Radioactive source holders - Dose rate through an aperture

Dose rate through an aperture

Dr Chris Robbins, Grallator

A source is sealed at the centre a solid spherical shield of radius \(r = 5 \space cm\). The shield material is such that the shield absorbs all energy of activity from the source. A hole of diameter \( a = 5 \space mm\) is drilled from the surface to the source as shown below.

Making the assumption that the source is small compared with the hole and so can be approximated as a point source, and assuming the energy released can only be through the aperture (i.e. paths that pass through a thin section of shield near the aperture are ignored), radiative energy will only be detected in a thin cone of subtended angle \(2 \theta\).

Calculating the power density

The energy released per second will radiate outwards and, in the absence of a shield be spread over the surface of a sphere. The power per unit area at a distance \(R\) from the source is given by the inverse square law: \[ P_a = \frac{P}{4 \pi R^2} \] where \(P_a\) is the power per unit area and \(P\) is the total power of the source. This assumes no absorption and energy is spread over the full solid angle of \(4\pi\) steradians. However, as shown in the diagram, only a small fraction of the solid angle is available for energy to escape. The solid angle available is that of a cone subtending the plane angle of \(2 \theta\), which is given by \[ \Omega=4 \pi \space \mathrm{sin}^2 \left( \frac{\theta}{2} \right) \] The fraction of the total energy radiated through the hole, \(f\) is given by the ratio \(\Omega:\Omega_{sp} \) where \(\Omega_{sp} =4\pi \), the solid angle of a sphere,i.e. \[ f=\frac{4 \pi \space \mathrm{sin}^2 \left( \frac{\theta}{2} \right)} {4\pi} = \mathrm{sin}^2 \left( \frac{\theta}{2} \right) \] The total power escaping the shield is then \(fP\). The power density a distance \(R\) along the dotted line above is still given by \(P_a = \frac{P}{4 \pi R^2}\) however, the area over which this applies is limited to that bounded by the rays depicted, i.e. and area \(A=\Omega R^2\).

Putting in numbers

The diagram gives \(r=5 \space cm\) and \( a = 5 \space mm = 0.5 \space cm\). The angle \(\theta\) is related to \(r\) and \(a\) by \[ \mathrm{sin} \left( \theta \right)=\frac{\frac{a}{2}}{r}=\frac{0.25}{5}=0.05 \] This is a small angle, and using the small angle approximation \( \mathrm{sin} \left( \theta \right) \approx \theta \), when \(\theta\) is in radians \[ \mathrm{sin} \left( 0.05 \right) \approx \theta \approx 0.05 \] Similarly, using the small angle approximation gives \[ f= \mathrm{sin}^2 \left( \frac{\theta}{2} \right) = \mathrm{sin}^2 \left( \frac{0.05}{2} \right) \approx {0.025}^2 = 6.25 \times 10^{-4} \] The above states that only about \(\frac{1}{1600}\) of the total radiative energy escapes the shielding sphere.

The solid angle \(\Omega\) is evaluated as \[ \Omega=4 \pi \space \mathrm{sin}^2 \left( \frac{\theta}{2} \right)=4\pi f \approx 4\pi \times 6.25 \times 10^{-4} = \frac{\pi}{400} \space \mathrm{steradians} \] so that the area irradiated at a distance R is given by \[ A=\Omega R^2=\frac{\pi R^2}{400} \] and recall the power density at the same distance is \[ P_a = \frac{P}{4 \pi R^2} \] Notice that the total power over the irradiated area, \(P_I\) is given by \[ P_I = P_aA = \frac{P}{4 \pi R^2} \times \frac{\pi R^2}{400} = \frac{P}{1600} \] i.e. a constant value that is the total power escaping the shield as discovered above.

Plotting the values of \(A\), \(P_a\) and \(P_I\) for \(R\) ranging from 5 cm to 200 cm and \(P=1\) gives the following.

Tabulating several values (note, only shown to three decimal places/significant figures - rounding may occur if replicated by hand):

Taking the irradiated are above, it is possible to construct the diameter of the area that will be irradiated, \(\Phi\) as a funcion of distance, given below.

At a distance of 200 cm = 2 m, a circle of diameter 20 cm centred on the dotted line in the diagram will be irradiated. Outside of this are there will be no irradiation. Such an area would, for example, cover a human chest but would not extend to the head, arms or legs.

Putting in more numbers

Assume now that the source is \(^{137}Cs\) with an activity of 35 MBq. To simplify matters we'll ignore the \( \beta \) decay component and only look at the 661.7 keV \(\gamma\) and calculate dose rate.

