Like the alchemist’s dream
of chemical transformation
Nuclear physics
has found the means
of transmutation
Nuclear Reactions
Besides his famous scattering of  particles off gold and lead foil,
Rutherford observed the transmutation:
14
7
N  He H  O
4
2
1
1
17
8
or, if you prefer
14
7
N    p 178O
Though a more compact form is often used:
14
7
N ( , p ) O
17
8
Target (Projectile, Detected Particle) Residual Nucleus
Whenever energetic particles
(from a nuclear reactor or an accelerator)
irradiate matter there is the possibility of a nuclear reaction
the projectile and emitted particle are
enclosed in brackets between the target and daughter nuclei
14
7
N  He H  O
4
2
1
1
17
8
or, if you prefer
14
7
N    p 178O
Though a more compact form is often used:
14
7
N ( , p ) O
17
8
Target (Projectile, Detected Particle) Residual Nucleus
Convenient as the bracketed part can be used by itself to
refer to a particular class of reactions like ( , p) or (n, g).
Classification of Nuclear Reactions
divided roughly into the two groups
•compound nucleus reactions and
•direct reactions.
Classification of Nuclear Reactions
direct reactions are of the types
•scattering reaction
projectile and scattered (detected) particle are the same
•elastic scattering
residual nucleus left in ground state
•inelastic scattering
residual nucleus left in excited state
Even inelastic proton scattering is strongly peaked
in the forward direction with the cross-section
only gradually changing with higher energies.
Ef , pf
Ei , pi

EN , pN
Ei  E f  EN ,recoil
 

pi  p f  pN ,recoil
The simple 2-body kinematics of scattering
fixes the energy of particles scattered through .
Number of particles with Ef
Measuring Ef at a fixed angle 
The predicted elastic scattering calculated
from the kinematics of a 2-body collision.
Only Ef
should be
observed
at .
8.0
8.5
9.0
9.5
10.0
Energy Ef of scattered particle (MeV)
Classification of Nuclear Reactions
Addition details on direct reactions
•inelastic scattering
individual collisions between the incoming projectile and a single
target nucleon; the incident particle emerges with reduced energy
• pickup reactions
incident projectile collects additional nucleons from the target
16
O + d  158O + 31H
(d, 3H)
8
Ca + 32He  40
Ca +  (3He,)
20
41
20
•stripping reactions
incident projectile leaves one or
more nucleons behind in the target
90
91
(d,p)
40 Zr + d  40Zr + p
3He,d)
23
3
24
(
11 Na + 2He  12Mg + d
•charge exchange reactions
A proton (neutron) enters the nucleus,
but emerges as a neutron (proton)
“exchanging charge” with one of the nucleons
Direct reactions are often described as “surface” reactions
• involving interactions with individual nucleons within the target
• occurring very rapidly (on the order of 10-22 seconds)
Restricted to highest energy projectiles (>20-MeV)
(from accelerators)
There is evidence for an altogether different mechanism
taking place at more moderate energies
• slower process – taking 10-16 to 10-18 seconds
involving an intermediate, unstable, short-lived
state outliving direct contact of projectile & target
• final products ejected isotropically
compared to the “forward-peaked” cross-section
we described for the direct processes
• for incident energies <20-MeV
governs all “natural” processes where the
projectiles are natural decay products
The most obvious
evidence for
long lived
intermediate states
in nuclear reactions
is the strongly
resonant nature
of nuclear
cross-sections.
This is illustrated
in the figure
at right which shows
the indium
total cross section
for neutrons.
The energy of these long lived states is defined to a few eV
and if we apply the uncertainty relation dEdt ~ h we see
this implies a lifetime of ~ 10-16 s which is very long
compared to the time it takes a nucleon to traverse a nucleus
~ 10-22 s.
Consider the time for a relativistic  to cross/pass through a
medium mass (A=125u) nucleus
2(41/ 3 r0  A1/ 3r0 ) / c  (2  1014 meters) /(3  108 m / sec)  1022 sec
Recall:
Direct reactions do not involve the formation of some
intermediate state; their characteristic time of interaction
is more like 10-22 s. Variations in their cross-sections,
as a function of energy, are spread over a few MeV.
What could possibly distinguish low from high energy collisions?
Let’s compare de Broglie wavelength for
1 MeV (“low” energy) proton
  h / p  4.13570 1015 eV  s / 2mE
  4.13570 1015 eV  s / 2(938MeV / c 2 )(1MeV )
23
8
14
  (9.54844 10 s)(3 10 m / s)  2.86 10 m  30 fm
5-10 MeV ’s (typical for nuclear reaction studies)
  4.13570 1015 eV  s / 2(3727MeV / c 2 )(10MeV )
 4.54 10 m  4 fm
15
100-MeV 
  A1/ 3r0
  4.13570 1015 eV  s / 2(3727MeV / c 2 )(100MeV )
  1.4 fm
  r0
High energy projectiles are better focused on
individual target nucleons
Their collisions are billiard ball-like
and result in
Direct Reactions.
Low energy projectiles cannot see anything finer
than the whole target nucleus
The resultant collisions are with the entire nucleus,
and the energy is shared across it.
The intermediate state (itself never directly
observed) is referred to as the
Compound Nucleus.
Cross Section
Total
,3n
,4n
,2n
,n
Alpha particle energy
The above figure shows the contributing cross-sections for the reactions
A
Z
A
Z
A
Z
X N  Z 2 X N 1  n
A 3
X N  Z 2 X N 1  2n
A 2
X N  Z 2 X N 1  3n
A1
A
X



