Formula for calculating dose rates from gamma emitting radioactive materials

A "rule of thumb" for estimating the dose rate

Dr Chris Robbins, Grallator

[Ionactive note: as detailed earlier, in the analysis which follows A (activity) has replaced M as used in the formula above so far discussed].

Consider a point source of \(\gamma\)-rays which produces monochromatic photons of energy \(E\) MeV. Further, assume that each decay produces only one photon, and that the activity of the source is \(A\) MBq, i.e. \(A\) million decays per second \(\Rightarrow\) \(A\) million photons per second. We want to estimate the dose rate to a person in \(\mu Gy hr^{-1}\) at a distance \(r\) metres from the source, \(D_{\mu Gy hr^{-1}} (r) \).

Recall the definition of the gray (Gy) as a dose unit is

\(\qquad\)1 gray is the absorption of 1 joule or energy by 1 kilogram of matter.

The dose rate from this is how many joules of energy are absorbed per unit time per kilogram of matter. Note, the final expression will be in \(\mu\) joules per unit time per kilogram of matter.

Calculating the power

The total energy released per second (the power) in \({MeV \cdot s^{-1}}\), \(P_{MeV \cdot s^{-1}}\), is given by the energy per photon multiplied by the number of photons produced per second, i.e.

\[P_{MeV \cdot s^{-1}} = AE\times 10^6\]

where the \(\times 10^6\) takes account of the statement that the production rate is given in MBq.

The dose rate we are calculating uses units of \(\mu J\) for energy and hours for time. The above power is converted to these units by noting that

\(\qquad 1\,J = 1.6\times 10^{-19}\, eV = 1.6\times 10^{-13}\, MeV\)

\(\qquad 1\,hr = 60\, s/min \times 60\, mins/hr = 3.6\times 10^{3}\, s\)

So that \[P_{J \cdot hr^{-1}} = AE\times 10^6 \times 1.6\times 10^{-13} \times 3.6\times 10^{3} = 5.76AE\times 10^{-4}\] Noting that \( 1\,J = 10^{6}\,\mu J \), the above is given in terms of \(\mu J hr^{-1}\) as \[P_{\mu J \cdot hr^{-1}} = 5.76AE\times 10^{-4} \times 10^6 = 5.76AE\times 10^{2} = 576AE\]

Calculating the power density per unit area

At a distance \(r\) from the source, the energy released is spread over the surface of a sphere of radius \(r\), which has an area of \(4 \pi r^2\). The power per unit area at a distance \(r\) is therefore given by \[P_{\mu J \cdot hr^{-1} \cdot m^{-2}} = \frac{576AE}{4\pi r^2} = \frac{144AE}{\pi r^2} \] This is just the inverse square law.

What happens on the way to the measuring point?

While travelling from the source to the measuring point a distance \(r\) m away, the \(\gamma\) photons can be absorbed by the air. There is a fixed probability of absorption per unit distance travelled so the attenuation resulting from travelling a distance \(r\) in the air is of the form \[ I = I_0 e^{-\mu_a r}\] where

\(\quad\) \(I_0\) \(\quad\) is the initial photon intensity

\(\quad\) \(I\) \(\quad\) is the photon intensity after travelling a distance \(r\)

\(\quad\) \(\mu_a\) \(\quad\) is the linear attenuation coefficient for air, units m\(^{-1}\)

The dimensionless quantity \(e^{-\mu_a r} \equiv \mathrm {exp} (-\mu_a r) \) is the attenuation factor; it gives the fraction of the initial photons which travel the distance \(r\) without being absorbed.

An alternative form of the above is to use the mass energy attenuation coefficient, which is the linear attenuation coefficient normalised to the density of the material, air in this instance. The attenuation factor in this case is given by the expression \[ f_a = \mathrm {exp} \left(-\left(\frac{\mu_a}{\rho_a}\right) \rho_a r \right) \] Using the values for air of \( ({\mu_a}/{\rho_a}) = 2.676 \times 10^{-3}\, m^2 kg^{-1}\) and \(\rho=1.293 \, kgm^{-3}\) gives \(f_a = \mathrm {exp} \left(-3.46 \times 10^{-3} r \right) \). For a distance \(r=10\) m, the attenuation factor is \(f_a = 0.966\), i.e. there is very little absorption. In the approximation that is being derived we will ignore absorption in air under the assumption that the dose rate is being approximated "close" to the source.

