Interactions with matter

Question 1

On a logarithmic graph, how should transmission vary with depth? Why, in reality, does it not look like that? (3 marks)

Answer 1

• Should be linear due to exp relationship, but isn’t due to x-ray beam containing a spectrum of energies.
• Line initially decreases more steeply due to preferential absorption of lower energies at shallow depth.
• Line the tends back close to the linear relationship as the beam harderns.

Question 2

When examining the transmission of x-rays, why is the mass attenuation coefficient used? (2 marks)

Answer 2

mass attenuation coefficient = (u/p)

This is a more convenient parameter as it takes into account the density of the material which also affects the transmission.

Question 3

What is the difference between attenuation and absorption? (2 marks)

Answer 3

• Attenuation = energy removed from the beam, could be due to absorption or scatter.
• Absorption = energy transferred from the beam to the material

Question 4

Explain the terms exposure and absorbed dose? (6 marks)

Answer 4

Exposure = X = dQ/dm = charge liberated per unit mass measured in C/kg, this is what is detected in an ionisation chamber. Only applies to air.

Absorbed dose = D = dE/dm = energy deposited per unit mass (J/kg)

Question 5

How can we work out the dose to air from exposure? (3 marks)

Answer 5

X = dQ/dm

Number of electrons = X/e

Absorbed dose = W*(X/e) where W is energy required to liberate an electron 33 eV

Question 6

What is air kerma? (3 marks)

Answer 6

ke released per unit mass = energy liberated in a mass, dm, and can be deposited either in or outside the mass.

It is equal to absorbed dose when the electrons ranges are much shorter than the dimensions of the detector (there is electronic equilibrium) and there is negligible bremstrahlung.