Interaction of Heavy Charged
Particles with Matter
BAEN-625
Advances in Food Engineering
Heavy charged particles
y
Charged particles other than the electron
and positron
Energy-loss mechanisms
y
A heavy particle traversing matter loses energy
primarily thru the ionization and excitation of
atoms
◦ Except in low velocities, it loses a negligible amount of
energy in nuclear collisions
y
y
y
The moving particle exerts electromagnetic
forces on atomic electrons and impart energy on
them
The transferred energy may be sufficient to knock
an electron out of an atom and thus ionize it
Or it may leave the atom in an excited state
Heavy charged particle
Can transfer only small fraction of its
energy in a single electronic collision
y Its deflection in the collision is negligible
y Thus it travels an almost straight path
thru matter,
y It loses energy continuously in small
amounts thru collisions with atomic
electrons
y
Maximum Energy Transfer in a
Single Collision
y
Assume
◦ the particle is moving rapidly compared to the
electron
◦ Energy transferred is large compared with the
BE (binding energy) of the electron in the
atom
◦ The electron is free and at rest
Maximum Energy Transfer in a Single
Collision
Maximum Energy Transfer in a
Single Collision
y
Since energy and momentum are conserved
1
1
1
2
2
MV = MV 1 + mv 12
2
2
2
MV = MV 1 + mv 1
y
Solving for V1:
( M − m )V
V1 =
M +m
1
1
4 mME
2
2
Q max = MV − MV 1 =
2
2
( M + m )2
E=MV2/2
Initial KE
Incident Particle is an Electron
y
y
Its mass is the same as that of the struck
particle, M = m
Qmax
4mME
4 MME
=
=
2
( M + m)
(M + M )2
Qmax
4M 2 E
=
=E
2
4M
Entire energy can be transferred in a single,
billiard-ball-type collision
Energy Transfer in a Single Collision if
Incident Particle is an Electron
Maximum Energy Transfer in a Single
Collision- Relativistic Expression
y
An electron is nonrelativistic as long as T is small
compared with the rest energy, mc2 = 0.511MeV
Q max
2 γ 2 mV 2
=
2
1 + 2γ m / M + m / M
γ = 1/ 1− β 2
β =V /c
2
Qmax in Proton Collision with Electron
Proton Kinetic
Energy, E
[MeV]
Qmax
[MeV]
Maximum %
Energy Transfer
100Qmax/E
0.1
0.00022
0.22
1
0.0022
0.22
100
0.0219
0.22
100
0.229
0.23
1000
3.33
0.33
10000
136
1.4
100000
1060
10.6
1000000
53800
53.8
10000000
921000
92.1
Elastic or Inelastic
y
y
y
y
y
Equations shown before for Qmax are kinematic
in nature
They follow from simultaneous conservation of
momentum and KE
The assumption made to calculate energy loss
was that the struck electron was not bound
Thus the collision being elastic
Charged-particle energy losses to atomic
electrons are, in fact, inelastic
Elastic Collision
Elastic collision
y
y
y
y
Both momentum and kinetic energy are conserved
This implies that there is no dissipative force acting
during the collision and that all of the kinetic energy of
the objects before the collision is still in the form of
kinetic energy afterward
For macroscopic objects which come into contact in a
collision, there is always some dissipation and they are
never perfectly elastic
In atomic or nuclear scattering, the collisions are typically
elastic because the repulsive Coulomb force keeps the
particles out of contact with each other.
Examples of Elastic Collision
y
y
For a head-on collision with a stationary object of equal
mass, the projectile will come to rest and the target will
move off with equal velocity, like a head-on shot with the
cue ball on a pool table.
This may be generalized to say that for a head-on elastic
collision of equal masses, the velocities will always
exchange.
Examples of Elastic Collision
y
In a head-on elastic collision where the projectile is much
more massive than the target, the velocity of the target
particle after the collision will be about twice that of the
projectile and the projectile velocity will be essentially
unchanged.
