Download Charged Particle Interactions with Matter: R Z M

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Charged Particle Interactions with Matter:
•
We will switch gears now that we know about transferring energy to charged
particles, let’s find out what happens next – how do these charged particles then
deliver the energy to matter as they slow down.
•
We will start with an important derivation that will then be generalized to
eventually work for electrons slowing down in condensed matter.
•
At first though, let’s start with a heavy charged particle with kinetic energy = T,
velocity=v with a lot of inertia going very quickly by an atom with an impact
parameter = b:
Mi
y
v
Zi
x
z
b
a
negatively charged
shells, with bound electrons:
− e, m
K
L
Rnuc
•
positively charged nucleus
The “heavy and fast” particle at the base of the arrow in the above figure has a
charge, Z i and a mass, M i , and a significant energy, γM i c 2 such that
γ ≡ 1 / 1 − v 2 / c 2 >> 1 (“fast” = “Born approximation”). The bound electron on the
other hand has only a charge, − e , a mass, m , and bound energy such that kinetic
energy, T, roughly equals its potential energy. The velocity is v ,(not frequency
here). In other words: mv 2 / 2 ~ e 2 / a0 where a0 ≈ h 2 / me 2 which gives
v 2 / c 2 ~ e 4 / h 2c 2 ~ (1 / 137) 2 , the fine structure constant, α , squared. Therefore,
the incident particle needs to have more kinetic energy than this: T > mc 2α 2 . This
implies that only about 53 keV is needed for the incident particle in this treatment.
Other generalization beyond these assumptions will also be possible. Note we are
using slightly different notations: T and m and z versus Z.
Lecture 9 MP 501 Kissick 2016
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Note that the above figure is not at all to scale!
-- The nuclear radius is Rnuc ≈ 1.4x10 −13 A 1/3 where A is the mass number (number
of protons plus the number of neutrons, an integer here, so not the same as A
before !). This is the “liquid drop model.”
-- The atom’s radius is about a ≈ 0.18βλ from Bohr’s theory, where λ is the
wavelength of the electron in an outer shell. Also, β = v / c , the particle speed
compared to the speed of light. This is much much larger than the nuclear radius!!
(like a baseball (nucleus) in Camp Randall stadium (atom)). Note that there is a
velocity dependence to all of this, and its just order of magnitude validity anyway.
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We will be categorizing reactions according to b:
-- soft collisions: b >> a : about half of what is happening, continuous energy loss
with many atoms at once – interactions with orbiting electrons. Leads to
excitations.
-- hard collisions: b ≈ a : also near half of the interactions, but large events that
can lead to the liberation of other charges with their own trajectories: we will
carefully need to account for that! The liberated charges are outer shell orbiting
electrons. Gives ionizations with resulting excitations.
-- nuclear electric field interactions: b ≈ Rnuc : This results in elastic collisions
(mostly) and bremsstrahlung (only 2-3% mostly for electrons). Note that only light
particles, like electrons, will show any significant bremsstrahlung.
(-- nuclear interactions: b < Rnuc : Nuclear reactions will only occur here.)
•
The quantity we want to get is the stopping power = rate of energy loss for
distance travelled into the medium:
•
dT
.
dx
We will also define the mass stopping power, and it is usually divided into soft and
hard collisional losses (that come together later we will see), and an additional term
for radiative (bremsstrahlung) losses:
dT  dT   dT 
 +

=
ρdx  ρdx c  ρdx  r
Lecture 9 MP 501 Kissick 2016
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The first measurements of stopping power were made by Mme. Curie in 1900:
(image from: http://www.srim.org/SRIM/History/)
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Classical Derivation of the Mass Collisional Stopping Power:
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Return to the figure at the start of this lecture, and consider the momentum
impulse as we integrate in time the rate of momentum change (Gaussian units here):
r
v dP
 r 1 r r
F=
= −e E + v × B 
dt
c


•
However, only the electric field part is important since v / c ~ α as described in the
appendix. For the geometry given above, the kicks in x and y are given by:
E x = − Zeγvt
E y = − Zeγb
•
(b
(b
2
2
+ γ 2v 2t 2
+ γ 2v 2t 2
)
3/ 2
)
3/ 2
One then integrates from – to + infinity in time while noticing that the kick in E x
vanishes because its an odd function. Therefore only the y-kick comes into the
momentum change:
Lecture 9 MP 501 Kissick 2016
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∞
∆P = ∆Py = ∫ dt eE y (t )
−∞
=2
=
•
Ze 2b ∞
d (γvt )
∫
2
0
v
b + (γvt ) 2
(
)
3/ 2
2 Ze 2
bv
This is like an impulse with Py _ initial = 0 . The energy lost is simply obtained then by
(∆P) 2 / 2m to arrive at the following very important result for the interaction with
a single electron:
∆T (b) =
( ∆P ) 2 2 Z 2 e 4  1 
=
 
2m
mv 2  b 2 
•
Notice right away that this energy loss is independent of the incident particle
mass, M i !!
•
Notice also that the energy loss is also proportional to ~ 1/v2 !! This is a very
important result that remains dominant as we proceed further …
•
However, that was only for a given b, but in our problem, we have a whole range of b
values! Therefore, we need to integrate over that range in b, impact parameter.
•
First identify the number of electrons in the annulus below of area = 2πbdb .
db
Mi
v
b
y
x
Zi
z
dx
•
It is [(2πbdb) ⋅ (dx)] ⋅ ρ ⋅ ( N A z / A) = number of electrons in this differential annulus.
Note: the small z is the material atomic number here: opposite from most books.
Lecture 9 MP 501 Kissick 2016
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The differential energy loss experienced by the particle having an impulse with this
annulus is then:
dT = [(2πbdb) ⋅ (dx)] ⋅ ρ ⋅ ( N A z / A) ⋅
•
2Z 2 e 4  1 
 
mv 2  b 2 
Or, rearranging and noting that we need to integrate over all the possible b’s:
2 4
 dT 
 N Az  Z e