Total \(\gamma\) power from decay is \( P =\mathrm{Activity} \times \mathrm{Energy\,per\,decay} \), i.e., \[P= 35 \times 10^6 \times 661.7 \times 10^3 \times 1.602 \times 10^{-19}= 3.710 \times 10^{-6} \,\,\,\, W \,\, \mathrm{or} \,\, Js^{-1}\]

The power density at a distance \(R\) cm along the dotted line is given by \[ P_a = \frac{P}{4 \pi R^2} = \frac{3.710 \times 10^{-6}}{4 \pi R^2} \,\,Js^{-1}cm^2 \] The dose rate is given (approximately) by \(P_a \times \left(\mu_{en}/\rho \right) \), where in these units \(\left(\mu_{en}/\rho \right) \) is in \( cm^2g^{-1} \). For air, \(\left(\mu_{en}/\rho \right) = 2.929 \times 10^{-2}\,\,cm^2g^{-1} \). This gives an expression for the absorbed energy rate as \[ D_{Js^{-1}g^{-1}} = \frac{3.710 \times 10^{-6}}{4 \pi R^2} \times 2.929 \times 10^{-2} = \frac{1.087 \times 10^{-7}}{4 \pi R^2} \,\,Js^{-1}g^{-1} \] Using the definition of \(1\,Gy = 1\,Jkg^{-1}\) which requires scaling up from per gramme to per kilogram (multiply by 1000), and converting the absorption time from seconds to hours (multiply by 3600) gives \[ D_{Gyhr^{-1}} = \frac{1.087 \times 10^{-7}}{4 \pi R^2} \times 1000 \times 3600 = \frac{0.391}{4 \pi R^2} \,\,Gyhr^{-1} \] Note, \(R\) is still in cm. At a distance of \(R=5\,\, cm\), i.e. at the aperture mouth the dose rate is \[ D_{Gyhr^{-1}} = \frac{0.391}{4 \pi \times 5^2} = 1.245 \times 10^{-3}\,\,Gyhr^{-1} \] For \(\gamma \) the absorption quality factor is 1, so \(1\,Gy \equiv 1\,Sv\), giving \[ D_{Gyhr^{-1}} = 1.244 \times 10^{-3}\,\,Gyhr^{-1} = 1.245\,\,mSvhr^{-1} \]

The above dose rate only applies at the centre dotted line of the aperture. As the distance increases along this line the dose rate falls in accordance with the inverse square law, but this only applies to a small area centred on the dotted line. This table above is modified to show dose rate and the applicable area as shown below.

Is this right?

The expressions above make the assumption that radiation can only emerge from the opening in the shielding. While this is true for, for example, visible light and an opaque shielding material, the situation would not quite be the same for, for example, \(^{137}Cs\) which is a \(\beta^{\,-} / \gamma \) emitter. In this case the \( \beta \) and \(\gamma\) rays would be attenuated by different amounts depending on the radiation type and how much sheilding material they passed through prior to exit. The net result is that instead of a hard circle of irradiation, there would be a wider area affected, with the power density falling outside of the centre area.This is analogous to the umbra and penumbra of a solar eclipse (except we're measuring 'brightness' rather than 'darkness').

It is reasonably simple to calculate the absorption fraction to apply as a function of the angle of the beam from the centre-line using simple geometry.

A \(\gamma\) photon travels a distance \(r\) to escape the shield. For an angle of travel relative to the hole centre-line, \(\alpha\), and \(\alpha\) being large enough that the path intersects the shielding, the path will be of length \(q\) in air, and length \(p\) in the shielding material. The relationship between \(\alpha \) and \(q\) is given by \[ \mathrm{sin} \left( \alpha \right) = \frac{a/2}{q} \Rightarrow q=\frac{a/2}{\mathrm{sin} \left( \alpha \right)} \] It is obivous that \(p+q=r\) so that \[ p=r-q = r-\frac{a/2}{\mathrm{sin} \left( \alpha \right)} \] From this, the attenuation factor due to linear absorption, \(a_f\) is given \[ a_f=\mathrm{exp} \left( -\lambda p \right) \] for an attenuation coefficient \(\lambda\). For the example of \(^{137}Cs\) and a lead shield, \(\lambda \approx 1 \,\, cm^{-1}\). Using the shield and hole dimensions above allow the following plot and brightness diagram to be constructed. The different path lengths through the shielding material results an a 'fuzzy' edge to the area of irradiation.

By changing to a logarithmic scale and extending the angle from -90 to +90 degrees from the hole centre line it is easy to see that the tenth value shielding thickness occurs at about \(5^ \circ \) (actually about \(5.3^ \circ \)) and two tenth value thicknesses occur at about \(40^ \circ \) (actually about \(38.7^ \circ \)). The half value thickness occurs at about \(3.3^ \circ \), which is just larger than the angle of \(2.87^ \circ \) at which first contact between the photon path and the shielding occurs.

To put these angles into context, consider an average 1.8 m person standing with their torso on the centre line of the aperture. The dose rate will be a function of the inverse square law distance and the attenuation of any shielding material the photon path passes through, as shown below

The analysis also assumes a point source; in reality, the source may be extended. Under the assumption that only 'line-of-sight' radiation can escape, this will given another mechanism to produce an umbra/penumbra analogous effect due to the various escape paths.Such a geometry will increase the area irradiated, as shown below

In actuality there will be effects from radiation attenuation near the apperture, and the extended nature of the source (and other considerations such as build-up). Mathematical descriptions could be constructed for these effects, or numerical simulation using Monte Carlo techniques could be used. But that's for another note...