SECTION C: NUCLEAR RADIATION AND NUCLEAR ENERGY LOSS PROCESSES
In this section we discuss decay and transmutation processes in nuclei (including α, β, and γ decay, as well as
fission and fusion processes), using the previous models where appropriate. Also we discuss the interaction of
beams of nuclei with matter, focusing on the processes that govern energy loss and penetration distances.
C.1 Radioactive decay equations
Before considering specific mechanisms of decay, it is useful to consider some general properties. We start with
a simple one-stage decay process, where the product of decay of the radioactive nucleus is stable, e.g., a 14C6
nucleus can decay into 14N7 (through beta decay) over several thousands of years.
If N denotes the number of radioactive nuclei at time t, and λ is the probability of decay per unit time, then the
rate of decrease (known as the activity) is
dN
!
= ! N , where λ is the probability of decay constant
dt
The unit of activity is the Bequerel (1 Bq = 1 disintegration per second). Rearranging and integrating the above
equation gives
!N$
N dN
t
implying
and so
N = N 0 e! !t .
=
"
!
dt
ln
# & = ' !t
! N0 N
!0
N
" 0%
The decay constant can also be expressed in terms of the half-life (denoted by t1/2), which is the time for half of
the nuclei to decay. Therefore exp(−λt1/2) = 0.5 , which gives
ln 2
0.693
t1/2 =
=
!
!
Also the mean life (denoted by τ) can be found as
"
"
" !2
1
! "t
! = # 0 t N 0 e! "t dt
N
e
dt
=
=
#0 0
!1
"
"
More generally, we might have a chain of decays if the initial product of a decay is itself radioactive and can
undergo a further decay (usually with a different decay constants). An example is
79 38
Sr (t1/2 ~ 2 min) → 79Rb37 (t1/2 ~ 23 min) → 79Kr36 (t1/2 ~ 35 hr) → 79Br35
For a 2-chain decay of the form A → B → C with decay constants λA and λB and C being stable, the rate of
decay equations are
dN
dN
! A = !A N A ,
! B = !B N B ! !A N A
dt
dt
The solutions when λA ≠ λB are then easily found to be (e.g., can be verified by substitution):
!A
N A (t) = N 0 exp(!!A t) ,
N B (t) = N 0
[exp(!!At) ! exp(!!Bt)]
!B ! !A
The results for NA, NB, and NC (= N0−NA−NB) will be discussed further in class.
(
)(
)
C.2 Alpha decay
Energy release in α decay
The α decay process is
( A, Z )
!
( A " 4, Z " 2)
+
4
He 2
(4He2 = α particle)
In terms of the energy balance, where Q is the energy release and B is related to the binding energy, we have
Zmp c 2 + (A ! Z )mn c 2 ! B(A, Z ) = Zmp c 2 + (A ! Z )mn c 2 ! B(A ! 4, Z ! 2) ! B(4, 2) + Q
After cancelling terms and solving for Q we get
Q = B(A ! 4, Z ! 2) ! B(A, Z) + 28.3 MeV ,
where the 28.3 MeV represents the experimental value of B for the α particle.
19 The decay process is favoured if Q > 0. We can substitute for the other binding energies using the liquid drop
model (semi-empirical mass formula) to calculate Q. This leads to the general conclusion that α decay will be
favoured for large nuclei (i.e., large A and Z). Specifically, it is predicted to occur if Z is greater than about 66
(assuming nuclei that are near the bottom of the mass parabola).
The experimental data broadly confirm this prediction, but there are several exceptions. Also there are some
anomalies with regard to the rate of decay (sometimes it is several orders of magnitude small than expected by
the above the above simple theory). To understand this, we need a QM theory.
Quantum theory of α decay
This would involve solving Schrödinger’s equation using spherical polar coordinates and assuming a
spherically symmetric potential V(r), like in the case of the Shell model.
However, the potential will be different in the present case. It will be made up of:short-range nuclear interactions (potential well) + Coulomb interaction term VC(r).
When the α particle has formed and moves apart from the parent nucleus, we have 2 nuclei with charges (Z–2)e
and 2e. The Coulomb repulsion will give
2(Z ! 2)e 2
where r = separation
VC (r) =
4!" 0 r
The minimum value of r (denoted by rs) will be when the two spheres are just touching:
with r0 = 1.1 fm
rs = r0 (A ! 4)1/3 + r0 41/3
The maximum value of VC(r) will occur at this separation.
For example, taking the case of α decay in 209Bi83, we get
rs = 8 fm
and
VC(rs) = 28 MeV
For comparison, the experimental value for the energy release in this case is Q = 3 MeV.
