Measuring Radiation Dosage — Question 6

Question 6

Layered attenuation multiplies transmissions and adds optical thicknesses

For an ideal narrow photon beam use transmission $I/I_0=e^{-\mu x}$. It passes through 2.00 cm of material 1 with $\mu_1=.500\,\mathrm{cm^{-1}}$, then 5.00 cm of material 2 with $\mu_2=.200\,\mathrm{cm^{-1}}$. Treat the downstream detector dose rate as proportional to transmitted intensity.

1. Find total transmission and dose rate if the unattenuated value is $10.0\,\mu\mathrm{Gy/min}$.

2. Find additional thickness of material 2 that would halve the transmitted intensity again.

Question 6 - Solution

Layered attenuation multiplies transmissions and adds optical thicknesses

(a) Method and calculation

$$\frac I{I_0}=e^{-\mu_1x_1}e^{-\mu_2x_2}=e^{-(\mu_1x_1+\mu_2x_2)}.$$$$\boxed{I/I_0=e^{-2}=.1353,\quad\dot D=1.353\,\mu\mathrm{Gy/min}}.$$

(b) Interpretation and further calculation

$$e^{-\mu_2\Delta x}=1/2\Rightarrow\Delta x=\ln2/\mu_2.$$$$\boxed{\Delta x=.693147/.200=3.466\,\mathrm{cm}}.$$

Physical check

This is a narrow-beam model without scattered-photon buildup or changes in spectrum. It is a calculation exercise, not a complete shielding design.