Energy absorption rate at measuring point

At the measuring point we wish to calculate the energy absorbed per unit time per kg of matter (air, tissue, etc.). The attenuation factor for travelling through air has been given above. For travelling a distance \(x\) m through arbitrary matter the attenuation factor is \[ f_m = \mathrm {exp} \left(-\left(\frac{\mu_m}{\rho_m}\right) \rho_m x \right) \] As \(f_m\) gives the fraction of photons that pass through a distance \(x\) without being absorbed, then \(1-f_m\) must give the fraction that are absorbed over a distance \(x\). Applying this to the power density per unit area gives the energy absorption per unit area per unit time for matter thickness \(x\) of \[ \begin{align} P_{\mu J \cdot hr^{-1} \cdot m^{-3}} &= \frac{144AE}{\pi r^2 } \left(1- \mathrm {exp} \left(-\left(\frac{\mu_m}{\rho_m}\right) \rho_m x \right) \right) \end{align} \] Note, the attenuation factor may be found to be nearly one, however, as we are interested in the energy absorption rate we do cannot ignore absorption as we did for air as this would result in zero dose!

Simplifying things

The expression \(\mathrm{exp}(z) \) can be written as the Maclaurin series expansion \[ \mathrm{exp}(z) = 1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \] For the case where \(z\) is small \( \mathrm{exp}(z) \approx 1+z \). In our problem we wish to approximate \( \mathrm{exp}(-z)\) which is simply given by \( \mathrm{exp}(-z) \approx 1-z \). Using this approximation \[ \begin{align} \left(1- \mathrm {exp} \left(-\left(\frac{\mu_m}{\rho_m}\right) \rho_m x \right) \right) &\approx \left(1-\left( 1 - \left(\frac{\mu_m}{\rho_m}\right) \rho_m x \right) \right) \\ \\ &= \left(\frac{\mu_m}{\rho_m}\right) \rho_m x \end{align} \] Using this approximation gives \[ P_{\mu J \cdot hr^{-1} \cdot m^{-3}} = \frac{144AE}{\pi r^2 } \left(\frac{\mu_m}{\rho_m}\right) \rho_m x \] Reminder, this is valid if \(\left(\frac{\mu_m}{\rho_m}\right) \rho_m x \) is small.

What value of \(x\) to use?

The dose rate approximation we want is in \(\mu Gy hr^{-1}\), so, from the definition of a Gy, the value of \(x\) should be chosen to give 1 kg of matter. Using the relationship that \(\rho = \frac{mass}{volume}\), the volume of matter required to give 1 kg is \(V_m = \frac{1}{\rho_m} \, m^3\). The power density above is given per unit area, so the value of \(x\) required as a depth over this unit area to give a volume of \(\frac{1}{\rho_m} \, m^3\) is approximately \(\frac{1}{\rho_m} \, m\), if this value is small enough that the surface areas of the upper and lower volume element faces are approximately equal. Using this gives an expression for the absorption factor as \[ \left(\frac{\mu_m}{\rho_m}\right) \rho_m x = \left(\frac{\mu_m}{\rho_m}\right) \rho_m \frac{1}{\rho_m} = \left(\frac{\mu_m}{\rho_m}\right) \] The energy absorption per unit area per unit time for 1 kg of matter is then given by \[ D_{\mu Gy hr^{-1}} \equiv P_{\mu J \cdot hr^{-1} \cdot kg^{-1}} = \frac{144AE}{\pi r^2 } \left(\frac{\mu_m}{\rho_m}\right) \] This is the dose rate expression we are seeking.

Using values of \( ({\mu_m}/{\rho_m}) = 3.40 \times 10^{-3}\, m^2 kg^{-1}\) and \(\rho=1000 \, kgm^{-3}\) as 'typical' values for tissue gives \(x = \frac{1}{1000} \,m\) and an absorption factor of \[ \left(\frac{\mu_m}{\rho_m}\right) = 3.40 \times 10^{-3} \] This value is small and so the approximation to remove the exponential function is valid. Also, the value of \(x\) is small, so the approximation of the volume being a small regular prism with a base of unit area is also valid.