The overall form of the total potential is something like:
Clearly, if Q < VC(rs)
(as in the quoted example of 209Bi83),
then classically the α particle which is
formed inside the well cannot escape
⇒ no α decay !
V(r)
VC(rs)
Q
r 0 rs However, according to QM the α
particle has a non-zero probability of
tunneling through the barrier
⇒ α decay can occur (if Q > 0),
but with very low probability in some
cases, depending on the height and
width of the barrier.
rc Width of tunneling barrier
~ rc − rs
≈ 75 − 8 fm ≈ 67 fm
for decay of 209Bi83 to occur.
This value is many times the nuclear radius and it makes the tunneling probability (and hence α decay) almost
negligibly small in this case.
A proper calculation would involve solving Schrödinger’s equation ⎯ this is complicated unless V(r) is
further simplified (to be discussed in class).
20 C.3 Beta and gamma decay
Beta decay
We have already discussed the different forms of beta decay (β - decay, β + decay, and electron capture) in
section B, where we mainly used the Liquid Drop model. This was done in terms of the mass parabola and
conditions for the different types of decay to be energetically possible were deduced involving the nuclear
masses. For example, we had the diagrams on pages 9 and 10 for A = 135 and 140 respectively.
Going further, there are some estimates to calculate the transition rates in beta decay (using an analog of the
Fermi golden rule in QM, but doing a relativistic version of the theory). Also the role of the neutrino and
antineutrino in the decay will be reconsidered later in discussing conservation laws (specifically, for parity).
Gamma decay
When nuclei are produced in reactions or decay processes, they are often found to be initially in excited states.
These can then emit energy to decay to the ground state (or lowest-energy state) of the same nucleus. Typically
the energy spacing is of order ~ 50 keV (<< binding energy per nucleon ~ 8 MeV), and this amount of energy
can be emitted as a single energetic photon (a γ “particle” of electromagnetic radiation).
The conservation of energy for this γ decay gives
M *c 2 = M c 2 + E! + Erecoil
where M* and M are the masses for the excited and ground state of the nucleus, E is the energy of the emitted γ,
and there is a recoil energy term for the nucleus (so that momentum is conserved). We know, by conservation of
linear momentum, that the momentum of the recoiling nucleus is of magnitude
E!
precoil =
c
2
E!2
precoil
Erecoil =
=
and so
(where the non-relativistic approximation is good)
2M
2Mc 2
This leads to a quadratic equation for the γ-ray energy of the form
E!2 + 2Mc 2 E! ! 2M (M * ! M )c 4 = 0
γ
and the solution may then be found.
A more complete theory of gamma decay (to find transition rates and decay times, etc) can based on either
using classical electromagnetism or (much better) using nonrelativistic QM.
C.4 Fission and fusion
Fission
This is a process whereby a nucleus rapidly divides into 2 approximately equal masses. (Technically, this is
binary fission; fission decay into 3 or more heavy masses can occur but is extremely rare).
We can work out a condition for fission to occur spontaneously by using the liquid drop model. For simplicity,
we focus on fission into 2 equal masses:(A, Z) ! ( 12 A, 12 Z) + ( 12 A, 12 Z)
taking A and Z to be even
"
a Z 2 a (A ! 2Z )2 %
Energy before =
(Zmp + Nmn )c 2 ! $a1 A ! a2 A 2/3 ! 3 1/3 ! 4
'
A
A
#
&
2
"
a3 ( 12 Z ) a4 ( 12 A ! Z )2 %
2
2/3
1
1
Energy after
=
(Zmp + Nmn )c ! 2 $a1 ( 2 A) ! a2 ( 2 A) !
!
'
( 12 A)1/3
( 12 A) &
#
21 So the energy release is
Q = ! 0.26a2 A 2/3 + 0.37
This condition becomes
Z2
A
>
0.26a2
0.37a3
a3 Z 2
A1/3
> 0
for fission to occur
after substituting the standard values for a2 and a3.
! 16
An improved version of the theory (due to Bohr and Wheeler) leads to the modified condition that
Z2
> 50
A
The main two factors playing a role in the improved theory are:
(i) The shape of the stability curve. As discussed previously the stability curve for nuclei looks like:
N = Z line N Coordinates for nucleus undergoing fission Coordinates for fission into equal parts If the nucleus undergoing fission lies on the
stability line, it is clear that because of the
curvature the products of fission into equal
fragments will not be on the line and will have
an excess of neutrons.
This leads to a likelihood of further decay (e.g.,
by β- decay), changing the energy balance
equation. Also fission into equal fragments is no
longer necessarily the optimum.
