Dose
Energy Gained
• Particles lose energy in matter.
• Eventually energy loss is due to ionization.
• An important measure is the amount of energy
gained by the material as a particle passes or stops.
Energy Transferred
• Energy transferred describes the kinetic
energy gained by charged particles.
• Energy imparted is the energy lost by
charged particles.
particle energy
10
MeV
5.4
3.6
energy transferred
9.0
energy imparted
2.8
4.2
2.0
3.3
2.8
1.1
0.1
2.8
1.1
Energy Imparted
• Energy transferred
– Radiant energy into a
volume from uncharged
particles
– Radiant energy out tpo
uncharged particles (not
brem or annihilation)
– Energy change from mass
nonrad
Etr  ( Rin )unc  ( Rout )unc
 Q
• Energy imparted
– Start with energy
transferred
– Energy in from charged
particles
– Energy out from charged
particles
nonrad
E  ( Rin )unc  ( Rout )unc
 ( Rin ) chg  ( Rout ) chg   Q
Kerma
• Kerma is the energy transferred
per unit mass.
– Kinetic Energy Released
per unit MAss
dEtr
K
dm
• Radiative kerma is the energy
loss per mass due to brem and
annihilation.
rad
dRunc
Kr 
dm
• Collision kerma subtracts the
reradiated photons.
– Net energy transferred per
mass
dEtrnet
KC  K  K r 
dm
Absorbed Dose
• Absorbed dose or dose is the
energy imparted per unit mass.
• Like kerma dose is based on
mean changes in energy.
• Two units are used.
– 1 gray (Gy) = 1 J / kg
– 1 rad = 100 erg / g (older)
dE
D
dm
J 107 erg
1 Gy 

kg
103 g
erg
 10 4
 100 rad
g
Lethality
• Lethality refers to the likelihood
that a dose will be fatal.
– Cell death
– Whole body death (see
graph at right)
Lethality %
• Dose can be compared to
physical effects.
Dose (cGy)
Federation of American Scientists
Equilibrium
• Measuring the relationship
between energy transferred and
imparted requires equilibrium
conditions.
• Radiation equilibrium
( Rin )chg  ( Rout )chg
( Rin )unc  ( Rout ) unc
E  Q
• Charged particle equilibrium
– Looser requirement
E  ( Rin )unc  ( Rout )unc   Q
Exposure
• Exposure is defined by the
ionization produced by photons.
– Gammas and X-rays
– Charge per unit mass in air
• The original unit is 1 esu / cm3
of dry air at STP (1928).
– Roentgen (R)
– R = 2.58 x 10-4 C / kg
(1962)
Useful Conversion
• Show that the original roentgen
is equivalent to the modern one.
• Look up constants:
– Density of air at STP is
0.001293 g / cm-3
– 1 esu = 3.34 x 10-10 C
• 3.34 x 10-10 C / 1.293 x 10-6 kg
= 2.58 x 10-4 C / kg
Gamma Rate
• Assume a point source of
gammas.
– Activity C
– Average photon energy E
• Consider a sphere or radius r.
– Fluence through sphere F
– Mass energy absorption
men/r = 2.7 x 10-3 m2/kg
• Find the dose and exposure rate.
– 1 R = 0.0088 Gy
• Rate of energy release:
dE
 CE
dt
CE
 
• Fluence rate: F
4r 2
• Dose rate:
 men  CE men
D  F
r 4r 2 r
1.27 10 6 (m 2s/BqJ) CE

Gy/s
2
r
• Exposure rate:
2
0
.
5
(m
s/BqJ) CE
X 
R/h
2
r
Radiation Factor
• The effect of radiation on tissue
depends on the LET as well as
the dose.
– Higher LET is more
damaging.
• Radiation has a weighting
factor based on particle.
– Factor WR or Q
– Updated in 1991
• In terms of LET
– LET L (keV / mm in water)
– < 10; WR = 1
– 10 – 100; WR = 0.32L – 2.2
– > 100; WR  300 / L
• In terms of particle
– e, g, m; WR = 1
– n; WR = 5 – 20
– p; WR = 5
– a; WR = 20
Restricted Stopping Power
• Secondary electrons ejected
from atoms are delta rays.
– Deltas deliver most of the
dose
– Energy less effective if it’s
too high
• Restricted stopping power
measures the energy lost up to a
limit D.
D
dE

 m  QP (Q)dQ
Qmin
dx
Typical Problem
• Estimate a cutoff value for
irradiating 300 Å viruses.
Answer
• Most energetic delta range
should not exceed 300 Å.
– Find range limit 3 x 10-6 cm
• Range in water
– 500 eV is 2 x 10-6 g/cm2
– 1 keV is 5 x 10-6 g/cm2
• Estimate cutoff at 700 eV.
Equivalent Dose
• The equivalent dose is a
measure that combines the type
of radiation and dose.
HT  WR D
• Unit is Sievert (Sv)
– 1 Gy equivalent
• Older unit is rem
– Roentgen equivalent man
– 1 rad equivalent
– 100 rem = 1 Sv
• Natural doses
– Cosmics: 0.3 mSv / yr
– Soil: 0.2 mSv / yr
– Radon: 2 mSv / yr
– Total natural: 3 mSv / yr
• Environmental hazards
– Flying at 12 km: 7 mSv / hr
– Chest x-ray: 0.1 mSv
– Mammogram: 1 mSv
– CT scan: 20 mSv
Neutron Interactions
• Neutrons present a
unique situation for
dose determination.
– No interaction
with atomic
electrons
– Cross sections
vary with target
nucleus
Neutron Scattering
• Elastic scattering from nuclei is
the most important process for
neutron energy loss.
– Assume classical collision
– Set M = 1 and compare for
nuclei
Qmax  12 MV 2  12 MV f2
4mM
2
1

(
MV
)
2 2
( M  m)
4mMEn

( M  m) 2
• Nucleus
– 1H
– 2H
– 4He
– 9Be
– 12C
– 16O
– 56Fe
– 118Sn
– 238U
Qmax/En
1.000
0.889
0.640
0.360
0.284
0.221
0.069
0.033
0.017
Neutron Dose Equivalent
• Neutron weighting factors were
variable.
– WR = 5-20.
• The factors can be determined
from assumed elastic scattering.
Example
• Find WR for 2-MeV neutrons.
– Average recoil p is 1 MeV
– Stopping power for 1 MeV
p in water is 270 MeV/cm
– Equal to 27 keV/mm
WR  0.32L  2.2
WR  0.32(27)  2.2  6.4
Neutron Threshold
• Inelastic collisions result in a
nuclear reaction.
– Many are endothermic
– Requires extra energy
• Change in rest energy
• For example 32S(n,p)32P:
– Eth = 0.957 MeV
– 32P  32S + b– Ebmax = 1.71 MeV
– T = 14.3 days
– Used to detect exposure
E1  E3  E4  Q
Q  M1  M 2  M 3  M 4
• Conservation of energy and
momentum
p1  p3  p4
 M3  M4 

E1  Eth  Q
 M 3  M 4  M1 
Neutron Activation
• Time dependence of activity
from neutron capture is based
on exposure and decay.
– Constant rate of fluence F
– Minimal loss of target NT
dN 
 FN T  N
dt
N
 N
F
T

1  e 
 t
Typical Problem
• A 3-g sample of 32S is irradiated
with fast neutrons at 155 cm-2s1. The cross section is 0.200
barn. What is the maximum
activity?
Answer
• The number of target nuclei,
3
NT 
N A  5.64 10 22
32
• The maximum is for large t.
 N  1.75 Bq
N max  F
T