# Presentation: 062960

## Interaction of  Gamma-Rays - General Considerations

**uncharged**
**transfer of energy**
**creation of fast electrons**

## 
photoelectric

Compton

pair production

photodisintegration

**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:**

## Slide 4

## Slide 5

## Photoelectric Effect

## Photoelectric Effect

**but ******** can also be expressed as a work function, a constant term "eW****o****" which varies from material to material   ******** eV****S **** = h******** - eW****o**
**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

**hV****th**** = eW****o**
**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**

**Photoelectric Effect**

## 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:

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

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

**Photoelectric Effect**

## 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"

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

**Photoelectric Effect**

## 
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

**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**

## Compton Scattering

_**3 equations**_**:  momentum conservation (2),**
**  ****	****	****      ****energy conservation**

**γ**

**Compton Scattering**

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

**must treat electron relativistically**

**Compton Scattering**

## while for photons

**energy**

**K =(p****0**** – p****1****)c**

**Compton Scattering**

## momentum:

**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 Scattering**

## wavelength

**the difference in energy E****o**** - E****1**** = K is the kinetic energy of the electron**

**Compton Scattering**

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

**this energy K****max**** is called the Compton edge **
**another form for this equation which uses photon energies instead of wavelengths is:**

**Compton Scattering**

## Compton Scattering

**at high incident energies E****o**** the back scattered photon approaches a constant energy**

**Compton Scattering**

##    0.511 MeV    θ =   90o
   0.255 MeV    θ = 180o 

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

**  ******** ****0.255 MeV    θ = 180o **
**in this limit we find that ********0**** « ******** ≈ ********, so that the energy:**

**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

**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

**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 **
**σ****pp**** ~ Z****2**
**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

**Pair Production**

## Pair Production

**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

**the nucleus carries away an appreciable amount of the incident photon's momentum, but because of its large mass, its recoil kinetic energy, k ≈ p****2****/2m****o****, 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****+****c****2**** + m****-****c****2**** = k****+**** + k****-**** + 2m****o****c****2**
**since the positron and the electron have the same rest mass;**
**	****	****m****o**** = 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

**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:

**where k *****l *****is the propagation vector:**
**2***** *****(k) = 2********/******

**Annihilation**

## Annihilation

**Annihilation**

## Annihilation

**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:

**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:**

**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 

**absorber nucleus captures a ********-ray and in most instances emits a neutron:**
**9****Be( ********,n) ****8****Be**
**important for high energy photons from electron accelerators**
**cross-sections are « total cross-sections **

**Photodisintegration**

## 
total attenuation coefficient  

**total attenuation coefficient  ******

** ****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 

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

**Combined Effects**

## Linear Attenuation and Absorption Coefficients for Photons in Water

## Slide 35

## 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
			thickness
		t = absorber thickness
		I = -ray intensity transmitted
		 = attenuation coefficient

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

**where:**
**	****	****I****0**** = ********-ray intensity at zero absorber**
**	****	****	****thickness**
**	****	****t = absorber thickness**
**	****	****I = ********-ray intensity transmitted**
**	****	******** ****= attenuation coefficient**
**if the absorber thickness is measure in cm, then ******** is called *****linear attenuation coefficient***** (********l****) having dimensions "per cm"**

**Exponential Absorption**

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

**      ****where ******** is the density of the absorber**

**Exponential Absorption**

## Exponential Absorption

**Exponential Absorption**

**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?

**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?**

**Exponential Absorption**

## 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

**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**

**Exponential Absorption**

## 

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

**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***

**Exponential Absorption**

## Linear Attenuation Coefficients

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

**Exponential Absorption**

## if we have a 1.0 MeV  - ray:

**compute the density thickness at 0.1 MeV**

**Exponential Absorption**

## 

at 1.0 MeV

**at 1.0 MeV**

**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**

**Exponential Absorption**

## Slide 47

## Slide 48

## 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

**the mass attenuation coefficient of lead (density = 11.35 g/cm****3****) is 0.0708 cm****2****/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

**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**

**Photon Interactions - Problem**

## 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 

**a. the linear attenuation coefficient ******** is obtained by multiplying the mass attenuation coefficient by the density **
****** ****= (0.0708 cm****2**** g****-1****) ******** (11.35 g cm****-3**** = **_**0.804 cm**__**-1**_**.****	**
**b. if I****o**** photons are incident on the lead  slab and I photons penetrate it without  interacting, then the fraction not interacting is given by:**
**	****	****       ****I/I****o****  = exp[- ******** x]**
**	****   ****	****             ****= exp[(-0.804 cm****-1****)(1 cm)] = 0.448 **

**Photon Interactions - Problem**

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

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

x = ln2/μ = 0.693/0.804 = 0.862 cm

**substituting I/I****o**** = 1/2 in eq. (1) above and rearranging yields:**
**	**
**x = ln****2****/μ = 0.693/0.804 = **_**0.862 cm**_

**Photon Interactions - Problem**

## 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 

**	****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/I****o**** = 1/e **

**Photon Interactions - Problem**