Interaction of Gamma-Rays - General Considerations uncharged transfer of energy creation of fast electrons

Interaction of Gamma-Rays - General Considerations

uncharged

transfer of energy

creation of fast electrons

photoelectric

Compton

pair production

photodisintegration

Interaction of Gamma-Rays - Types

Photoelectric Effect

Einstein, 1905, as part of his Nobel prize winning paper on the photon theory of light - a prediction which was later verified experimentally in detail

photon absorbed by atom, goes into excited state and ejects an electron with excess kinetic energy:

smax eVK

electron ejected of energy binding

photon

hEelectron k



- h = eV

C 10 1.6 = e

potentialor voltage stopping = V

s

19

s

but can also be expressed as a work function, a constant term "eWo" which varies from material to material eVS = h - eWo

where: h (the slope) remains constant for all material being equal to Planck's constant; 6.6 10-34 J-s

there exists a threshold frequency:

hVth = eWo

below this threshold photons will not have sufficient energy to release even the least tightly bound electrons

cross-section drops abruptly at the K-edge because below there is insufficient energy to overcome the binding and the K-shell electrons no longer participate

the L-edge is really 3 energies due to fine level splitting


K-shell binding energies vary from 13.6 eV (H); 7.11 keV (Fe); 88 keV (Pb); 116 keV (U); if hν < Ek, only L and higher shell electrons can take part

photoelectric effect favored by low-energy photons and high Z absorbers; cross-section varies as:

)(hZ

3

4

PE

strong Z-dependence makes Pb a good x-ray absorber, usually followed by Cu and Al in a layered shield


as vacancy left by the photoelectron is filled by an electron from an outer shell, either fluorescence x-rays or Auger electrons may be emitted

the probability of x-ray emission is given by the "fluorescent yield"

90Zfor 0.965

8Zfor 0.005

created vacancy

emitted photonsyield tfluorescen


wave interpretation predicts that when electromagnetic radiation is scattered from a charged particle, the scattered radiation will have the same frequency as the incident radiation in all directions

the scattering of electromagnetic radiation from a charged particle is viewed as a perfectly elastic billiard ball

Compton Scattering

4 unknowns: E1, , K,

3 equations: momentum conservation (2), energy conservation

θ

E1 P1

K, P e-

e-

Eo Po

before after

γ

Compton Scattering

K = (m - mo)c2 difference between the total energy E of the moving particle and the rest energy Eo (at rest)

2o

2o

2o

)c m - m ( = K

cm - mc = E - E = K

must treat electron relativistically

42o

2222o

2 c m + c p = ) cm + K ( = E

Compton Scattering

while for photons

cE = p

cE = p 1

1o

o

cm+ K + E = cm + E 2o1

2oo

energy

K =(p0 – p1)c

Compton Scattering

momentum:

2

10

2

1

2

0 pcospp2pp

p0 – p1 cos = p cos

square and add

p1 sin = p sin

Compton Scattering

substitute these expressions for K1, p2 into relativistic electron energy expression to get (after manipulation):

where Δ is the shift that the scattered photon undergoes

the wavelength is usually measured in multiples of "Compton units", the ratio of:

Compton the to Ehc wavelength the

)cos1(cm

h

0

01

Compton Scattering

wavelength

1 =

)MeV ( E

.511 =

E

cm =

cm

hcE

hc

=

m 10 2.43 = cm

hc =

2e

2e

o

122

eC

) cos - 1 ( + 1

) cos - 1 ( E = E - E = K o1o

the difference in energy Eo - E1 = K is the kinetic energy of the electron

Compton Scattering

the min. electron energy corresponds to min. scattered photon energy (θ = 180º) so that

1« for E + 1

E = 2 + 1

2 E = K o

21o

omax

this energy Kmax is called the Compton edge

another form for this equation which uses photon energies instead of wavelengths is:

Compton Scattering

) cos -1 ( ) cm /E ( + 1

E = E :or

cmE =

) cos - 1 ( + 1

1 =

E

E

2o

2o

o

o

1

at high incident energies Eo the back scattered photon approaches a constant energy

