ei qr / 


0
(b).
0

d   (r )r 2 sin 
0
4 
F (q ) 
 (r )r sin( qr / )dr

0
Zeq
2
(31)
(32)
If the nuclear charge distribution is like a step function
shown
  0 , r  a
0,
ρ0
ra
please show that
3 sin x  x cos x
F (q ) 
x3
2
a
where
qa
x

(33)
2. If a nucleus is seen as a point charge without any extensive
distribution then we have the form factor simply as 1.
 (r )  Ze 3 (r )
The point charge is located at the origin.
1
i q r / 
F (q ) 

(
r
)
e
dV

Ze
1
3
i qr / 

Ze

(
r
)
e
dV

Ze
2
   3 (r )ei qr /  dV
1
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.
q  p'  p
(qr / )  1
q  2 p sin(  / 2)
4 
F (q ) 
 (r )r sin( qr / )dr

0
Zeq
2
(32)
sin( qr /  ) is a rapidly changing periodic function.
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
4. Low-q transfer measurement shows the mean square average
of the nuclear charge extension.
qr 1  qr 
sin( qr / ) 
  
 6  
3
4
F (q ) 
Zeq 0

2
 qr  1  qr 3 
 (r )r      dr
   6    
1 q2
2
1

r

2
6
4  2
2
r 
(
r
)

(
r
)
r
dr

0
Ze
2
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.
t
ρ
 (r ) 
a
0
ra
1  exp 

 d 
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.
The measured differential scattering crosssection 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 , r  a
0,
ra
D = 2a = 2  4.1 = 8.2 F
θ ≈ 24°
Te = 450 MeV →  = 2.76 F
The first maximum occurs at


  sin 1 1.22

2.76 
-1 


sin
1
.
22


  24
D
8.2 

Here shows results from some nuclei with momentum
transfer q = 800 MeV/c. It is apparent that our WoodsSaxon 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.
a  1.18 A1/ 3  0.48 F
d  0.55  0.07 F
Where the ± indicates the range of values found in
nuclei with A > 40. Below A = 40 there are marked
changes in thickness with A.
Problems (I)
1. Show that, for a spherically symmetric charge distribution,
r2
dF (q 2 )
 2
2
dq q 2  0
6
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
F (q 2 ) 
3sin( qa /  )  (qa /  ) cos( qa /  )
(qa /  )3
Problems (II)
3. Find the form factor for a charge distribution
 (r ) 
0e  r / a
r
4. An electron of momentum 330 MeV/c is scattered at an angle
of 10° by a calcium nucleus. Assuming no recoil, find the
momentum 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.
§ 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
α-particles
3
tritons 1
H
4
2
He
deuterons
2
1
H
helions 23 He
and even the long-lived elementary particles….
protons
α-particles
3
tritons 1
H
4
2
He
deuterons
2
1
H
helions 23 He
and even the long-lived elementary particles….
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.
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
V (r )  Vc (r )
The Coulomb potential for proton only
 Vf1 (r )
The nuclear potential well
 iWf 2 (r )
The imaginary potential representing
absorption of the incident nucleon
2
   1 df LS

 VLS 
LS
 m c  r dr
The spin-orbit interaction
1. V, W and VLS are expected to and do vary with energy.
1
f (r ) 
2. The functions f1(r), f2(r), fLS(r) are usually

r  a
1

exp

taken to have the familiar Woods-Saxon form


 

 d 
f (r ) 
1

r  a
1  exp 


 d 

For f1(r), f2(r), and fLS(r)
Roughly We have the following results
a = 1.2A1/3 F
d = 0.75 F
Nuclear matter distribution
a = 1.18A1/3 ± 0.48 F
Nuclear charge distribution
d = 0.55 ± 0.07 F
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 ~