Measuring Radiation Dosage — Question 9

Question 9

Ionization charge measures deposited energy through the mean pair energy

An ideal gas ionization chamber collects charge $Q_c=1.60\,\mathrm{nC}$ without recombination. Its gas mass is $m_g=10.0\,\mathrm{mg}$. Mean energy per ion pair is W=34.0 eV. Use $e=1.60\times10^{-19}\,\mathrm C$ and $1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm J$.

1. Find the number of ion pairs and the energy deposited in the gas.

2. Find the gas absorbed dose and explain why it does not directly equal a tissue dose without further calibration.

Question 9 - Solution

Ionization charge measures deposited energy through the mean pair energy

(a) Method and calculation

$$\boxed{N_{\mathrm{pairs}}=Q_c/e=10^{10}}.$$$$\boxed{E_g=N_{\mathrm{pairs}}W=10^{10}(34.0)(1.60\times10^{-19})=5.44\times10^{-8}\,\mathrm J}.$$

(b) Interpretation and further calculation

$$\boxed{D_g=E_g/m_g=5.44\times10^{-8}/10^{-5}=.00544\,\mathrm{Gy}=5.44\,\mathrm{mGy}}.$$$$\boxed{D_{\mathrm{tissue}}\ne D_g\text{ without a material-response conversion}}.$$

Physical check

Collected charge counts one elementary charge per ion pair in the external circuit, not twice that charge. W is a mean expenditure per pair, larger than a single ionization threshold.