Nuclear Fuel Cycle

Q: The effective dose rate to a person standing on a surface contaminated with gamma emitters is measured as 2 μSv/h at a height of 1 m above the ground. Using the extended-area-source formula introduced in Section 6.14 (Worked Example 3),

$\dot{D}(h) = \Gamma C \cdot \ln\left[\frac{A \times W + h^2}{h^2}\right]$

where $\Gamma$ is the specific gamma constant, $C$ is the activity per unit area (Bq/m$^2$), $A = 10$ m is the effective length of the contaminated area, $W = 5$ m is the width, and $h$ is the measurement height, show that the dose rate at a height of 0.25 m (e.g. the head of a small dog or the legs of a child) is greater than 2 μSv/h but less than 7.5 μSv/h.

Answer:

Step 1: Find $\Gamma C$ using the measurement at $h = 1$ m.

$2 = \Gamma C \cdot \ln\left[\frac{10 \times 5 + 1^2}{1^2}\right] = \Gamma C \cdot \ln(51) = \Gamma C \times 3.932$

$\Gamma C = \frac{2}{3.932} = 0.509 \ \mu\text{Sv/h}$

Step 2: Calculate the dose rate at $h = 0.25$ m.

$\dot{D}(0.25) = 0.509 \cdot \ln\left[\frac{10 \times 5 + 0.25^2}{0.25^2}\right] = 0.509 \cdot \ln\left[\frac{50.0625}{0.0625}\right] = 0.509 \cdot \ln(801.0)$

$\dot{D}(0.25) = 0.509 \times 6.686 = 3.40 \ \mu\text{Sv/h}$

Result: 2.0 μSv/h < 3.4 μSv/h < 7.5 μSv/h — QED.

Physical interpretation: Reducing the height by a factor of 4 (1.0 m → 0.25 m) only increases the dose rate by a factor of ~1.7, not the factor of 16 that an inverse-square (point-source) law would predict. This is because the contamination is an extended planar source, not a point source: the dose rate depends logarithmically on height, not quadratically. The same formula applied in Section 6.14 to a contaminated floor inside a decommissioned facility also covers the analogous yellowcake-road case at the mining end of the cycle.