Home / CIE AS & A Level Physics 9702: Topic 24: Medical physics- Unit : 24.2 Production and use of X-rays Study Notes

CIE AS & A Level Physics 9702: Topic 24: Medical physics- Unit : 24.2 Production and use of X-rays Study Notes

A Level PhysicsA Level MathsA Level ChemistryA Level Biology

Concise Summary Notes- All Topics

PRODUCTION & USE OF X-RAYS

Production of X-rays

  •  X-rays are short wavelength, high-frequency part of the electromagnetic spectrum
    •  They have wavelengths in the range $10^{-8}$ to $10^{-13} \mathrm{~m}$
  •  X-rays are produced when fast-moving electrons rapidly decelerate and transfer their kinetic energy into photons of EM radiation

Producing X-rays

  • At the cathode (negative terminal), the electrons are released by thermionic emission
  • The electrons are accelerated towards the anode (positive terminal) at high speed
  • When the electrons bombard the metal target, they lose some of their kinetic energy by transferring it to photons
  • The electrons in the outer shells of the atoms (in the metal target) move into the spaces in the lower energy levels
  • As they move to lower energy levels, the electrons release energy in the form of $\mathbf{X}$-ray photons
  • When an electron is accelerated, it gains energy equal to the electronvolt; this energy can be calculated using:

$
\mathbf{E}_{\max }=\mathbf{e V}
$

  •  This is the maximum energy that an X-ray photon can have
  • Therefore, the maximum X-ray frequency $f_{\max }$, or the minimum wavelength $\lambda_{\min }$, that can be produced is calculated using the equation:

$
\mathrm{E}_{\max }=\mathrm{eV}=\mathrm{hf}_{\max }=\frac{h c}{\lambda_{\min }}
$

Maximum frequency: $\mathrm{f}_{\max }=\frac{e V}{h}$
Minimum wavelength: $\lambda_{\min }=\frac{h c}{e V}$

  •  Where:
    •  $\mathrm{e}=$ charge of an electron (C)
    •  $\mathrm{V}=$ voltage across the anode (V)
    •  $\mathrm{h}=$ Planck’s constant $(\mathrm{J}$ s)
    •  $c=$ speed of light $\left(\mathrm{m} \mathrm{s}^{-1}\right)$

Worked example: Production of X-rays
A typical spectrum of the X-ray radiation produced by electron bombardment of a metal target is shown below.

         

Explain why:
a) A continuous spectrum of wavelengths is produced.
b) The spectrum has a sharp cut-off at short wavelengths.

Answer/Explanation

Part (a)

    • Photons are produced whenever a charged particle is accelerated towards a metal target
    •  The wavelength of the photons depends on the magnitude of the acceleration
    •  The electrons which hit the target have a distribution of accelerations, therefore, a continuous spectrum of wavelengths is observed

Part (b)

    •  The minimum wavelength is equal to

$
\lambda_{\min }=\frac{h c}{E_{\max }}
$

    • This equation shows the maximum energy of the electron corresponds to the minimum wavelength
      •  Therefore, the higher the acceleration, the shorter the wavelength
    •  At short wavelengths, the sharp cut-off occurs as each electron produces a single photon, so, all the electron energy is given up in one collision

Using X-rays in Medical Imaging

  •  X-rays have been highly developed to provide detailed images of soft tissue and even blood vessels
  • When treating patients, the aims are to:
    • Reduce the exposure to radiation as much as possible
    •  Improve the contrast of the image

Reducing Exposure

  • X-rays are ionising, meaning they can cause damage to living tissue and can potentially lead to cancerous mutations
  •  Therefore, healthcare professionals must ensure patients receive the minimum dosage possible
  • In order to do this, aluminium filters are used
    • This is because many wavelengths of X-ray are emitted
    • Longer wavelengths of $\mathrm{X}$-ray are more penetrating, therefore, they are more likely to be absorbed by the body
    •  This means they do not contribute to the image and pose more of a health hazard
    •  The aluminium sheet absorbs these long wavelength X-rays making them safer

Contrast & Sharpness

  • Contrast is defined as:

The difference in degree of blackening between structures

  •  Contrast allows a clear difference between tissues to be seen
  •  Image contrast can be improved by:
    • Using the correct level of X-ray hardness: hard X-rays for bones, soft X-rays for tissue
    •  Using a contrast media

Sharpness is defined as:
                                         How well defined the edges of structures are

  •  Image sharpness can be improved by:
    • Using a narrower X-ray beam
    • Reducing X-ray scattering by using a collimator or lead grid
    •  Smaller pixel size

ATTENUATION OF X-RAYS IN MATTER

Attenuation of X-rays in Matter

  •  Bones absorb X-ray radiation
    •  This is why they appear white on the X-ray photograph
  • When the collimated beam of X-rays passes through the patient’s body, they are absorbed and scattered
  •  The attenuation of X-rays can be calculated using the equation:

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

  • Where:
    •  $10=$ the intensity of the incident beam $\left(\mathrm{W} \mathrm{m} \mathrm{m}^{-2}\right)$
    • $I=$ the intensity of the reflected beam $\left(\mathrm{W} \mathrm{m} \mathrm{m}^{-2}\right)$
    •  $\mu=$ the linear absorption coefficient $\left(\mathrm{m}^{-1}\right)$
    • $x=$ distance travelled through the material $(\mathrm{m})$
  • The attenuation coefficient also depends on the energy of the X-ray photons
  • The intensity of the X-ray decays exponentially
  • The thickness of the material that will reduce the X-ray beam or a particular frequency to half its original value is known as the half thickness

Worked example: Attenuation of X-rays in matter

A student investigates the absorption of X-ray radiation in a model arm. A cross-section of the model arm is shown in the diagram.

                             

Parallel X-ray beams are directed along the line MM and along the line BB. The linear absorption coefficients of the muscle and the bone are $0.20 \mathrm{~cm}^{-1}$ and $12 \mathrm{~cm}^{-1}$ respectively.
Calculate the ratio:
intensity of emergent $X$-ray beam from model intensity of incident $X$-ray beam on model for a parallel X-ray beam directed along the line
a) $\mathrm{MM}$
b) $\mathrm{BB}$
and state whether the X-ray images are sharp, or have good contrast.

Answer/Explanation

Part (a)

Step 1:

Write out the known quantities
Linear absorption coefficient for muscle, $\mu=\mathbf{0 . 2 0} \mathbf{~ c m}^{-1}$
Distance travelled through the muscle, $x=\mathbf{8 . 0} \mathbf{~ c m}$

Step 2:
Write out the equation for attenuation and rearrange

$
I=I_0 e^{-\mu x}
$
$
\frac{\text { intensity of emergent } X-\text { ray beam from model }}{\text { intensity of incident } X-\text { ray beam on model }}=\frac{I}{I_0}=e^{-\mu \mathrm{x}}
$

Step 3:
Substitute in values and calculate the ratio

$
\frac{I}{I_0}=\mathrm{e}^{-(0.20 \times 8)}=0.2
$

Part (b)
Step 1:

Write out the known quantities
Linear absorption coefficient for bone, $\mu_b=12 \mathbf{c m}^{-1}$
Distance travelled through the muscle, $x_m=\mathbf{4 . 0} \mathbf{~ c m}$
Distance travelled through the bone, $x_b=\mathbf{4 . 0} \mathbf{~ c m}$

Step 2:
Write out the equation for attenuation for two media and rearrange

$
\frac{I}{I_0}=e^{-\mu_m x_m} \times e^{-\mu_b x_b}
$

Step 3:
Substitute in values and calculate the ratio

$
\frac{I}{I_0}=\mathrm{e}^{-(0.20 \times 4)} \times \mathrm{e}^{-(12 \times 4)}=6.4 \times 10^{-22} \approx 0
$

Step 4:

Write a concluding statement
Each ratio gives a measure of the amount of transmission of the beam
A good contrast is when:

  • There is a large difference between the intensities
  •  The ratio is much less than 1.0

Therefore, both images have a good contrast

COMPUTED TOMOGRAPHY SCANNING

Computed Tomography Scanning

  •  A simple X-ray image can provide useful, but limited, information about internal structures in a 2D image
  •  When a more comprehensive image is needed, a computerised axial tomography (CAT or CT) scan is used
  • The main features of the operation of a CT scan are as follows:
    • An X-ray tube rotates around the stationary patient
    •  A CT scanner takes X-ray images of the same slice, at many different angles
    •  This process is repeated, then images of successive slices are combined together
    •  A computer pieces the images together to build a 3D image
    •  This 3D image can be rotated and viewed from different angles

Advantages & Disadvantages of CAT Scans

  • Advantages:
    • Produces much more detailed images
    • Can distinguish between tissues with similar attenuation coefficients
    • Produces a 3D image of the body by combining the images at each direction
  • Disadvantages:
    •  The patient receives a much higher dose than a normal X-ray
    • Possible side effects from the contrast media

Worked example
 An X-ray image is taken of the skull of a patient. Another patient has a CT scan of his head. By reference to the formation of the image in each case, suggest why the exposure to radiation differs between the two imaging techniques.

Answer/Explanation

X-ray
The simple X-ray image involves taking a single exposure
This produces a single $2 D$ image
CT scan
The $\mathrm{CT}$ scan requires taking several exposures of a slice from many different angles
This is then repeated for different slices before being combined together to build a 3D image
This involves taking a much greater exposure than the simple X-ray

Scroll to Top