Unit 10 : Worked Problems

Unit 10 : Worked Problems#

Worked Problem : Medical Radioisotopes#

A \( ^{99}Mo \) generator for \( ^{99m}Tc \) initially has an activity of 20 GBq and contains no \( ^{99m}Tc \). Technetium is extracted from the generator 24 hours after production, and again 24 hours after that. Assuming that the extraction is 100% efficient (an exaggeration, but not too much of one—the specification says >90%), what are the activities of the two technetium extractions?

The half-life of \( ^{99}Mo \) is 66.0 hours, and that of \( ^{99m}Tc \) is 6.0 hours. The branching ratio of the decay from \( ^{99}Mo \) to \( ^{99m}Tc \) is 86.8% (the remaining 13.2% of the time it decays directly to \( ^{99}Tc \)).

Solution

For the \( ^{99}Mo \), we know that:

\[N_{Mo}(t) = N_{Mo,0} e^{-\lambda_{Mo} t},\]

where \( \lambda = \ln(2)/t_{1/2} \). The activity is therefore:

\[A_{Mo}(t) = -\frac{dN_{Mo}}{dt} = \lambda_{Mo} N_{Mo,0} e^{-\lambda_{Mo} t} = \lambda_{Mo} N_{Mo}(t).\]

For the \( ^{99m}Tc \), we must account for the atoms gained by the decay of \( ^{99}Mo \), and those lost by the decay of \( ^{99m}Tc \). Therefore:

\[\frac{dN_{Tc}}{dt} = f \lambda_{Mo} N_{Mo} - \lambda_{Tc} N_{Tc},\]

where \( f = 0.868 \) is the branching ratio to \( ^{99m}Tc \). Rearranging:

\[\frac{dN_{Tc}}{dt} + \lambda_{Tc} N_{Tc} = f \lambda_{Mo} N_{Mo}.\]

This is a standard form, which can be solved using an integrating factor \( F(t) \):

\[F(t) = e^{\lambda_{Tc} t}.\]

After solving and substituting, the activity of \( ^{99m}Tc \) is given by:

\[A_{Tc}(t) = \frac{f \lambda_{Tc} A_{Mo,0}}{\lambda_{Tc} - \lambda_{Mo}} \left( e^{-\lambda_{Mo} t} - e^{-\lambda_{Tc} t} \right).\]

Using the given data:

  • \( \lambda_{Tc} = \ln(2) / 6.0 = 0.1155 \, \text{hr}^{-1} \),

  • \( \lambda_{Mo} = \ln(2) / 66.0 = 0.0105 \, \text{hr}^{-1} \).

For \( t = 24 \, \text{hr} \):

\[A_{Tc}(24 \, \text{hr}) = \frac{0.868 \times 0.1155 \times 20}{0.1155 - 0.0105} \left( e^{-0.0105 \times 24} - e^{-0.1155 \times 24} \right) = 14 \, \text{GBq}.\]

For \( t = 48 \, \text{hr} \), reset the clock at the previous extraction:

\[A_{Mo} = 20 \times e^{-0.0105 \times 24} = 15.5 \, \text{GBq}.\]

Replacing \( 20 \) with \( 15.5 \) in the equation gives:

\[A_{Tc}(48 \, \text{hr}) = 11 \, \text{GBq}.\]

Worked Problem : Gamma Cameras#

The collimator in a gamma camera typically consists of a lead sheet of thickness \( l \), with holes of diameter \( d \) separated by “septa” of thickness \( t \):

  • a. What is the resolution of the camera, expressed in terms of \( l \), \( d \), and \( z \), where \( z \) is the distance between the object and the front of the collimator?

  • b. A particular collimator has \( d = 1.40 \, \text{mm} \), \( t = 0.18 \, \text{mm} \), and \( l = 24.7 \, \text{mm} \). If used to image a tumor at \( z = 40 \, \text{cm} \), calculate its resolution and efficiency.

Solution

The maximum angle accepted by the collimator is:

\[\theta \approx \frac{d}{l}.\]

At a distance \( z \), the resolution at the object is:

\[\Delta x = \theta (l + z) = \frac{d (l + z)}{l}.\]

For the given dimensions:

\[\Delta x = 30.9 \, \text{mm}.\]

Efficiency is given by:

\[\eta \approx \frac{d^4}{12 l^2 (d + t)^2}.\]

Substituting:

\[\eta = 2.1 \times 10^{-4}.\]