Z (ii) The effect of a Coulomb barrier. There will be a Coulomb barrier for the same general reasons as in the
case of α decay, BUT the details are different because the different nuclei are highly distorted in shape just
before and after fission.
These distortions modify the surface area term (proportional to a2) and the Coulomb energy term (proportional
to a3) in the liquid drop model. This changes the condition for Z2/A.
All the previous discussion was for “spontaneous fission”, i.e., it occurs because the energetics of the decay is
favourable. However, we can also have “induced fission”, which refers to fission induced by supplying the
nucleus with extra energy (such as by bombardment with neutrons as in a nuclear reactor for power generation).
For example, consider the fission process for 235U92. We have Z2/A = (92)2/235 = 36 , which is less
12.1 Basic Properties of Fission
161
235 92
than the usual threshold of 50 and so fission does not occur spontaneously. However,
if theProcesses
U is bombarded
with a ’slow’ neutron having KE ~ 1 MeV, then induced fission can occur:
Barrier energy VC = neutron capture
energy + lllustration
1 MeV ~ of6 or
7 MeV. Shapes During
Schematic
Nuclear
236 92
When the neutron has been captured,the
we have
an
unstable
nucleus
of
U
(often called a “compound nucleus),
Fission Process
-14
which then decays into 2 fragments in a time of ~ 10 s.
n + *oo*Ge
O*O-ca)
Stage 1 Stage 2 Stage 3 Stage 4 Stage 5 At t = 0
we have Stage 1 consisting of the n and the 235U
-14
At t ~ 10 s we have stages 2, 3, and 4 where the compound 236U forms, splits (with prompt n and γ emission)
thestage
context
the liquicl
model,
drop
fission
canβ, be
viewed
as the fragAt t > 10-10 s weIn
have
5 whereofunstable
fragments
move
apart (with
further
γ and
n emission).
mentation of a drop as illustrated in Figure 12.4. Any oscillations in the shape
In some nuclei (A, Z) that are less massive (therefore slightly more stable) than the above example, we might
of the drop will tend to grow and to cause rhe drop to fragmenr, if the cleforhave Stages 1, 2, and 3 occur, to temporarily form (A+1, Z) in an excited state. Then,
instead of Stages 4 and 5,
mation from spherical symmetry makes the drop more stable. tWe can calculate
22 the energy associated with nuctear deformation (or stretching as shown in the
figure) in terms of nuclearr binding energies as
Fisure.'
2.3
I ffli?T151'5i,i3?'"'#33ft"Ji::JJ",;:T
(solid
Semiempirical Mass Formula
curve)
rr"""
where sis a def
elliptical, then e
volume of the nu
equate the volum
we
emission
giveof(A+1,
Z) inshell
its stable
ground state. This process is called “radiative capture”
structure.
effect
the nuclear
curve γ
illustrates
theto
Thewould
dotted have
volume
of the s
of the n.
@@
N N
:
\
il
il
z t\
il
O
6
il
=
+
+ +
45
c)
40
.9 35
(g 30
-o
c
,o 25
a
.9.
TL
\ NI
\ =I
il
@
+
il
50
The figure on the left shows the calculated
potential barrier (plotted versus A) for nuclear
fission to take place. It illustrates why we
onlyd and b a
where
get fission at large enough A.
tively. The ecce
O
o
€=
\I
N
il
@
il
il
t\
=
@
N
+,
il
e
T
I
I
V
t
i
3
il
15
The smooth-looking solid curve is found using
a simple theory based on the Liquid Drop
model, whereas the dotted curve is based on the
Shell model and includes the additionalUsing the semie
stability effects associated with the magic
ated with the de
numbers.
be manifested in
mula. The surfac
=
10
+
t
0
100
150
A
where Re is rela
soid relative to t
Fusion
Reprinted from Sfilliam D. Myers and Wladyslaw J. Swiatecki, "Nuclear masses and
Nuclear Physics 81 (1966),1-60, Copyright 1966 with pern.rission
deformations,"
Fusion
is effectively
the inverse process to fission, since it corresponds mainly to two nuclei combining to form
from Elsevier Science.
a larger nucleus. The energy conditions for this to occur are rather restrictive, and it is of interest only for the
smaller nuclei. Recall the earlier figure (in Section A) for the binding energy per nucleon:
This has a peak at around A ~ 55, so the process of
combining two much lighter nuclei to form a larger
nucleus (with A less than about 55) will increase
the binding energy per nucleon. The consequent
release of energy if this occurs can in principle be
used as an energy source.
The main problem is how to achieve this with
sufficient efficiency, since there is a Coulomb
potential barrier to overcome (by analogy with the
case of α decay considered earlier.
