(a) Distinguish between an X-ray image of a body structure and a CT scan. X-ray image: CT scan:
Q#7 (Past Exam Paper – June 2014 Paper 42 Q11)
(a) Distinguish between an X-ray image of a body structure and a CT scan.
X-ray image:
CT scan:
[5]
(b) Data for the linear absorption (attenuation) coefficient μ of X-ray radiation of energy 80 keV are given in Fig. 11.1.
metal μ / mm-1
aluminium 0.46
copper 0.69
Fig. 11.1
A parallel X-ray beam is incident on a copper filter, as shown in Fig. 11.2.
Fig. 11.2
The intensity of the incident beam is I0.
(i) Calculate the thickness of copper required to reduce the intensity of the emergent beam to 0.25 I0. [2]
(ii) An aluminium filter of thickness 2.4 mm is now placed in the X-ray beam, together with the copper filter in (i).
Calculate the fraction of the incident intensity that emerges after passing through the two
filters. [2]
(iii) Express your answer in (ii) as a gain in decibels (dB). [3]
Solution:
(a)
An X-ray image is a flat / shadow / 2D image regardless of the depth of the object.
The image from a CT scan is built up from (many) images at different angles. The image is three-dimensional and the image can be rotated (and viewed at different angles).
(b)
(i)
I = I0 exp(-μx)
0.25I0 = I0 exp(-0.69x)
{Since the beam is incident on the copper filter, we need to consider the linear attenuation coefficient of copper.
0.25 = exp (-0.69x)
ln 0.25 = - 0.69 x
Thickness x = ln 0.25 / -0.69}
Thickness, x = 2.0 mm
(ii)
{For a beam incident on a medium of thickness x1 and linear attenuation coefficient μ1, the transmitted intensity I is given by
I = I0 exp(-μ1x1)
Fraction of incident intensity that emerges: I / I0 = exp(-μ1x1)
For a beam incident on 2 media, one of thickness x1 and linear attenuation coefficient μ1, and the other of thickness x2 and linear attenuation coefficient μ2, the transmitted intensity I is given by
I = I0 exp(-μ1x1) × exp(-μ2x2)
Fraction of incident intensity that emerges: I / I0 = exp(-μ1x1) × exp(-μ2x2)
From the previous part,
For copper, I = 0.25I0 giving I / I0 = exp(-μ1x1) = 0.25}
For aluminium, I / I0 = exp (-0.46×2.4) = 0.33
{I / I0 = exp(-μ1x1) × exp(-μ2x2)}
Fraction of incident intensity = 0.33 × 0.25 = 0.083
(iii)
{When expressing an answer in dB, we need to take the log of the fraction.}
Gain (in dB) = 10 lg(I/I0)
Gain (in dB) = 10 lg(0.083)
Gain (in dB) = (-) 10.8 dB
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