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Saturday, February 9, 2019

An X-ray source is placed on one side of a metal plate, as shown in Fig. 11.1.


Question 5
(a) An X-ray source is placed on one side of a metal plate, as shown in Fig. 11.1.


Fig. 11.1

The intensity of the X-ray beam is measured at points A and B.

State two reasons, other than absorption of X-ray photons in the metal, for the intensity at point A to be different to that at point B. [2]


(b) A specimen of muscle and bone undergoes X-ray examination. Parallel beams of X-rays are incident on the specimen, as shown in Fig. 11.2.


Fig. 11.2

The specimen has a total thickness of 4.0 cm. One section contains a bone of thickness
1.5 cm.

Data for the linear absorption (attenuation) coefficient μ for the bone and for the muscle in the specimen are given in Fig. 11.3.

μ / cm-1
bone                3.0
muscle            0.27

Fig. 11.3

(i) Calculate the ratio
                                 intensity of X-ray beam incident on the specimen    .
intensity of X-ray beam emerging from the specimen

for the beam passing through

1. the 4.0 cm thickness of muscle alone, [2]
2. the bone and the muscle. [2]

(ii) Using your answers in (i), suggest and explain whether an X-ray image of this specimen is likely to have good contrast. [2]


Reference: Past Exam Paper – November 2015 Paper 43 Q11

Solution:
(a)
e.g.
scattering {of the X-ray beam} (in the metal)
the beam is not parallel (not just “A closer than B”)
reflection (from metal)
diffraction in the metal/lattice


(b)
(i)
1.
{I = I0 exp(-μx)
Ratio = I0 / I
I / I0 = exp(-μx)
So, I0 / I = exp(μx)
Ratio = exp(μx) }
ratio = exp(μx)
ratio = exp(0.27 × 4.0)
ratio = 2.94 (2.9)        

2.
{The beam passes through 1.5 cm of bone and (4.0 – 1.5 =) 2.5 cm of mucle. As the beam passes through each of them, it gets attenuated.
ratio = exp(μx)
Through 2.5 cm muscle only: ratio = exp(0.27×2.5) = 1.96
Through 1.5 cm bone only: ratio = exp(3.0×1.5) = 90
To obtain the overall ratio for the beam going through both the muscle and bone, we multiply the 2 ratios.}
ratio = exp(0.27 × 2.5) × exp(3.0 × 1.5)
ratio = 1.96 × 90
ratio = 177 (180)                                            


(ii)
Each ratio gives a measure of the transmission {of the beam}.                   
Since the ratios (in (i)) are very different, the image will have a good contrast.

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