Generalizing the previous expression for the potential barrier height, we now have
e 2 Z1Z 2
which occurs when the separation of the nuclei is R1+R2
VC (max) =
4!" 0 R1 + R2
Here Zi and Ri denote the atomic number and radius for nucleus i (= 1, 2). Using
Ri = rs Ai1/3 ! rs (2Zi )1/3
and then taking A1 ~ A2 ~ A and putting in numerical values for the constants, we eventually estimate
VC (max) ~ A5/3 / 8 MeV
Therefore, fusion is most likely to occur for very small A, and in such cases the barrier height is less than about
1 MeV.
23 How should this amount of energy be supplied so that fusion can occur? One way would be high-energy
collisions (this turns out to be very inefficient because they are mainly elastic collisions and so there is not
much energy transfer).
Another way is to have very high thermal energies kBT, as in stellar systems. For 1 MeV of energy this
corresponds to T ~ 1010 K, which is much larger than in stellar interiors (typically 107 K). Actually, 107 K is
high enough, because there is a Maxwellian distribution of velocities (and energies) and there will be enough
energetic particles in the Maxwellian “tail” of this distribution.
The main ongoing practical problem in the use of fusion technology is the containment of the high-temperature
plasma combined with achieving the high-enough temperatures.
Some of the simplest fusion reactions involve deuterons d (= 2H1), e.g.,
d + p → 3He2 + γ
(Q = 5.5 MeV)
3 1
d + d → H + p
(Q = 4.0 MeV)
C.5 Energy loss for charged particles in matter
The energy range of the particles is typically in a range of ~ few keV up to ~ 10MeV (sometimes more). As a
charged particle passes through matter, it will generally interact and lose energy.
The processes by which this occurs will depend on the particle kinetic energy T, as well as on the type of
particle and the target material. We divide the discussion into two cases:
♦ Heavy particles, e.g, protons, α particles
♦ Light particles, e.g., electrons, positrons
Applications are in medical physics (e.g., ionizing effects of heavy particles in cell tissue for cancer studies) and
in solid state physics.
Energy loss by collisions for heavy particles
Consider the process of a proton p passing through a thin sheet of metal (such as Al) and the probability of
transmission:
The main process of energy loss for p is “collision” with the
Al sheet
electrons in the Al (via the Coulomb interaction). If the electrons
are assumed to behave as if free, they will acquire recoil energy
during the collision and this will represent energy lost by p.
p
We use a non-QM treatment, and also initially we ignore
relativistic effects.
The general collision process is approximately as depicted
below:Electron, mass me charge −e Charged heavy
particle
b
Coulomb force
x
The incident heavy particle has charge ze , mass M , speed v , and kinetic energy T . Also b = impact parameter.
Magnitude of Coulomb force ~
ze 2
, which acts for a time (the “collision time”) ~
4!" 0 b 2
b
v
24 ze 2
4!" 0 bv
For a better calculation, we would write down the instantaneous force and integrate with respect to time. Then
the result becomes
ze 2
(only the numerical factor is changed)
2!" 0 bv
Therefore the recoil KE imparted to the electron is
2
1 " ze 2 %
z 2e4
!T (b) =
=
$
'
2me # 2!" 0 bv &
8! 2" 02 b 2 v 2 me
Now we need to integrate over b to get the total energy loss. Consider a cylindrical shell between b and b + db :
So,
momentum imparted to electron
~
force × time ~
Let there be n atoms per unit volume in
the target metal, so there are nZ electrons
per unit volume.
v
No. of electrons in the cylindrical shell
= nZ 2πb db dx
dx
Total energy loss (= −ΔT ) in length dx =
So,
!
dT
dx
=
z 2 e 4 nZ
4!" 02 v 2 me
"
bmax
bmin
nZ "
db
b
bmax
bmin
!T (b) 2! b db dx
and the final result is
"b %
L = ln $ max '
# bmin &
This quantity −dT/dx is known as the “specific energy loss” or the “absolute stopping power”, and is often
denoted by S(T).
!
dT
dx
=
z 2 e 4 nZ
L
4!" 02 v 2 me
with
We still need to specify bmin and bmax to determine the L factor:
For bmin, we use a value deduced from the Uncertainty Principle. In the centre-of-mass frame of the electron
plus heavy particle, speed of electron ~ v , and so Δb mev ~ ħ or Δb ~ ħ/mev for the uncertainty in
b.
Therefore, we take bmin ~ ħ/mev .
For bmax, we use the fact that the electrons are not really free, but mostly are bound to a particular atom (with an
ionization energy I ). We can use this and the Uncertainty principle to define a time τ by τ ~ ħ/ I (This is
just ~ the period of orbital motion of the electron in a Bohr atom).