Examples of Elastic Collision
y
In a head-on elastic collision between a small projectile
and a much more massive target, the projectile will
bounce back with essentially the same speed and the
massive target will be given a very small velocity.
Inelastic Collision
y
y
y
Perfectly elastic collisions are those in which no
kinetic energy is lost in the collision.
Macroscopic collisions are generally inelastic
and do not conserve kinetic energy, though of
course the total energy is conserved.
The extreme inelastic collision is one in which
the colliding objects stick together after the
collision
Inelastic Collision
Single-Collision Energy-Loss Spectra
y
y
y
Details about charged-particle penetration are
embodied in the spectra of single-collision energy
losses to atomic electrons
The collisions by which charged particles transfer
energy to matter are inelastic
KE is lost in overcoming the BE of the struck
electrons
Single-Collision Energy-Loss Spectra
y
The ordinate gives the
probability density
W(Q)
W(Q)dQ is the
probability that a
given collision will
result in an energy
loss between Q and Q
+ dQ
0.06
W(Q) (eV-1)
y
In liquid water
50-eV electrons
0.04
0.00
0
5-MeV protons
150-eV electrons
50
Energy Loss Q (eV)
100
Single-Collision Energy-Loss Spectra
y
For fast particles (speed > orbital
speed)
In liquid water
◦ Similarities in the region from 1070eV
For slow charged particles
◦ The energy-loss spectra differ
from one another
◦ The time of interaction is longer
than for fast particles
◦ The BE is more important
◦ Energy losses are closer to Qmax
◦ Slow particle excites atoms
instead of ionizing them
y
A minimum energy Qmin >0 is
required for excitation or
ionization of an atom
50-eV electrons
W(Q) (eV-1)
y
0.06
0.04
1-MeV protons
150-eV electrons
0.00
0
50
Energy Loss Q (eV)
100
Stopping Power
The average linear rate of energy loss of a
heavy particle in a medium [MeV/cm]
y Also referred as linear energy transfer
(LET) of the particle
y
Stopping Powers
y
y
Can be calculated from energy-loss spectra
For a given type of charged particle at a given energy, the
SP is given by
◦ The probability μ per unit distance of travel that an electronic
collision occurs
◦ The average energy loss per collision, Qmax
Qavg = ∫
Qmax
Qmin
QW (Q)dQ
Qmax
dE
−
= μQavg = μ ∫ QW (Q)dQ
Qmin
dx
[MeV/cm]
[1/cm]
[MeV]
Stopping Power-Semi Classical
Calculation
ze
Y
V
Coulomb force
r
k 0 ze 2
F =
r2
b
θ
Fx
Representation of the sudden
Collision of a heavy charged
Particle with an electron,
Located at the origin XY
Fy
m -e
X
Stopping Power-Semi Classical
Calculation
y
The total momentum imparted to the electron is the
collision is:
∞
∞
∞ cos θ
p = ∫ Fy dt = ∫ F cos θdt =k0 ze 2 ∫
dt
−∞
−∞
−∞ r 2
t = 0 ( time the heavy particle cross Y - axis)
cosθ = b/r
∞ cos θ
∞ b
∞
dt
dt
dt
b
=
=
2
2
∫−∞ r 2
∫0 r 3
∫0 (b 2 + V 2t 2 ) 2 / 3
∞
⎡
⎤
t
2
=
= 2b ⎢ 2 2
2 2 1/ 2 ⎥
⎣ b (b + V t ) ⎦ 0 Vb
2k0 ze 2
p=
Vb
p 2 2k 0 z 2 e 4
Q=
=
2m mV 2b 2
Stopping Power-Semi Classical
Calculation
y
In traversing a distance dx in a medium having a uniform density of n
electrons per unit volume
y
The heavy particle encounters 2πnb db dx electrons at impact
parameters between b and b + db
y
The energy lost to these electrons per unit distance traveled is 2πnQb
db
y
The total linear rate energy loss is:
Qmax
4πk02 z 2 e 4 n bmax db
dE
−
