Measuring Radiation Dosage — Question 1

Question 1

Absorbed dose is deposited energy divided by absorbing mass

A laboratory absorber of mass 5.00 g retains $2.00\,\mu\mathrm J$ from a radiation pulse. Use $1\,\mathrm{Gy}=1\,\mathrm{J/kg}$ and $1\,\mathrm{rad}=.0100\,\mathrm{Gy}$.

1. Find absorbed dose in gray, milligray and rad.

2. If the same retained energy is uniformly distributed through 20.0 g instead, find the new dose and explain the mass dependence.

Question 1 - Solution

Absorbed dose is deposited energy divided by absorbing mass

(a) Method and calculation

$$\boxed{D=E_{\mathrm{abs}}/m=2.00\times10^{-6}/.00500=4.00\times10^{-4}\,\mathrm{Gy}}.$$$$\boxed{D=.400\,\mathrm{mGy}=.0400\,\mathrm{rad}}.$$

(b) Interpretation and further calculation

$$\boxed{D\prime=2.00\times10^{-6}/.0200=1.00\times10^{-4}\,\mathrm{Gy}=.100\,\mathrm{mGy}}.$$$$E_{\mathrm{abs}}\text{ fixed}:\quad\boxed{D\propto1/m,\quad D\prime/D=1/4}.$$

Physical check

Dose refers to energy actually retained in the specified material. Emitted source energy and incident energy need not equal deposited energy.