nuclear reactions - university of saskatchewan
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Nuclear Reactions
Nuclear and Particle reactions.
Can be written like a chemical reaction.
e.g. = pion Superscripts (if given) indicate charge.
e.g. there are three types of pions +, o, .
e.g.
This is an example of a decay. Follows the same rules of nuclear reactions but with only
one particle in the initial state.
In this example: = e = electron.
e = electron neutrino. A bar over the top of a particle explicitly indicates an anti-particle.
0 np
epn
Nuclear Reactions
Examples of Nuclear Reactions:
Nuclear reactions often written,
A(a,b)B
e.g.
np 5
12
76
12
6 NC HHeLi 3
1
4
2
6
3 n
Target Nucleus Undetected
reaction
products Projectile Detected
reaction
products
N),C( 1212 np N),(O 1516 p N),(O 1416 np
Nuclear Reactions
Some Definitions:
Scattering: The incident particle is also one of the reaction products.
Before: After:
Elastic Scattering: The incident particle and target are unchanged.
i.e. A(a,a)A
Inelastic Scattering: The target nucleus is changed or in a different energy state.
e.g. A(a,a)A* (the * indicates an excited state of A.)
a A ? a
Nuclear Reactions Conserved quantities in “low energy” nuclear reactions.
“low energy” means:
There is not enough energy to create new particles (via E = mc2).
1. Mass-Energy
2. Momentum
3. Angular Momentum
4. Parity (only if the Weak interaction is not involved – will discuss later)
5. # of nucleons
6. Charge [Actually, provided the weak interaction is not involved, 5 & 6 imply that
the # of protons and # of neutrons are both conserved.
The weak interaction can allow a proton to be converted to a neutron and vice-versa – more about this later.]
Nuclear Reactions
e.g.
Let and recall
Conservation of nucleons:
27 + 1 = 4 + A A = 24
Conservation of charge:
13 + 0 = 2 + Z Z = 11
Z = 11 is Sodium
Therefore
?),(Al27
13 n
XA
Z? He4
2
Na? 24
11
Cross Section
The cross section is related to the probability of a nuclear reaction.
e.g. the probability of the reaction
Experimental arrangement might be:
We will assume:
Thickness of target t is small, and, P, the probability of a reaction occurring in the thickness t is also small.
This assumption means that the reduction in the number of incident particles as they pass through t is negligible.
N),(C 1312 p
p
t
carbon target
Cross Section Let:
N0 = Number of incident protons.
NR = Number of reactions that occur.
n = number density of target nuclei.
Now: P = probability of a reaction.
Clearly we must have:
Where is a constant of proportionality.
So
is called the Cross Section.
It is proportional to the reaction probability with target specific parameters removed.
PNNR 0
t
nP
tnP
tnNNR 0
Cross Section
For a Thick Target. Thickness t.
x
N(x)
dx 0 t
dNR = number of reactions in the thickness dx
N(x) dNR
dxnxNdNR )(
N(x) = number surviving after passing
through distance x.
The change in N(x) through the distance dx is
dxnxNdNdN R )(
N0
N0 = number incident at x = 0
(We must assume that the only reason that a particle is removed
from the incident beam is if it undergoes the reaction in question.)
Cross Section
Satisfied by:
The number of reactions in a total thickness t is
dxnxNdNdN R )(
nxeNxN 0)(
)1(
)(
0
0
nt
R
eN
tNNN
P = probability of a reaction in thickness t.
We can write:
Where,
= mean free path
= Radiation Length.
(Mean distance of the incident particle before there is a
reaction.)
/0)( xeNxN
n
1
PN0
nxNdx
dN)(
Cross Section Note that:
= cross section for removing an incident projectile from the particle beam.
= Total Cross Section.
There may be many different possible reactions that could remove an incident particle from the beam.
e.g.
These are called Reaction Channels
Each reaction channel will have a different probability.
i = cross section for reaction channel i.
= Partial Cross Section.
Then
etc...
N
OO
15
1516
p
n
i
i
Cross Section If Ni = number of reactions for reaction channel i.
Then, Ni = N0Pi (Pi = probability for channel i)
Clearly,
So,
If we have a “thin target” , nt is small.
This is true if
Then since
)1( ntiii ePP
)1(0
ntii eNN
nt
1
ntNN ii 0
= mean free path
(Same as the thin target expression we had
before. i.e. we can ignore the attenuation of the
incident beam.)
xe x 1
Cross Section Sometimes…
instead of a beam passing
through a target,
we have a small target
immersed in a flux of incident
particles.
