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.