This time can be compared with the “collision time” ~ b / v , so that
if b / v << ħ / I , this is a “fast” collision
⇒ electron behaves same as if free ( able to recoil)
if b / v >> ħ / I , this is a “slow” collision ⇒ electron behaves same as if bound to atom (cannot recoil).
We choose bmax such that b / v = ħ / I giving bmax = v ħ/ I .
! m v2 $
!b $
Therefore,
L = ln # max & = ln # e &
" bmin %
" I av %
where Iav is an average value of I for all electrons.
Comments:
(1) S(T) is independent of M (mass of incident particle), but it does depend on its charge factor z and speed v.
25 1
1
at non-­‐relativistic speeds (v << c).
!
2
v
T
The general behavior (showing also the high-energy relativistic correction) is like:- S(T)
Neglecting the weak logarithmic dependence, S(T ) !
Increase due to
relativistic terms
T
(2) The range R of penetration of the heavy charged particle into the metal target can be worked out from
dT
z 2 e 4 nZ
dv
!
=
L(v) = ! Mv
2 2
dx
4!" 0 v me
dx
2
(using T = ½Mv
⇒
dT = Mv dv)
R
4!" 02 me M vi v 3dv
R = ! dx =
Range is
! v f L(v)
0
z 2 e 4 nZ
The usefulness of this expression is limited by its breakdown at very low velocities (because of difficulties in
fixing the lower limit of integration for the speed), but it is good for comparing R for different cases.
Also, there are other loss mechanisms that become significant at low v, adding to the uncertainty in the above
expression for R (see the text-book).
(3) At very high incident energies, apart from the relativistic corrections for the collision formula, there is the
additional mechanism of Cerenkov radiation. This is an electromagnetic process that can occur at high enough
energies when a charged particle passes through a dense medium of refractive index n0 (n0 > 1).
If the particle speed v > c/n0 (the speed of light in the medium), the particle can radiate energy by forming a
coherent wave front (“shock wave”)
A
c/n0
C
angle
θ
Consider the particle moving along the straight line
AB with speed v. A wave front spreads out from A
with speed c/n0 and from the later instantaneous
positions of the particle along AB.
A coherent wavefront (Huygen’s construction) is
formed as BC.
Since we must have cosθ < 1 and v < c, a condition is
that
v
B
The energy loss is typically of order 2 keV/cm (for a
proton), which is usually much smaller than for
collision loss.
Energy loss for light particles (electrons and positrons)
26 Energy loss will still occur by the previous process of collision loss (via the Coulomb interactions), but there are
other mechanisms that can become important and may dominate.
One of these mechanisms is radiation loss of energy when a charge is accelerated (in the strong electromagnetic
fields close to other atomic nuclei. From Maxwell’s equations it can be shown that the radiation loss per unit
time for an accelerated charge q is proportional to q2a2, where a is the acceleration.
Now, for a given electromagnetic field,
a ∝ 1/m
where m is the mass.
Comparing this mechanism for electrons and protons,
2
" mp %
Radn loss for e!
~ $ ' ~ 10 6
Radn loss for p
# me &
Thus radiation losses are much more important for lighter particles, especially at higher values of the kinetic
energy T and for target materials with higher Z.
In practice, the radiation losses for electrons and positrons require special conditions and usually occur by
(a) Bremsstrahlung (radiation involving strong electric field close to a nucleus)
OR
(b) Radiation in a high magnetic field (e.g., in a synchrotron)
OR
(c) Cerenkov radiation (high v in a medium, as discussed)
The special conditions arise as a consequence of satisfying conservation of both energy and momentum for the
emission of radiation:
photon pph, Eph electron p1, E1
θ
φ
“after” “before” electron p2, E2
We now show that it is impossible for this process to occur in isolation. We have:
p1 = p2 cos ! + p ph cos"
and
p2 sin ! = p ph sin "
from momentum conservation
E1 = E2 + E ph
and
from (relativistic) total energy conservation
In addition, we have
and
E12
=
p12 c 2 + me2 c 4 , E22 =
E ph = p ph c
p22 c 2 + me2 c 4
for each electron state
for the photon
Eliminating angle φ from the above equations eventually leads to
1/2
! me2 c 2 $
E1
cos! =
= #1+ 2 &
> 1
p1c
p1 %
"
This gives a contradiction ⇒ The conservation laws cannot be satisfied in isolation.
Therefore, we need an electric or magnetic field [as in cases (a) and (b) above] or a dense medium [as in (c)].
27