Note, for air the expression \[ \begin{align} \left(\frac{\mu_m}{\rho_m}\right) \rho_m x = \left(\frac{\mu_m}{\rho_m}\right) \rho_m \frac{1}{\rho_m} &= 2.676 \times 10^{-3} \times 1.293 \times \frac{1}{1.293} \\ \\ &= 2.676\times 10^{-3} \end{align} \] is also small, however \(x=\frac{1}{1.293}\) is not small so inverse square effects could become important. In this case, the correct volume element should be a frustum of a pyramid or a cone, and the value of \(x\) will be reduced to account for this. This is not considered here, so the dose rate derived here will apply more correctly to relatively dense matter such as tissue.

The dose rate rule of thumb expression

Using a fixed tissue absorption factor of \( ({\mu_m}/{\rho_m}) = 3.40 \times 10^{-3}\), the final expression for the dose rate rule of thumb in units of \(\mu Gy hr^{-1}\) is \[ \begin{align} D_{\mu Gy hr^{-1}} &= \frac{144AE}{\pi r^2 } \times 3.40 \times 10^{-3} \\ \\ &=\frac{306AE}{625 \pi r^2 } \\ \\ \end{align} \] A further simplification can be made by taking the gross approximation that \(\frac{306}{625 \pi} \approx \frac{1}{6} \) to the nearest simple fraction so that the rule of thumb dose rate becomes \[ \begin{align} D_{\mu Gy hr^{-1}} &= \frac{AE}{6 r^2 } \end{align} \] For \(A\) in MBq, \(E\) in MeV and \(r\) in metres.

What happens if multiple photons with multiple energies are emitted?

It is not unusual for a \(\gamma\) source to produce a spectrum of photon energies. In this case, the energy should be averaged over all the photon energies possible through \[ E = \sum_i p_i E_i \] where \(p_i\) is the probability that a photon of energy \(E_i\) is released during the decay. In some cases the evaluation of the above can be quite involved and multiple photons can be released. For example, \(^{60}Co\) has two decay paths:

99.88% of decays give two \(\gamma\) photons of energy 1.1732 MeV and 1.3325 MeV respectively

0.12% of decays give one \(\gamma\) photon of energy 1.3325 MeV

In this case the total photon energy released per decay is

\[E = (0.9988 \times 1.1732) + (0.9988 \times 1.3325) + (0.0012 \times 1.3325) = 2.5043 MeV\]

Warning, this is only valid if \( ({\mu_m}/{\rho_m})\) is a fixed value for all photon energies, as in the above assumption!

Other units

The gray (Gy) is converted to sieverts (Sv) by multiplying by a quality weighting factor \(W_R\). For X-rays and \(\gamma\) rays \(W_R=1\) so the above dose rate in \(\mu Sv hr^{-1}\) is simply given by multiplying the dose rate in \(\mu Gy hr^{-1}\) by 1 to give \[ \begin{align} D_{\mu Sv hr^{-1}} &= \frac{AE}{6 r^2 } \end{align} \] For \(A\) in MBq, \(E\) in MeV and \(r\) in metres.

Does the rule of thumb work?

The rule of thumb expression has been derived using a number of assumptions and simplifications. By far the biggest of these are that a fixed value of the absorption factor, \( ({\mu_m}/{\rho_m}) = 3.40 \times 10^{-3}\) is used and \(\frac{306}{625 \pi} \approx \frac{1}{6} \). For the latter, \(\frac{306}{625 \pi} = \frac{1}{6.41665\, \dots} \) so even if the absorption factor is appropriate, the rule of thumb expression will tend to over-predict values by about 7%.

It is also important to note that in reality the absorption factor varies with photon energy and the matter that is absorbing it, so the rule of thumb is only technically valid for a source with an energy release being absorbed by matter with an absorption coefficient of \( ({\mu_m}/{\rho_m}) = 3.40 \times 10^{-3}\). In fact this value is slightly high for most tissues but reasonably flat over an energy range of 0.05 to 2 MeV. For generic values you should revert to the full expression \[ D_{\mu Gy hr^{-1}} = \frac{144AE}{\pi r^2 } \left(\frac{\mu_m}{\rho_m}\right) \] For lower values of \( ({\mu_m}/{\rho_m})\), the above could approximate to \(D_{\mu Sv hr^{-1}} = \frac{AE}{7 r^2 }\) or \(D_{\mu Sv hr^{-1}} = \frac{AE}{8 r^2 }\), etc.; the actual best rule of thumb depends on circumstances.