1cm

Efor

cos1cm

)cos1(cm

E1

E)(E

2

0

0

2

0

2

o

0

01

Compton Scattering

0.511 MeV θ = 90º

0.255 MeV θ = 180º

in this limit we find that 0 « ≈ , so that the energy:

0

1

1 oft independen is hchc

E

Compton Scattering

In a Compton experiment an electron attains kinetic energy of 0.100 MeV when an x-ray of energy 0.500 MeV strikes it. Determine the wavelength of the scattered photon if the electron is initially at rest

A 13 = MeV0.400

A MeV10 12.4 =

E

hc =

MeV0.400 = E

MeV0.100 + E = MeV0.500

) (k E = E

E = E

3-

e

finalinitial

210.

Compton Scattering Problem

in the process of pair production the energy carried by a photon is completely converted into matter, resulting in the creation of an electron-positron pair

σpp ~ Z2

since the charge of the system was initially zero, 2 oppositely charged particles must be produced in order to conserve charge

Pair Production

in order to produce a pair, the incident photon must have an energy of the pair; any excess energy of the photon appears as kinetic energy of the particles

E = h

heavy nucleus Mo

before pair production:

Pair Production

after pair production:

E+ = m+c2

E- = m-c2

E = Moc2 + k

p+

p-

p

Pair Production

pair production cannot occur in empty space

the nucleus carries away an appreciable amount of the incident photon's momentum, but because of its large mass, its recoil kinetic energy, k ≈ p2/2mo, is usually negligible compared to kinetic energies of the electron-positron pair

thus, energy (but not momentum) conservation may be applied with the heavy nucleus ignored, yielding:

h = m+c2 + m-c2 = k+ + k- + 2moc2

since the positron and the electron have the same rest mass;

mo = 9.11 x 10-31 kg

Pair Production

the inverse of pair production can also occur

in pair annihilation a positron-electron pair is annihilated, resulting in the creation of 2 (or more) photons as shown

at least 2 photons must be produced in order to conserve energy and momentum

Annihilation

in contrast to pair production, pair annihilation can take place in empty space and both energy and momentum principles are applicable, so that:

21finalinitial

212

ofinalinitial

k 2

h + k

2

h = vm + vm or P = P

hv + hv = k + k + c2mor E = E

where k l is the propagation vector:

2 (k) = 2/

Annihilation

before pair annihilation

E+ = m+c2 = moc2 + k+

E- = m-c2 = moc2 + k-

Annihilation

after pair annihilation

h1

h2

Annihilation

problem:

how many positrons can a 200 MeV photon produce?

the energy needed to create an electron-positron pair at rest is twice the rest energy of an electron, or 1.022 MeV; therefore maximum number of positrons equals:

positrons 195 pair

positron1

MeV 1.022pair 1

MeV) 200(

Annihilation

absorber nucleus captures a -ray and in most instances emits a neutron:

9Be( ,n) 8Be

important for high energy photons from electron accelerators

cross-sections are « total cross-sections

Photodisintegration

total attenuation coefficient

ppcsPEt + + =

in computing shielding design the above equation is used

this is the fraction of the energy in a beam that is removed per unit distance of absorber

Combined Effects

the fraction of the beam's energy that is deposited in the absorber considers only the energy transferred by the photoelectron, Compton electron, and the electron pair

energy carried away by the scattered photon by Compton and by annihilation is not included

1.021.02 - hf

+ + = ppcePEe

Combined Effects

Linear Attenuation and Absorption Coefficients for Photons in Water

Exponential Absorption

due to the different interaction of -rays with matter, the attenuation is different than with α or particles

intensity of a beam of photons will be reduced as it passes through material because they will be removed or scattered by some combination of photoelectric effect, Compton scattering and pair production

reduction obeys the exponential attenuation law:

I = Ioe-t

where:I0 = -ray intensity at zero absorber

thicknesst = absorber thicknessI = -ray intensity transmitted = attenuation coefficient

if the absorber thickness is measure in cm, then is called linear attenuation coefficient (l) having dimensions "per cm"


if the absorber thickness "t" is measured in g/cm2, then (m) is called mass attenuation coefficient (m) having dimensions cm2/g

32m

1l /cmg /gcm = cm

where is the density of the absorber


through passes 70.5%

0.705 =

e = e = I

I

mm) (5.0 ) (0.07mm-x-

o

1


what percent of an incident x-ray passes through a 5 mm material whose linear absorption is 0.07 mm-1?

a monochromatic beam of photons is incident on an absorbing material

if the intensity is reduced by a factor of 2 by 8 mm of material, what is the absorption coefficient?

1

mm8o

o

mm 0.0866 = for solve

e I = 2I


Half-Value Thickness (HVT)

thickness of absorber which reduces the intensity of a photon beam to 1/2 its incident value

find HVT of aluminum if = 0.070 mm-1

mm 9.9 =x

e = 2

1

x ) mm (0.07- 1


Atomic Attenuation Coefficient a

fraction of an incident -ray beam that is attenuated by a single atom, or the probability that an absorber atom will interact with one of the photons

where a is referred to as a cross-section and has the units barns

section-cross cmacroscopi =

) ( section-cross cmicroscopi =

cm 10 = barn 1

l

a

224


Linear Attenuation Coefficients

what is the thickness of Al and Pb to transmit 10% of a 0.1 MeV -ray?

) less times 137 ( Pbfor cm 0.0385 =t and

) Al( cm 5.3 =t t 0.435 = 10 ln

e 10

1 = e =

I

I

Pbfor 59.7 =

for Al cm 0.435 =

cmt ) cm -(4.35t-

o

1

11

1

(0.435 cm-1)


if we have a 1.0 MeV - ray:

) less times4.67 ( ) (Pb cm 2.97 =t

) (Al cm 13.86 =t

compute the density thickness at 0.1 MeV

23d

23d

g/cm 0.435 = cm 0.0385 g/cm 11.34 = Pb t

g/cm 14.3 = cm 5.3 g/cm 2.7 = Alt


at 1.0 MeV

2d

2d

g/cm 33.6 = Pb t

g/cm 37.4 = Alt

this shows that Pb is only slightly better on a mass basis than Al

however for low energy photons Pb is much better

in general, for energies between 0.8 5 MeV almost all materials, on a mass basis, have approximately the same -ray attenuating properties


1-MeV photons are normally incident on a 1-cm lead slab

the mass attenuation coefficient of lead (density = 11.35 g/cm3) is 0.0708 cm2/g and the atomic weight is 207.2

Photon Interactions - Problem

calculate the linear attenuation coefficient

what fraction of 1-MeV photons interact in a 1-cm lead slab?

what thickness of lead is required for half the incident photons to interact?

calculate the mean free path


Solution

a. the linear attenuation coefficient is obtained by multiplying the mass attenuation coefficient by the density

= (0.0708 cm2 g-1) (11.35 g cm-3 = 0.804 cm-1.

b. if Io photons are incident on the lead slab and I photons penetrate it without interacting, then the fraction not interacting is given by:

I/Io = exp[- x]

= exp[(-0.804 cm-1)(1 cm)] = 0.448


the fraction of photons interacting is then: 1 - 0.448 = 0.552

c. substituting I/Io = 1/2 in eq. (1) above and rearranging yields:

x = ln2/μ = 0.693/0.804 = 0.862 cm


d. the mean free path (MFP) is the average distance that an incident photon travels before interaction and is the reciprocal of the attenuation coefficient:

MFP = 1/μ = 1/0.804 = 1.24 cm

this is also numerically equal to the "relaxation length“

the relaxation length is the shield thickness needed to attenuate a narrow beam of monoenergetic photons to 1/e (= 0.368) of its original intensity and can be derived by substituting I/Io = 1/e


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Interaction of Gamma-Rays - General Considerations uncharged transfer of energy creation of fast electrons