Transcript Particle Interactions - NIU - Northern Illinois University

Particle Interactions
Heavy Particle Collisions
• Charged particles interact in matter.
– Ionization and excitation of atoms
– Nuclear interactions rare
• Electrons can lose most of their energy in a single
collision with an atomic electron.
• Heavier charged particles lose a small fraction of
their energy to atomic electrons with each collision.
Energy Transfer
• Assume an elastic collision.
– One dimension
– Moving particle M, V
– Initial energy E = ½ MV2
– Electron mass m
– Outgoing velocities Vf, vf
• This gives a maximum energy
transfer Qmax.
MV  MVf  mvf
1
2
MV 2  12 MVf2  12 mv2f
Vf 
M m
V
M m
Qmax  12 MV 2  12 MV f2

4m M 1
4m E
2
(
MV
)

( M  m) 2 2
M
Relativistic Energy Transfer
• At high energy relativistic
effects must be included.
2g 2 m V 2
Qmax 
1  2gm / M  m2 / M 2
• This reduces for heavy particles
at low g, gm/M << 1.
Qmax  2g 2 m V 2  2g 2  2 m c2
Qmax  2(g 2  1)m c2
Typical Problem
• Calculate the maximum energy
a 100-MeV pion can transfer to
an electron.
• mp = 139.6 MeV = 280 me
– problem is relativistic
• g = (K + mp)/ mp = 2.4
• Qmax = 5.88 MeV
Protons in Silicon
• The MIDAS detector measured
proton energy loss.
– 125 MeV protons
– Thin wafer of silicon
MIDAS detector 2001
Linear Energy Transfer
• Charged particles experience multiple interactions
passing through matter.
– Integral of individual collisions
– Probability P(Q) of energy transfer Q
Q 
Qmax
Qmin
QP(Q)dQ
• Rate of loss is the stopping power or LET or dE/dx.
– Use probability of a collision m (cm-1)
Qmax
dE

 mQ  m  QP (Q )dQ
Qmin
dx
Energy Loss
• The stopping power can be
derived semiclassically.
– Heavy particle Ze, V
– Impact parameter b
• Calculate the impulse and
energy to the electron.
kZe2 b
p   Fy dt   2
dt
r r
y
Ze
b
p  kZe  2
dt
(b  V 2t 2 )3 / 2
2
V
b
2kZe2
p
Vb
r

kZe2
F 2
r
Fx
Fy
m, e
x
p 2 2k 2 Z 2 e 4
Q

2m m V 2 b 2
Impact Parameter
• Assume a uniform density of
electrons.
– n per unit volume
– Thickness dx
• Consider an impact parameter
range b to b+db
– Integrate over range of b
– Equivalent to range of Q
• Find the number of electrons.
2pnb(db)dx
• Now find the energy loss per unit
distance.
dE
4pnk 2 Z 2 e 4 db

 2pn  Qbdb 
b
dx
m V2
dE 4pnk 2 Z 2e 4 bmax


ln
2
dx
mV
bmin
Stopping Power
• The impact parameter is related
to characteristic frequencies.
• Compare the maximum b to the
orbital frequency f.
– b/V < 1/f
– bmax = V/f
• Compare the minimum b to the
de Broglie wavelength.
– bmin = h / mV
• The classical Bohr stopping
power is
dE 4pnk 2 Z 2e 4 m V 2


ln
2
dx
mV
hf
Bethe Formula
• A complete treatment of stopping power requires
relativistic quantum mechanics.
– Include speed b
– Material dependent excitation energy
dE 4pnk 2 Z 2e 4


dx
m c2  2
 2m c2  2
2
ln




2
I
(
1


)