 = 4π 

2
 A  mv
 ρdx  c
∞
∫ (db / b)
b =0
•
One can immediately see we are heading for a logarithmic divergence !
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Aside: a word about units, since you may have been wondering. In terms of mass=m,
length=l, and time=t, and noting that we are working in Gaussian units
([e]=statcoulombs): e 2 = qe2 / 4πε 0
 T  ml 2 l 2
m1/ 2l 3 / 2  N A z e 4  1 ml 4
[e] =
∴
⋅ 2=
⇔ = 2 ⋅
2
t
m
 ρx  t
 A mv  m t
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However, the matter at hand is how to perform the integral. Instead of the full
integral, let’s go from a bmin to a bmax:
2 4
 dT 
 N A z  Z e  bmax 

ln

 = 4π 

2
ρ
dx
A
mv
b



c
 min 
•
To determine the distance of minimum energy transfer, bmax, it is a real headache
with quantum mechanics of energy transfers with bound electron shell states …
•
In 1913, N. Bohr, using many bound oscillating electrons ( ωi ) with individual
oscillator strengths ( f ) to define an average oscillator: ω =
∑f
i
ln ωi / z arrived
at his final form after some complicated Bessel function math:
Bc ≡
γ 2 mv 3
bmax
= (1.123)
bmin
Zie2 ω
[
Z 2 ze 4
 dT 
ln (Bc ) − β 2 / 2
 = 4π (n )
2
dx
mv


in 
]
Lecture 9 MP 501 Kissick 2016
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Where n is number density of atoms. This was the published result and is known as
the Bethe formula as are many variations on this theme. Many people have modified
and added to this. Note the relativistic correction starting to appear. The factor
1.123 is from 1.123 = 2/exp(0.577…), and 0.577 … is Euler’s constant.
o
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However, we will leave this complexity now, but see Appendix for more.
Let’s separate the collisional mass stopping power into two terms: one for hard
collisions, and one for soft collisions. We will start out with an arbitrary energy
cutoff, H, that will separate the two by an impact parameter or equivalently in this
case, an energy. If σ s and σ h are cross-sections for soft and hard collisions
respectively, then:
T ' max
H

 dT   dTs   dTh   N A z  
dσ s
dσ h
T ' dT '+ ∫
T ' dT '

 = 
 + 
 = 
 ∫
dT
'
 ρdx c  ρdx c  ρdx c  A  T ' min dT '

H
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The quantity T’ here is the energy transferred by the fast charged particle to
electrons. The quantities σ s and σ h are the cross-section for soft and hard
collisions respectively.
•
The quantity H here is somewhat arbitrary, but it should be about the value of the
minimum energy at which the electron is ejected for hard collisions. The values of
T’max are as follows:
-- T 'max = T / 2 for electrons
-- T ' max = T for positrons*
* Note, we can always differentiate between this as
the fast charged particle and any electron,
but above we cannot.


1 + (2 Mc 2 / T )

 for a heavy charged particle, mass = M.
-- T ' max = T 
2 2
1
+
(
M
+
m
)
c
/(
2
m
/
T
)


If M>>m, and T<<Mc2, then it reduces to T 'max ≈ 2mc 2
•
Note and remember: (γ − 1) = T / Mc
2
β2
≈ 2mv 2
2
1− β
& γ = 1/ 1− β 2
Lecture 9 MP 501 Kissick 2016
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Soft Collision Mass Stopping Power:
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If the energy transferred, T’, is less than the ionization potential, only excitations
can occur.
•
(If it is larger, then ionizations can occur, but they will also have excitations as
well, and the energy must be larger than H.)
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Define I = mean excitation potential of the absorbing medium, includes both
excitations and ionizations. See Attix, appendix B.1 and B.2. There is a lot of
uncertainty to I, but at least it is in the logarithm which reduces the sensitivity to
that uncertainty.
•
I is approximately proportional to z, the atomic number of the medium. Here, Z is
the fast particle charge. Again, small z is the material. Opposite from most
sources
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Derived by Bethe using the Born approximation (kinetic energy is much greater than
an electrons potential (orbital) energy), and that approximation is roughly
β >> zZ / 137 , then,
2
 dTs 
N zZ

 = 2πr02 mc 2  A  2
 A β
 ρdx c
  2mc 2 β 2


H  − β 2 
ln 2
2
  I (1 − β ) 



 z


Z 2    2mc 2 β 2
2
2

= (0.1535)
⋅ 2  ln 2
H
−
β
( MeV /( g / cm ))
2

β 
A
I (1 − β ) 
 {

 material {   
particle 

Where, 2πr02 mc 2 N A = 0.1535( MeV /( g / cm 2 ))
 z Z2 
k ≡ (0.1535) ⋅ 2 ( MeV /( g / cm 2 ))
A β 
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For ease later, we can define:
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The above applies to electrons, positrons, and heavy charged particles. Soft
collisions are most or at least half of the collisions.
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Next lecture: hard collisions.
Lecture 9 MP 501 Kissick 2016