Indirectly ionising radiation

Question 1

What is the effective atomic number?

Answer 1

The atomic number of an element which would interact in the same way as the composite material

Question 2

What is the equation of effective atomic number?

Answer 2

Zbar = (sum(alphan.Zn^2.94))^(1/2.94)

Question 3

Describe Compton scatter

Answer 3

The incident photon has energy»binding energy so it interacts with a sea of free electrons. It creates scattered photons, bremsstrahlung, and delta rays through elastic collisions.
The incoming photon interacts with an electron, depositing some of its energy with the electron and both are scattered.

Question 4

What is the energy of the photon after Compton scatter?

Answer 4

hv’ = hv0(1/(1+alpha(1-cos(phi)))

Question 5

What is the energy transferred in Compton scatter?

Answer 5

Etr = hv0((alpha(1+cos(phi)))/(1+alpha(1-cos(phi))))

Question 6

What is alpha in Compton scatter?

Answer 6

alpha = hv0/m0c^2

Question 7

What is cos(theta), the angle of electron scatter, given by in Compton scatter?

Answer 7

cos(theta) = (1+alpha)tan(phi/2)

Question 8

In Compton scatter, at low and high energiess which particle gets the majority of the energy?

Answer 8

low = photon
high = electron

Question 9

What does Compton scatter depend on?

Answer 9

The number of free electrons = Na Z/A

Question 10

At high energies in Compton scatter, what direction are the photon and electron scattered in?

Answer 10

Forward direction

Question 11

What is the mass energy transfer coefficient?

Answer 11

u/density, linear attenuation coefficient with only a dependence on the composition of the medium and not the density.

Question 12

What is KERMA?

Answer 12

KERMA is the initial kinetic energy of all charged particles released as a result of interactions in mass dm. It includes bremsstrahlung, energy deposited outside the region of interest by particles created in it. Only defined for neutral particles

Question 13

What is the equation for KERMA?

Answer 13

KERMA = psi.(utr/density).e^(-up.x)

Question 14

What is the absorbed dose?

Answer 14

The total energy imparted by ionising radiation to mass dm. Considers energy deposited in dm by radiation produced outside the region dm, but not energy deposited outside dm by charged particles created inside dm.

Question 15

What is the equation for absorbed dose?

Answer 15

D = psi.(utr/density).e^(-up.x) dx

Question 16

At depth dx what is the fluence decreased by a factor of?

Answer 16

e^(-ue.dx), ue = electron linear attenuation coefficient

Question 17

What gives the fraction of energy re-emitted as bremsstrahlung?

Answer 17

g

Question 18

How does g affect the mass attenuation coefficient equation?

Answer 18

uen/density = utr/density (1-g)

Question 19

What are the requirements for measuring absorbed dose?

Answer 19

In an infinite medium

Beyond Dmax and build-up region

Question 20

What is the equation for the Bragg Cavity Theory?

Answer 20

Dmed = JW/e . Dgas
Dmed = dQ/dm . W/e . Dgas

Question 21

What is the Fano theorem?

Answer 21

In a medium of given composition, exposed to a given flux of primary radiation, the flux of secondary radiation is also uniform and independent of density of the medium as well as density variations from point to point.

Question 22

What are the requirements of the Fano theorem?

Answer 22

Interaction coefficients are independent of density
Cavity in finite medium and is homogeneous
Uniform irradiation
Thickness of chamber wall»range of secondary electrons
Chamber wall needs the same electron density as medium.

Question 23

What is the equation for dose from the Fano theorem?

Answer 23

Dmed = J (Wair/e) Smed,gas (uen/density)med,gas

Question 24

What is the linear attenuation coefficient?

Answer 24

A measure of non-attenauted primary beam after it passes through an absorber, u
It is a measure of the energy removed from a beam

Question 25

How is the linear attenuation coefficient derived?

Answer 25

Expose an absorber of thickness dx to primary beam. Place a detector at point P such that it only measures unattenuated primary beam. At P, dN is proportional to N0.dx. So dN=-uN0dx.

Question 26

What is the cross section of interaction?

Answer 26

sigma = probability of intersection/unit particle fluence

Question 27

How can linear attenuation coefficient be derived from the cross section?

Answer 27

dN=B.sigma.N0.dx, so u = B.sigma where B is the number of interaction centres per unit volume and B=Na.density/Molar mass.

Question 28

What is the energy transfer coefficient?

Answer 28

The energy removed from the photon beam which becomes kinetic energy for electrons in the absorber. utr = Etr/hv0 .u, where Etr is the average energy transferred into KE per interaction and Etr/hv0 is the average fraction of energy transferred to KE.

Question 29

How is the energy absorption coefficient defined?

Answer 29

The energy deposited in medium, uen = utr(1-g)

Question 30

How is the mass attenuation coefficient defined?

Answer 30

u/density, so it is u which only depends on the composition of the medium and not the density.

Question 31

What is the process of the Photo-Electric Effect?

Answer 31

An incoming photon has enough energy to ionise an inner shell electron. A higher shell electron then drops to the lower energy state, emitting a characteristic x-ray.

Question 32

How are Auger electrons produced?

Answer 32

The characteristic x-ray which is produced in the photo-electric effect ionises another electron in the atom.

Question 33

What is the mass attenuation coefficient for the photo-electric effect proportional to?

Answer 33

tau/density prop 1/hv0^3 prop Z^3

Question 34

What is the mass energy transfer coefficient for the photo-electric effect?

Answer 34

tauk/density = tau/density . (hv0-hvc)/hv0 = tau/density . (1-hvc/hv0) where hvc is the average energy transferred to KE.

Question 35

What is the equation for exposure in a gas?

Answer 35

X = dQ/dm

Question 36

Describe how a free air ionisation chamber works

Answer 36

A beam from a source is aimed through a diaphragm of known area into a region of free air. A high polarising voltage is applied. Ions are produced and are attracted to the electrodes, losing all of their energy before they reach there. Guard electrodes ensure that only ions created in the measuring volume are collected. The distance between the plates must be greater than double the range of the secondary electrons.

Question 37

Derive KERMA from exposure in a free air ionisation chamber.

Answer 37

X = Q/A.L.density = psi(uen/density)(e/W) uen/density = utr/density . (1-g) So, KERMA = X/1-g (W/e)

Question 38

What requirement allows KERMA = X(W/e) for superficial energies?

Answer 38

g<<1

Question 39

How is dose to medium found from dose to air?

Answer 39

Dmed = KERMA. (uen/density)med/(uen/density)air

Question 40

What are the requirements of making an ionisation chamber at superficial energies?

Answer 40

Need electronic equilibrium betwen air on the inside and graphite All the electrons produced in air are stopped by the graphite Photo-electric effect dominates so air equivalence between the layers is needed. A measuring electrode and insulator need to be introduced.