X N 1  4n
Z
N
Z 2
A
and so on…
Cross Section
Total
,3n
,4n
,2n
,n
Alpha particle energy
With increasing energy
• more and more neutrons are likely to be knocked free
• it becomes less and less likely for an individual neutron to be freed
This is NOT explicable as an -nucleon collision, the  shares its
energy across the nucleus, elevating it to some excited state.
E
V=0
R
Consider a neutron (or -particle)
with energy E
within a square well potential.
V = V0
At r = R there is a large confining force
which as we saw in Problem Set #1 can produce reflections.
u
Let’s consider the simple ℓ=0 case, where we can substitute:  
2
r
d uin
2
m
2
with u satisfying:  
 Kuin  0 for K  2 ( E  V0 )
2

dr
inside the well.
u

r
2
inside the well:

d uin
dr
2
 Kuin  0
Since uinside must vanish at the origin:
uin  Cin sin Kr
2
outside the well:

d uout
dr
2
2m
K  2 ( E  V0 )

2
 k uout  0
2mE
k  2

2
For scattering off this potential, we want to consider
final states that are traveling outwards outside the well:
uout  Couteikr
for r > R
1 duin
1 duout

uin dr uout dr
Continuity requires
or
at r = R
K cot KR  ik
The energy of the inside states must be expressed as a complex number!
In scattering experiments we are not dealing with simple
“stationary states” with real-valued energy eigenvalues.
but with E = E0  i/2
( r, , , t )   ( r, , )e
iEt / 
  ( r , ,  ) e
  (r, , )e
 iE0t /  t / 2 
e
i ( E0 i / 2 ) t / 
( r, , , t )   ( r, , )e
iE0t /  t / 2 
e
A stationary state of energy E0
that decays!
because the probability of finding the particle in that state:
 e
2
t
These are “virtually bound” states or resonances.
For a short range or abrupt-sided potential there exist
quasi-bound or virtual single particle states
of positive energy.
Long range potentials (like the coulomb potential)
have no such states.
The projectile can be momentarily trapped in one of these excited
states, sharing its energy through interactions with the nucleons
inside the nucleus: raising some of them into excited states, itself
dropping into lower states.
This is the formation of the many
particle excited state
which is the compound nucleus.
At this stage all memory of the original mode of excitation is lost.
At a later time when a decay
occurs, the energy is once more
concentrated in a single particle.
The Optical Model
To quantum mechanically describe a particle being absorbed,
we resort to the use of a complex potential
in what is called the optical model.
Consider a traveling wave moving in a potential V then
this plane wavefunction is written
ψ e
ikx
where
k  8 m( E  V ) / 
2
2
If the potential V is replaced by V + iW then k also becomes complex
and the wavefunction can be written
ik1x  k2 x
and now here
ψe
e
k  k1  ik 2
ik1x  k2 x
ψe
e
now describes a traveling wave of decreasing amplitude:
 *  e
 i ( k1 ik2 ) x
e
i ( k1 ik2 ) x
e
A decreasing amplitude
means that the transmitted particle
is being absorbed.
2 k2 x
In most cases V >> W.
To make an estimate of the mean free path
(the “attenuation length”) we will assume that condition:
k  8 m( E  V  iW ) / 
2
2
 8 m( E  V ) /   1  iW /( E  V )
2
2
 8 m( E  V ) /  1  iW /[ 2( E  V )]
2
2
replacing the kinetic energy term (E - V) by the expression mv2/2 gives
k  4 m v / h 1  iW /( mv )
2
2
2
2
2
 (2mv / h){1  iW /( mv )}
2
k  2mv / h  i 2W / hv
 k1
 ik 2
2p/h = 2/
where  is the de Broglie wavelength of incident projectile
 *  e
 i ( k1 ik2 ) x
e
i ( k1 ik2 ) x
e
2 k2 x
the mean free path - the distance over which the intensity ( * )
is attenuated to 1/eth its initial value – is given by
1
h