Silicon Stopping Power
• Protons and pions behave
differently in matter
– Different mass
– Energy dependent
MIDAS detector 2001
Range
• Range is the distance a particle
travels before coming to rest.
• Stopping power represents
energy per distance.
– Range based on energy
• Use Bethe formula with term
that only depends on speed
– Numerically integrated
– Used for mass comparisons
R( K )  
K
0
1
 dE 

 dE
 dx 
1
R( K )  2
Z
1
R(  )  2
Z


0

K
0
dE
G( )
Mg (  )d M
 2 f ( )
G(  )
Z
Alpha Penetration
Typical Problem
• Part of the radon decay chain
includes a 7.69 MeV alpha
particle. What is the range of
this particle in soft tissue?
• Use proton mass and charge
equal to 1.
– Ma = 4, Z2 = 4
Ra (  ) 
M
Rp ( )  R p ( )
2
Z
• Equivalent energy for an alpha
is ¼ that of a proton.
– Use proton at 1.92 MeV
• Approximate tissue as water
and find proton range from a
table.
– 2 MeV, Rp = 0.007 g cm-2
• Units are density adjusted.
– Ra = 0.007 cm
• Alpha can’t penetrate the skin.
Electron Interactions
• Electrons share the same
interactions as protons.
– Coulomb interactions with
atomic electrons
– Low mass changes result
• Electrons also have stopping
radiation: bremsstrahlung
e
e
e
g
e
e
• Positrons at low energy can
annihilate.
Z
Beta Collisions
• There are a key differences between betas and heavy ions in matter.
– A large fractional energy change
– Indistinguishable  from e in quantum collision
• Bethe formula is modified for betas.

4pnk 2 e 4
 dE 

 
2 2
 dx  col m c 
 2m c2   2


ln

F
(

)


2
I


1  2
F ( ) 
2
 2

1


(
2


1
)
ln
2


 8

2 
14
10
4 

F ( )  ln 2 
23



24 
  2 (  2) 2 (  2)3 


K
mc 2
Radiative Stopping
• The energy loss due to
bremsstrahlung is based
classical electromagnetism.
– High energy
– High absorber mass
4nk 2 Z (Z  1)e4
 dE 

 
137mc2
 dx  rad
1

ln
2



3 
• There is an approximate
relation.

KZ  dE 
 dE 

 


dx
700
dx

 rad

col
-(dE/dx)/r in water
Energy
1 keV
10 keV
100 keV
1 MeV
10 MeV
100 MeV
1 GeV
MeV cm2 g-1
col
rad
126
0
23.2
0
4.2
0
1.87
0.017
2.00
0.183
2.20
2.40
2.40
26.3
Beta Range
• The range of
betas in matter
depends on the
total dE/dx.
– Energy
dependent
– Material
dependent
• Like other
measures, it is
often scaled to
the density.
Photon Interactions
• High energy photons interact with electrons.
– Photoelectric effect
– Compton effect
• They also indirectly interact with nuclei.
– Pair production
Photoelectric Effect
• A photon can eject an electron
from an atom.
– Photon is absorbed
– Minimum energy needed
for interaction.
– Cross section decreases at
high energy
g
e
Z
Ke  h  
Compton Effect
• Photons scattering from atomic
electrons are described by the
Compton effect.
– Conservation of energy and
momentum
g’
g

f
e
h  mc2  h   E
h h 

cos   P cos f
c
c
h 
sin   P sin f
c
h
h  
1
h
(1  cos )
2
mc
Compton Energy
• The frequency shift is
independent of energy.
• The energy of the photon
depends on the angle.
– Max at 180°
• Recoil angle for electron related
to photon energy transfer
– Small   cot large
– Recoil near 90°
h
 
(1  cos  )
mc
K
h (1  cos  )
mc 2 / h  1  cos 

h 

cot  1  2  tanf
2  mc 
Compton Cross Section
• Differential cross section can be
derived quantum mechanically.
– Klein-Nishina
– Scattering of photon on one
electron
– Units m2 sr-1
• Integrate to get cross section
per electron
– Multiply by electron density
– Units m-1
d
k 2e4   

 
d 2m2c 4   
  2pn 
2
 
2 
   sin  
  

d
sin d
d
Pair Production
• Photons above twice the
electron rest mass energy can
create a electron positron pair.
– Minimum  = 0.012 Å
• The nucleus is involved for
momentum conservation.
– Probability increases with Z
• This is a dominant effect at high
energy.
h  2mc2  K  K
g
e
e
Z
Total Photon Cross Section
• Photon cross sections are the sum of all effects.
– Photoelectric , Compton incoh, pair k
Carbon
Lead
J. H. Hubbell (1980)