When a number of photons can be released, as in the \(^{60}Co\) example above, the total dose rate should be calculated as the sum of the dose rates of each photon. Do not use the energy averaging above. \[ \begin{align} D_{\mu Gy hr^{-1}} &= \sum\limits_i p_i\frac{144AE_i}{\pi r^2 } \left(\frac{\mu_m}{\rho_m}\right)\bigg |_{E_i} \\ \\ &= \frac{144A}{\pi r^2 } \sum\limits_i p_i E_i \left(\frac{\mu_m}{\rho_m}\right)\bigg |_{E_i} \end{align} \] where \(p_i\) is as before and \( ({\mu_m}/{\rho_m}) | _{E_i} \) is the value of \( ({\mu_m}/{\rho_m})\) for a photon of energy \(E_i\).

For the \(^{60}Co\) example used above the following energies, probabilities and energy absorption coefficients for soft tissue are used.

i	\(E_i \,\,(MeV) \)	\(p_i\)	\( (\mu_m / \rho_m)\,\, (m^2/kg) \)	\( p_i E_i (\mu_m / \rho_m) |_{E_i} \)
1	1.1732	0.9988	\(2.9760 \times 10^{-3}\)	\(3.4873 \times 10^{-3}\)
2	1.3325	0.9988	\(2.9047 \times 10^{-3}\)	\(3.8659 \times 10^{-3}\)
3	1.3325	0.0012	\(2.9047 \times 10^{-3}\)	\(4.6446 \times 10^{-6}\)

\[\begin{align}D_{\mu Gy hr^{-1}} &= \frac{144A}{\pi r^2 } \sum\limits_i p_i E_i \left(\frac{\mu_m}{\rho_m}\right)\bigg |_{E_i} \\ \\&= \frac{144A}{\pi r^2 } \left( 3.4873 \times 10^{-3} + 3.8659 \times 10^{-3} + 4.6446 \times 10^{-6} \right) \\ \\&= \frac{144A}{\pi r^2 } \times 7.3578\times 10^{-3} \\ \\&\approx \frac{A}{2.83 r^2 }\end{align}\]

What happens if you work backwards from the above result to get an expression of the form \[ \begin{align} D_{\mu Gy hr^{-1}} &= \frac{AE}{N r^2 } \end{align} \] for an appropriate total energy \(E\) and integer \(N\)? The working above has calculated a single value for the total energy E for \(^{60}Co\) as \(E=2.5043 MeV\). Multiplying the actual value top and bottom by this value for \(E\) (keeping the symbol on the top and substituting the value on the bottom) gives \[ \begin{align} D_{\mu Gy hr^{-1}} &= \frac{A}{2.83 r^2 } \times \frac{E}{E} \\ \\ &=\frac{A}{2.83 r^2 } \times \frac{E}{2.5043} \\ \\ &\approx \frac{AE}{7 r^2 } \end{align} \] So, nearly the rule of thumb, but with a slightly larger denominator.

Can low energy photons be ignored?

From the expression \[ \begin{align} D_{\mu Gy hr^{-1}} &= \frac{144A}{\pi r^2 } \sum\limits_i p_i E_i \left(\frac{\mu_m}{\rho_m}\right)\bigg |_{E_i} \\ \\ \end{align} \] it is noted that the contribution to the total dose rate from a photon of energy \(E_i\) is proportional to \( p_i E_i (\mu_m / \rho_m) |_{E_i} \). The example above is for a case where all the possible photons are of broadly similar energy, with similar energy absorption coefficients, so contributions are dominated by the first two photons in the list. However, there are sources, such as \(^{192}Ir\) that produce \(\gamma\) photons over a wide range of energies from 9 keV to 612.5 keV. It may be tempting to ignore low energy photons in the calculation, but this should be done with caution as, although a 9 keV photon has about 1/68 the energy of a 612.5 keV photon, its energy absorption coefficient is several orders of magnitude higher, as shown in the following graph.