2k2 4W
since this gives a (mean) distance where
h
 v t
4W
the mean free time - the time by which the intensity ( * )
is attenuated to 1/eth its initial value
h
t 
4W
ψ e e
ik1x
h
t 
4W
 ik2 x
The SHELL MODEL we relied on earlier has a potential depth ~40 MeV.
If the attenuation distance is of the same order as the nuclear radius
(3-6 fm) then W  6-8 MeV.
To describe both diffraction and scattering phenomena
with the optical model
requires an imaginary potential of a few MeV.
NOTE: this is entirely consistent with the lifetime of the virtual
single particle state before absorption of about 1022 sec.
t  h /(4W )  4.13570  10
t  5.5  10
23
15

eV  sec/(4  6MeV ) 
~ 10-22
Writing the compound nucleus reaction as
a  X  C*  Y  b
the cross-section (which remember expresses a probability) can be expressed as
( probability of forming C*)  ( probability of decaying to Y  b)
 a b
 b 
 a  

the width
for decay to b
the total
decay width
the cross-section
for absorbing a
and forming
the compound nucleus
b is the “partial width”
 is the total width (ħ times the total decay probability).
The behavior of the cross-section with energy depends on
the relative sizes of  & the spacing between energy levels.
For low excitation of a nucleus the energy levels are
relatively well spaced and the cross-section exhibits
resonance while at higher energies of excitation
the width  will overlap several energy levels and
the cross-section varies much more slowly with energy.
This is the continuum.
The energy at which the resonance  continuum transition
occurs depends upon A.
For A ~ 20
~10 MeV
while for A ~ 200
~100 keV.
For well separated levels, an individual state will decay as exp(- t/ ħ)
RECALL: we write such a wave function as
  (r )e
 iE0t /  t /(2  )
e
the 1st exponential gives the normal oscillatory time dependence
of a wavefunction with E0 (here the energy above ground state).
The second exponential gives the decay of the state.
Since it is decaying, such a state is not a solution of Schrödinger’s
equation with a static potential (not a stationary state).
It can,however,be considered a superposition of such states
   A ( E )e
 iEt / 
dE
This convolution you’ll recognize as the Fourier transform!
The function A(E) can be obtained by Fourier transform which yields
( r )
A(E)  2
2
2
4 [( E  E0 )   / 4]
2
2
Giving the cross-section the form
C
a 
2
2
( E  E0 )  ( / 2)
The constant C depends upon the phase space available
to the incident particle
2
h a ( 2  1)
C
2
4p
where p is the momentum of the incident particle,
ℓ its orbital angular momentum, and
a the partial width for the decay back to the initial state.
All together
 ab
 (2  1)a b
1

2
2
4 ( E  E )  (  / 2)
0
2
where
  h/ p
This is the Breit-Wigner formula for a single level reaction cross-section.
For example - in the case of elastic scattering (a = b)
at the maximum of the resonance (E = E0) the cross-section is
 a (2  1)
elastic 
2

2
2
And if no other processes to compete with (a=)
elastic   (2  1) / 
2
the maximum possible elastic cross-section.
For the inelastic cross-section (b = a) at the maximum of its
resonance (E = E0)
2
 a (  a )(2  1)
inelastic 
2

 (2  1)
inelastic 
4
with a maximum value of
2
when a
= /2.
When the separation of energy levels is much smaller than the total width
there are many levels contributing to a given process.
This mixture of levels
described as a compound
nucleus.
The projectile is captured by the target forming an intermediate state
the compound nucleus
In a compound nucleus all sense of direction of the incident particle is lost
the produced particles are ejected in an essentially isotropic distribution
in the center of mass frame.
Furthermore
The subsequent decay of this intermediate state is largely independent
of its mode of formation. A given compound nucleus may be formed
by any of several reactions but the probability of a certain type of
final state is only dependent upon the amount of excitation energy.
19
9
Fp
19
10
20
10
19
9
18
Fp
9
O
14
7
N  3 Li
6
8
C

Be
6
4
12
10
B

B
5
5
10
F d
3
F

H
9
1
3
O

He
8
2
8
Ne  g
17
17
16
Ne  n
3
O

He
8
2
17
[
20
10
Ne]*
O
16
8
6
N

Li
7
3
14
7
N

Li
7
3
13
8
C

Be
6
4
12
9
C

Be
6
4
11
10
B

B
5
5
10
11
B

B
5
5
9
Here are plotted
the yields of
decay products from
the compound nucleus
64Zn
formed by
2 different routes
Note:
the relative cross-sections
for the different processes
63Cu(p,n)63Zn
and
60Ni(,n)63Zn
remain ~constant
across the
plotted  energies.