= 2πn ∫ Qbdb =
=
2
∫
Qmin
bmin b
dx
mV
dE 4πk02 z 2 e 4 n bmax
−
=
ln
2
dx
mV
bmin
Relativistic Stopping Power (Bethe’s
Equation)
y
The linear rate of energy loss to atomic electrons along
the path of a heavy charged particle in a medium is the
basic physical quantity that determines the dose that the
particle delivers in the medium
dE 4πk02 z 2 e 4 n ⎡ 2mc 2 β 2
2⎤
−
=
−β ⎥
⎢ln
2 2
2
dx
mc β ⎣ I (1 − β )
⎦
k0 = 8.99 ×109 Nm2C −2
z = atomic number of the heavy particle;e = magnitudeof electron charge;
n = number of electronsper unit volume in the medium
m = electron rest mass; c = speed of light; β = V/c = speed of the particle relative to c
I = mean excitation energy of the medium
Stopping Power
y
y
y
y
Depends only on the charge ze and velocity β of the
heavy particle
The relevant properties of the medium are its mean
excitation energy I and the electronic density n
m is the mass of the target atomic electrons
Units: MeV/cm, mass stopping power-[MeV cm2/g]
Stopping Power, general
y
For any heavy charged particle in any medium
dE 5.09 × 10 −31 z 2 n
[ F ( β ) − ln I eV ], MeV/cm
−
=
2
β
dx
1.02 × 10 β
2
F( β ) = ln
−
β
1− β 2
6
2
Mass Stopping Power
y
Useful quantity because it express the rate of energy loss of the
charged particle per g/cm2 of the medium traversed
In gas –dE/dx depends on pressure, but –dE/ρdx does not
y MSP does not differ greatly for materials with similar atomic
composition (primarily light elements)
y
y
For 10 MeV protons the MSP of H2O is 45.9 MeV cm2/g and for
C14O10 44.2 cm2/g, however for Pb(Z=82) the MSP is 17.5 cm2/g
y
Heavy elements are less efficient on a g/cm2 basis for slowing down
heavy charged particles (many of their electrons are too tightly
bound in the inner shells to participate effectively in the absorption
of energy)
Mean Excitation Energies
Can be calculated from SP equation
y It is the material parameter describing the
ability of the target system to absorb
energy
y Empirical expressions:
y
⎧19.0eV , Z = 1
⎪
I ≅ ⎨11.2 + 11.7 Z eV , 2 ≤ Z ≤ 13
⎪52.8 + 8.71Z eV, Z > 13
⎩
Mean Excitation Energies
y
y
When a material is a compound or mixture, the SP can
be calculated by simply adding the separate
contributions from the individual components
If there is Ni atoms/cm3 of an element with atomic
number Zi and mean excitation Ii:
n ln I = ∑ N i Z i ln I i
i
Total # of electrons/cm3 in the material
(n=ΣNiZi)
Example
y
Calculate the mean excitation energy of H2O
H ( Z = 1), I H = 19.0eV
O( Z = 8), I O = 11.2 + 11.7 × 8 = 105eV
Ni Zi
2 ×1
1× 8
ln 105 = 4.312
ln I =
ln I i =
ln 19.0 +
n
10
10
I = 74.6eV
Table for Computation of SP’s
y
Use of table to facilitate the computation of SP
for heavy charged particles
dE 5.09 ×10 −31 z 2 n
−
=
[ F ( β ) − ln I eV ], MeV/cm
2
β
dx
1.02 ×106 β 2
2
F( β ) = ln
−
β
1− β 2
Table for Computation of SP’s
Proton KE
[MeV]
β2
F(β)
0.01
0.000021 2.179
0.06
0.000128 4.873
0.1
0.000213 5.161
0.4
0.000852 6.771
1
0.002129 7.685
6
0.01267
9.753
10
0.02099
9.972
60
0.1166
11.96
100
0.1834
12.16
SP of Water for Protons
y
y
y
y
y
Protons, Z = 1
MW of water = 18.0g/mol
Number of electrons/molecule, n = 10 ele/mol
1 m3 of water = 106 g
Density of electrons is
6
10
g
23
n = 6.02 × 10 ×
× 10 = 3.34 ×10 29 m −3
18.0 g
SP of Water for Protons
y
y
Ln Iev = 4.3212
SP of water for a proton of speed β is:
dE 0.170
−1
[
]
−
=
F
(
)
−
4
.
