Attenuation of X-rays (AQA A Level Physics): Revision Note

Attenuation of X-rays in matter

• When a collimated beam of X-rays passes through a patient’s body, the X-ray photons are absorbed and scattered

• Different materials absorb X-rays by different amounts

  • For example, bones absorb a large proportion of X-ray photons which is why they appear bright white on an X-ray image

• As the X-rays pass through a material, the intensity of the beam is found to decay exponentially

  • This decrease in intensity is known as attenuation

• The attenuation of X-rays can be calculated using the equation:

$I=I_0{e}^{-\mu x}$

• Where:

  • $I_0$ = the intensity of the incident beam (W m^-2)

  • $I$ = the intensity of the transmitted beam (W m^-2)

  • $\mu$ = the linear absorption coefficient (m^-1)

  • $x$ = distance travelled through the material (m)

• The linear attenuation coefficient $\mu$ is defined as:

The fraction of X-rays removed per unit thickness of the material for a specified energy of the X-rays

• The value of $\mu$ depends on the density of a substance and the energy of the X-ray photons 

  • The greater the density of a material, the greater the value of $\mu$ 

  • For example, bone absorbs a greater proportion of X-rays than soft tissue due to its higher density 

Half-value thickness

• Similar to half-life in radioactivity, a material's ability to absorb X-rays is known as its half-value thickness

• The half-value thickness of a material can be defined as:

The thickness of the material which will reduce the intensity of X-rays to half its original level for a specified energy of the X-rays

• If the half-value thickness is $x=x_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, then intensity has a value of $I=\frac{I_0}{2}$, so substituting this into the attenuation equation gives:

$\frac{I_0}{2}=I_0{e}^{-\mu x_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

$\frac{1}{2}=e^{-\mu x_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

• Taking natural logarithms of both sides gives:

$\ln \frac{1}{2}=-\mu x_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\ln 2=\mu x_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• Hence, the half-value thickness of a substance is given by:

$x_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln 2}{\mu }$

Differential tissue absorption

• The amount of attenuation of a beam of X-rays depends on:

  • the density of the absorbing tissue

  • the energy of the X-ray photons

• The linear attenuation coefficient $\mu$ of an absorber is proportional to the density ρ of the absorbing substance

  • The higher the density of a material, the more X-ray energy that it absorbs

  • This is because the photons interact with more atoms, or a larger mass of atoms, in the same volume

• Therefore, dividing the value of $\mu$ of a material by its density gives a constant value for that particular substance

• This constant is known as the mass attenuation coefficient

${\mu }_m=\frac{\mu }{\rho }$

• Where:

  • ${\mu }_m$ = mass attenuation coefficient (m² kg^–1)

  • $\mu$ = linear attenuation coefficient (m^–1)

  • $\rho$ = density of the absorbing material (kg m^–3)

• The mass attenuation coefficient of a substance describes how easily a beam of X-rays of a certain energy can penetrate it

  • The greater the mass attenuation coefficient, the stronger the absorption of X-rays by the material

  • The lower the mass attenuation coefficient, the greater the penetration of X-rays through the material

• Photons of energies less than 30 keV are absorbed by soft tissue and bone

  • Therefore, these photons are removed from the X-ray beam by placing a suitable metal filter (e.g. lead or tin) in the path of the X-ray beam

• Photons of energies between 30 keV and 100 keV are absorbed more readily by bone than by soft tissue

  • This is because the elements in bone have higher atomic numbers than the elements in soft tissues so bone can absorb photons in this energy range more readily

  • Therefore, these photons are used to distinguish between soft tissue and bone

• Photons of energies greater than 100 keV are absorbed more equally in all types of tissue, including bone

  • This means they produce no distinction between any tissues

  • Therefore, these photons are not used in diagnostic X-ray imaging

Attenuation in different elements

• The graph of mass attenuation coefficient and X-ray photon energy for elements with different values of atomic number Z shows that:

  • elements with lower Z values tend to absorb a lower proportion of X-rays

  • elements with higher Z values tend to absorb a greater proportion of X-rays

• The table below shows the composition of different substances and the effect of atomic number on attenuation