The incident beam is
described by the intensity or
Flux, 0, which has units of
particles per unit area per unit
time.
We wish to calculate the rate
of reactions in the target.
N0
t
0
Cross Section
We will assume that the thin target
approximation is valid.
So, in some time interval t, the
number of reactions in the target is
Here
The rate of reactions in the target is
The volume of the target is
So
0
ntNNR 0
tAN 00
t
A
ntAt
nttA
t
ntN
t
NR R
0
00
AtV
nVR 0
Cross Section
In this expression nV is just the total
number of target nuclei, N, in the
target sample.
So
The expression is correct even if the
target sample has an irregular shape.
nVR 0
0 V
(You can imagine splitting the volume up into an infinite
number of regular shapes, and adding them all.)
This formulation is sometimes more convenient.
NR 0
N
Cross Section
We can calculate n (the number of target nuclei per unit volume) from (for e.g.)
where = density of material (e.g. in g/cm3)
MA = Atomic or Molecular molar Weight
Note: mass in u = mass in grams that contains 1 mole
(1 mole # of 12C atoms in 12 g of 12C)
NA = Avogadro’s Number = 6.022 1023 per mole
It is often more convenient to characterize the thickness of a target by the product
t = mass per unit area seen by the incident particle beam. (Units of g/cm2)
A
A
NM
n
atom one of Mass
Density
tnNNR 0 nVR 0
Cross Section
Collider Experiments
Collisions between bunches of particles.
We are trying to measure the cross section for the reaction
X + Y Some specific final state
It is convenient in this case to express the rate of reactions, R,
in terms of a quantity called the Luminosity, L, defined by
R = L
L must have units (for example) of cm2.s1
L is related to the characteristics of the colliding beam
bunches.
particles X particles Y
Cross Section As a specific example: Let us suppose the colliding bunches
have the same area A and the particles are distributed
uniformly in each bunch.
The number of particles in each bunch are NX and NY.
If we think of bunch X as the beam hitting bunch Y,
the number of reactions when a pair of bunches collide is
Then so
particles X particles Y
A A
NX NY
YYXR tnNN
tY
Y
YY
At
Nn
A
NNN YX
R
Cross Section If the frequency of bunch collisions is f
then the rate of reactions is
So in this case the luminosity is
If we know the luminosity and we measure the rate of
reactions, then the cross section is simply
In practice we cannot make such simple assumptions about
the bunches. We would normally measure the rate for some
reaction for which we already know the cross section. That
would allow us to determine the luminosity. We can then use
that to determine an unknown cross section.
A
NfNfNR YX
R
A
NfNL YX
L
R
Cross Section
The luminosity determines the
rate of reactions we can
observe, so it determines how
small a cross section can be
measured.
A higher luminosity is better.
e.g. At the LHC the colliding
bunches are both protons.
The LHC is designed for
luminosities of about
1034 cm2s1.
Cross Section
If a cross section is expressed in a way that shows that it explicitly depends on some variable it is usually written as a Differential Cross Section.
e.g. The cross section may depend on the direction of the product particle. A(a,b)B
The probability that particle b will go in exactly the direction specified by and is infinitely small. So it makes no sense to talk about the cross section at a particular angle.
a
b
Cross Section
Total number of reactions is
The number of reaction products that hit a fictional detector of area dA at a distance r from the target is
d
d
dA
r ntNN 0
ntdNdN ),(0
Where ),( d = cross section for the reaction product
emerging in the solid angle subtended
by dA at the angles and .
Since ddrdrrddA sin)sin)(( 2
The solid angle subtended by dA is ddr
dAd sin
2
Cross Section We usually write the number of reactions per unit solid angle:
Where is the Differential Cross Section.
Units: e.g. b/sr, mb/sr
If an actual detector subtends a solid angle and d/d does not vary very much over the angles subtended by the detector, then we can make the approximation, that
the number hitting the detector is
ntd
dN
d
dN
),(0
d
d ),(
d
dNN
ntd
dNN
),(0
Cross Section
Note that:
or
Note that is not a “true” differential.
It is just describes how the cross section per unit solid
angle varies as a function of and .
