Question

A particle of ionizing radiation creates $4000$ ion pairs in the gas inside a Geiger tube as it passes through. What minimum energy was deposited, if $30.0\mathrm{e}\mathrm{V}$ is required to create each ion pair?

Short answer

Maximum energy deposited is obtained as: $0.12\mathrm{M}\mathrm{e}\mathrm{V}$.

Step-by-step solution

Define Radioactivity

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

Evaluating the minimum energy that was deposited

Number of ions is created as:

$n=4\times {10}^3\mathrm{i}\mathrm{o}\mathrm{n}$

The energy of the radiation in the Geiger tube is given as:

$E_{rad}=30.0\mathrm{e}\mathrm{V}$

The number of ion pairs, is obtained using the equation:

$\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{p}\mathrm{a}\mathrm{i}\mathrm{r}\mathrm{s}=\frac{\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{G}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{u}\mathrm{b}\mathrm{e}}{\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{z}\mathrm{e}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}}$

Rearranging the values and we obtain:

$\begin{array}{rl}E& =n\times E_{ion}\\ & =4\times {10}^{3}ion\times 30eV\\ & =1.2\times {10}^{5}eV\\ & =0.12MeV\end{array}$

Therefore, the minimum energy deposited was: $0.12\mathrm{M}\mathrm{e}\mathrm{V}$.