The table below gives values and dose rate contributions from all the possible photon emissions of \(^{192}Ir\). Note, the derived values below may show rounding effects due to the source values having more decimal places than displayed.

i	\(E_i \,\,(MeV) \)	\(p_i\)	\( (\mu_m / \rho_m)\,\, (m^2/kg) \)	\( p_i E_i (\mu_m / \rho_m) |_{E_i} \)	Dose rate contribution
1	0.0090	\(1.470 \times 10^{-2}\)	\(6.9051 \times 10^{-1}\)	\(9.1355 \times 10^{-5}\)	3.13%
2	0.0094	\(4.000 \times 10^{-2}\)	\(5.9586 \times 10^{-1}\)	\(2.2500 \times 10^{-4}\)	7.70%
3	0.0615	\(1.170 \times 10^{-2}\)	\(3.2031 \times 10^{-3}\)	\(2.3044 \times 10^{-6}\)	0.08%
4	0.0630	\(2.020 \times 10^{-2}\)	\(3.1440 \times 10^{-3}\)	\(4.0010 \times 10^{-6}\)	0.14%
5	0.0651	\(2.630 \times 10^{-2}\)	\(3.0651 \times 10^{-3}\)	\(5.2494 \times 10^{-6}\)	0.18%
6	0.0668	\(4.520 \times 10^{-2}\)	\(3.0047 \times 10^{-3}\)	\(9.0762 \times 10^{-6}\)	0.31%
7	0.0714	\(8.700 \times 10^{-3}\)	\(2.8558 \times 10^{-3}\)	\(1.7740 \times 10^{-6}\)	0.06%
8	0.0757	\(1.970 \times 10^{-2}\)	\(2.7304 \times 10^{-3}\)	\(4.0719 \times 10^{-6}\)	0.14%
9	0.1363	\(1.810 \times 10^{-3}\)	\(2.6965 \times 10^{-3}\)	\(6.6544 \times 10^{-7}\)	0.02%
10	0.2013	\(4.660 \times 10^{-3}\)	\(2.9454 \times 10^{-3}\)	\(2.7631 \times 10^{-6}\)	0.09%
11	0.2058	\(3.290 \times 10^{-2}\)	\(2.9571 \times 10^{-3}\)	\(2.0021 \times 10^{-5}\)	0.69%
12	0.2833	\(2.610 \times 10^{-3}\)	\(3.1316 \times 10^{-3}\)	\(2.3152 \times 10^{-6}\)	0.08%
13	0.2960	\(2.896 \times 10^{-1}\)	\(3.1563 \times 10^{-3}\)	\(2.7052 \times 10^{-4}\)	9.26%
14	0.3085	\(2.967 \times 10^{-1}\)	\(3.1721 \times 10^{-3}\)	\(2.9030 \times 10^{-4}\)	9.94%
15	0.3165	\(8.284 \times 10^{-1}\)	\(3.1796 \times 10^{-3}\)	\(8.3367 \times 10^{-4}\)	28.54%
16	0.3745	\(7.290 \times 10^{-3}\)	\(3.2293 \times 10^{-3}\)	\(8.8159 \times 10^{-6}\)	0.30%
17	0.4165	\(6.620 \times 10^{-3}\)	\(3.2526 \times 10^{-3}\)	\(8.9673 \times 10^{-6}\)	0.31%
18	0.4681	\(4.780 \times 10^{-1}\)	\(3.2631 \times 10^{-3}\)	\(7.3006 \times 10^{-4}\)	25.00%
19	0.4846	\(3.160 \times 10^{-2}\)	\(3.2662 \times 10^{-3}\)	\(5.0012 \times 10^{-5}\)	1.71%
20	0.4891	\(3.970 \times 10^{-3}\)	\(3.2670 \times 10^{-3}\)	\(6.3431 \times 10^{-6}\)	0.22%
21	0.5886	\(4.520 \times 10^{-2}\)	\(3.2556 \times 10^{-3}\)	\(8.6609 \times 10^{-5}\)	2.97%
22	0.6044	\(8.180 \times 10^{-2}\)	\(3.2520 \times 10^{-3}\)	\(1.6078 \times 10^{-4}\)	5.50%
23	0.6125	\(5.330 \times 10^{-2}\)	\(3.2484 \times 10^{-3}\)	\(1.0604 \times 10^{-4}\)	3.63%

The above table shows that the two lowest energy photons contribute more than 10% to the total dose rate because they are far more easily absorbed. In fact the two lowest energy photons contribute more to the total dose rate than the two highest energy photons, even though there are more than twice as many of the two highest energy photons produced as there are the two lowest energy photons.

Dr Chris Collins can be contacted at Grallator (opens in a new window).