4
),(d
d
d
ddd
dsin
),(
0
2
0
d
d ),(
Rutherford Scattering
Mildly generalize:
Projectile with charge +ze
Nucleus with charge +Ze
b
+ze
+Ze
We can derive the relation between the impact parameter b
and the scattering angle . The derivation uses conservation of angular momentum and the
impulse-momentum theorem.
Result is: b
p
2)2/tan(
where p = distance of closest approach of
projectile to nucleus when b = 0.
projectile
nucleus
As an example of a cross section calculation.
Rutherford Scattering
Using conservation of mechanical energy
p = distance of closest approach of
projectile to nucleus when b = 0.
T +ze +Ze p
T = kinetic energy of projectile
p
zZeT
2
04
1
T
zZep
2
04
1
So
T=0
Rutherford Scattering
b
m v
b
p
2)2/tan( Derivation of
D
Let point D be the position of closest approach to O.
Point X is an arbitrary point on the projectile path, at
a distance r from O and an angle from OD.
Using conservation of angular momentum about O:
O
X
r
dt
dmrmvb
2(initial) (at X)
F
Rutherford Scattering
Using the impulse-momentum theorem:
dtif Fpp
We consider just the component in the direction OD.
Then:
dtFmvmv cos))2/sin(()2/sin(
dtr
zZemv
cos
4
1)2/sin(2
2
2
0
From the angular momentum equation we can get the
relation between dt and d to change variables in the
integral.
i.e. dvb
rdt
2
When t = + , = ()/2
When t = , = ()/2
Rutherford Scattering
2/)(
2/)(
2
2
0
2
cos1
4)2/sin(2
dvb
r
r
zZemvSo:
Fortunately the r2 cancel since r also depends on .
2/)(
2/)(0
2
][sin)2(4
)2/sin(2
bT
zZe Using 2
21 mvT
]2/)(sin[]2/)sin[(2
)2/sin(2 b
p
Using T
zZep
2
04
1
))2/cos(()2/cos(2
b
p)2/cos(
b
p
b
p
2)2/tan( QED
Rutherford Scattering
When b < b0 then > 0
So, whenever the projectile passes through the area
the projectile is scattered by an angle greater than 0.
Therefore
0 b0
+ze
+Ze
projectile
nucleus
0
02
)2/tan(b
p
2
00)( b
)2/(tan
1
4)(
0
2
2
0
p
2
00)( b
Rutherford Scattering The area
is the same as the cross section for scattering by
angles greater than 0.
2
00)( b
Consider a projectile fired
randomly within an area A of
a target with thickness t.
t
A 2
0b
The probability that a projectile will be scattered by an angle
greater than 0 is
AP
nuclei) target (#
We assume a thin target so that target nuclei
are not shielded by an area in front of it.
A
nAt
ntP So Same as the definition for we had before
for a thin target.
Rutherford Scattering
So now we ask the question:
What is the area, d, the projectile must hit if
it is to be scattered between the angles 0
and 0 + d?
i.e into the solid angle
We want
0 b0
+ze
+Ze
projectile
nucleus
d
dd 0sin2
d
d
d
d
d
d
0
0 )(
d
d
dd
0
0 )(
d
00
0
sin2
1)(
d
d
Rutherford Scattering
)2/(tan
1
4)(
0
2
2
0
p
00
0
sin2
1)(
d
d
d
d
Doing the derivative, we get,
Using
21
0
20
32
0
0
)2/(cos
1)2/(tan)2(
4
)(
p
d
d
)2/(sin
)2/cos(
4
)(
0
3
0
2
0
0
p
d
d
Then: 00
3
0
2
sin2
1
)2/(sin
)2/cos(
4
p
d
d
Rutherford Scattering
T
zZep
2
04
1
)2/(sin
1
16
1
0
4
2
p
d
d
)2/(sin
1
16 4
2
0
2
T
zZe
d
d
Then with and dropping the subscript we get
This derivation shows us, for a specific case, the relationship between cross section and an area.
)2/cos()2/sin(2
1
)2/(sin
)2/cos(
8 000
3
0
2
p
d
d
So
)2/(sin
1
16
14
2
p
d
d
Cross Section
Total number of reactions is
The number of reaction products that hit a fictional detector of area dA at a distance r from the target is
d
d
dA
r ntNN 0
ntdNdN ),(0
Where ),( d = cross section for the reaction product
emerging in the solid angle subtended
by dA at the angles and .
Since ddrdrrddA sin)sin)(( 2
The solid angle subtended by dA is ddr
dAd sin
2