Define half-value thickness.

The half-value thickness in tissue for X-rays of a specific energy is 3.50 mm. Determine the fraction of the incident intensity of X-rays that has been transmitted through tissue of thickness 6.00 mm.

For X-rays of higher energy than those in (b), the half-value thickness is greater than 3.50 mm. State and explain the effect, if any, of this change on your answer in (b).

the distance after which the intensity of the incident X-rays gets reduced to half;

\(\mu  = \left( {\frac{{\ln 2}}{{{x_{\frac{1}{2}}}}} = \frac{{\ln 2}}{{3.50}} = } \right){\text{ }}0.198{\text{ m}}{{\text{m}}^{ - 1}}\);

\(\frac{I}{{{I_0}}} = ({e^{ - \mu x}} = ){\text{ }}{e^{ - 1.98 \times 6.00}}\);

\(\frac{I}{{{I_0}}} = 0.305\);

or

X-rays travel \(\frac{6}{{3.5}}{\text{ }}( = 1.71)\) half thicknesses;

\(I = {I_0}{\left[ {\frac{1}{2}} \right]^{{\text{1.71}}}}\);

\(\frac{I}{{{I_0}}} = 0.350\);

it will be larger;

because a larger half-value thickness implies a smaller attenuation coefficient and so a smaller reduction in intensity;

or

X-rays travel further before their energy is reduced by a given factor;

hence intensity is greater;

Most candidates were able to correctly define half-value thickness.

Many candidates had problems with exponentials.

Many candidates had problems with exponentials in (b) and so were not able to answer part (c) sensibly.