# Net energy transferred per mass Absorbed Dose

Document Sample

```					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.

2.8   1.1
5.4   4.2
10
particle energy     MeV   3.6
2.0
0.1

energy transferred         9.0

energy imparted           2.8   3.3   2.8   1.1
Energy Imparted
• Energy transferred                • Energy imparted
volume from uncharged               transferred
particles                         – Energy in from charged
– Radiant energy out tpo              particles
uncharged particles (not          – Energy out from charged
brem or annihilation)               particles
– Energy change from mass

Etr  (Rin )unc  (Rout )unc  Q
E  ( Rin )unc  ( Rout ) unc

 ( Rin ) chg  ( Rout ) chg   Q
Kerma
• Kerma is the energy transferred
per unit mass.
– Kinetic Energy Released           dEtr
K
per unit MAss                    dm

dRunc
loss per mass due to brem and     Kr 
dm
annihilation.

• Collision kerma subtracts the
KC  K  K r 
– Net energy transferred per                     dm
mass
Absorbed Dose
• Absorbed dose or dose is the
energy imparted per unit mass.
dE
D
• Like kerma dose is based on          dm
mean changes in energy.

• Two units are used.                      J 107 erg
1 Gy      
– 1 gray (Gy) = 1 J / kg               kg     103 g
– 1 rad = 100 erg / g (older)         erg
g
Lethality
• Dose can be compared to
physical effects.

• Lethality refers to the likelihood
that a dose will be fatal.

Lethality %
– Cell death
– Whole body death (see
graph at right)

Dose (cGy)

Federation of American Scientists
Equilibrium
• Measuring the relationship       • Radiation equilibrium
between energy transferred and
imparted requires equilibrium       (Rin )chg  ( Rout )chg
conditions.
( Rin ) unc  ( Rout ) unc

E  Q

• Charged particle equilibrium
– Looser requirement
E  (Rin )unc  (Rout )unc  Q
Exposure
• Exposure is defined by the         Useful Conversion
ionization produced by photons.    • Show that the original roentgen
– Gammas and X-rays                  is equivalent to the modern one.
– Charge per unit mass in air
• Look up constants:
• The original unit is 1 esu / cm3      – Density of air at STP is
of dry air at STP (1928).               0.001293 g / cm-3
– Roentgen (R)                       – 1 esu = 3.34 x 10-10 C
– R = 2.58 x 10-4 C / kg
(1962)                         • 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           • Rate of energy release:
gammas.                                dE
– Activity C                              CE
dt
– Average photon energy E                             CE
•  Fluence rate:F
4r 2
• Dose rate:
• Consider a sphere or radius r.         m
D  F en 
CE men
– Fluence through sphere F                 r 4r 2 r
– Mass energy absorption               1.27 106 (m2s/BqJ)CE
                        Gy/s
men/r = 2.7 x 10-3 m2/kg                        r 2

• Exposure rate:
• Find the dose and exposure rate.         0.5(m 2s/BqJ)CE

X                 R/h
– 1 R = 0.0088 Gy                              r 2
• The effect of radiation on tissue    • In terms of LET
depends on the LET as well as           – LET L (keV / mm in water)
the dose.                               – < 10; WR = 1
– Higher LET is more                   – 10 – 100; WR = 0.32L – 2.2
damaging.
– > 100; WR  300 / L

factor based on particle.            • In terms of particle
– Factor WR or Q                       – e, g, m; WR = 1
– Updated in 1991                      – n; WR = 5 – 20
– p; WR = 5
– a; WR = 20
Restricted Stopping Power
• Secondary electrons ejected        Typical Problem
from atoms are delta rays.         • Estimate a cutoff value for
– Deltas deliver most of the         irradiating 300 Å viruses.
– Energy less effective if it’s   • Most energetic delta range
too high                          should not exceed 300 Å.
– Find range limit 3 x 10-6 cm
• Restricted stopping power          • Range in water
measures the energy lost up to a        – 500 eV is 2 x 10-6 g/cm2
limit D.
– 1 keV is 5 x 10-6 g/cm2
dE      D
     m  QP(Q)dQ               • Estimate cutoff at 700 eV.
dx      Qmin
Equivalent Dose
• The equivalent dose is a         • Natural doses
measure that combines the type      – Cosmics: 0.3 mSv / yr
of radiation and dose.              – Soil: 0.2 mSv / yr
H T  WR D                       – Radon: 2 mSv / yr
– Total natural: 3 mSv / yr
• Unit is Sievert (Sv)
– 1 Gy equivalent               • Environmental hazards
• Older unit is rem                   – Flying at 12 km: 7 mSv / hr
– Roentgen equivalent man          – Chest x-ray: 0.1 mSv
– 1 rad equivalent                 – Mammogram: 1 mSv
– 100 rem = 1 Sv                   – 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      • Nucleus    Qmax/En
neutron energy loss.                   – 1H      1.000
– Assume classical collision          – 2H      0.889
– Set M = 1 and compare for           – 4He     0.640
nuclei
– 9Be     0.360
Qm ax  1 MV 2  1 MV f2
2        2                – 12C     0.284
4mM                           – 16O     0.221
            ( 1 MV 2 )
( M  m) 2 2                    – 56Fe    0.069
4mME n                         – 118Sn   0.033

( M  m) 2                      – 238U    0.017
Neutron Dose Equivalent
• Neutron weighting factors were     Example
variable.                          • Find WR for 2-MeV neutrons.
– WR = 5-20.                          – Average recoil p is 1 MeV
– Stopping power for 1 MeV
• The factors can be determined            p in water is 270 MeV/cm
from assumed elastic scattering.       – Equal to 27 keV/mm

WR  0.32 L  2.2

WR  0.32 (27 )  2.2  6.4
Neutron Threshold
• Inelastic collisions result in a     • Change in rest energy
nuclear reaction.
Q  M1  M 2  M 3  M 4
– Many are endothermic
– Requires extra energy             • Conservation of energy and
momentum
• For example 32S(n,p)32P:                E1  E3  E4  Q
– Eth = 0.957 MeV
p1  p3  p4
– 32P  32S + b-
– Ebmax = 1.71 MeV
– T = 14.3 days                                      M3  M4 
M M M 
E1  Eth  Q          
– Used to detect exposure                            3   4   1
Neutron Activation
• Time dependence of activity     Typical Problem
from neutron capture is based   • A 3-g sample of 32S is irradiated
on exposure and decay.             with fast neutrons at 155 cm-2s-
– Constant rate of fluence F      1. The cross section is 0.200

– Minimal loss of target NT       barn. What is the maximum
activity?
 FNT  N
dt                         • The number of target nuclei,
3
FN T
                         NT      N A  5.64 1022
N

1  e 
 t                32
• The maximum is for large t.
N m ax  FN T  1.75 Bq


```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 18 posted: 3/7/2012 language: English pages: 18