31
MeVcm
β
dx
β2
at 1 MeV - Table - β 2 = 0.00213; F ( β ) = 7.69
0.170
dE
[7.69 − 4.31] = 270MeVcm−1
−
=
dx 0.00213
SP of Water for Any Particle
y
Previous table can be used for any other particle
– get F(β) and β2 by using the following
relationship:
Tpartic
Tproton
=
M partic
M proton
Example
y
What is the SP for an 10-MeV alpha particle in
water
Talpha
Tproton
4
= =4
1
Tproton = 10 MeV / 4 = 2.5MeV
find F( β ) and β from Table 5.2
calculate - dE/dx
2
Range
y
y
y
Of a charged particle is the distance it travels before
coming to rest
The reciprocal of the stopping power gives the distance
traveled per unit energy loss
So, R(T) of a particle of kinetic energy T is:
−1
⎛ dE ⎞
R (T ) = ∫ ⎜ −
⎟ dE
dx ⎠
0⎝
T
T
1
dE
R (T ) = 2 ∫
z 0 G(β )
Range
y
For a heavy particle:
M
R( β ) = 2 R p ( β )
z
The proton range
Particle’s velocity
Example
y
Use Table 5.3 to find the range of an 80-MeV
3+ ion in soft tissue
He
2
M
R( β ) = 2 R p ( β )
z
2
z = 4, M = 3
3
R( β ) = R p ( β )
4
Example
3
R( β ) = R p ( β )
4
80
= 26.7 MeV
Proton Energy =
3
interpolation in Table 5.3
R p = 0.705cm − 2
3
R ( β ) = 0.705 = 0.529 gcm − 2
4
0.529cm / ρ soft tissue
Ranges in cm of protons, alpha, and electrons
in air at STP
105
Range in air (cm)
104
y
electrons
103
For alpha particle at
15oC and 1 atm:
R = 0.56 E , E < 4
protons
102
R = 1.24 E − 2.62,
Alpha
particles
101
y
100
10-1
10-2
10-1
100
101
Energy (MeV)
102
103
4< E <8
R in cm and E in MeV
Slowing-Down Time
y
We can use the SP formula to calculate the rate at which a heavy
charged particle slows down
y
The time rate of energy loss, -dE/dt can be expressed in terms of the
SP by:
dE
⎛ dE ⎞⎛ dt ⎞
⎛ dE ⎞
−
= −⎜
⎟⎜ ⎟ = V ⎜ −
⎟
dt
⎝ dx ⎠⎝ dx ⎠
⎝ dx ⎠
for proton with T = 0.5MeV in water
dE
−
= 4.19 × 1011 MeVs −1
dt
Slowing-Down Time
y
A rough estimation can be made of the time it takes a
heavy charged particle to stop in matter, if one assumes
that the slow-down rate is constant
T
T
τ≈
=
− dE / dt V (−dE / dt )
for proton with T = 0.5MeV in water
τ ≈ (0.5MeV ) /(4.19 ×1011 MeVs −1 )
τ ≈ 1.2 ×10 −12 s
Limitations of Beth’s Equation
y
y
It is valid at high energies as long as γm/M<<1 holds (e.g.
up to ~ 106MeV for protons
At higher energies it needs to consider
◦ Forces on the atomic electrons due to the particle’s spin and
magnetic moment
y
It is based on the assumption that the particle moves
much faster than atomic electrons
◦ At low energies it fails because the term 2mc2β2/I becomes
negative (given a negative value for stopping power)
y
many more (see text)