Question

Suppose a particle of ionizing radiation deposits\(1.0\,{\rm{MeV}}\)in the gas of a Geiger tube, all of which goes to creating ion pairs. Each ion pair requires\(30.0\,{\rm{eV}}\)of energy. (a) The applied voltage sweeps the ions out of the gas in\(1.00\,{\rm{\mu s}}\). What is the current? (b) This current is smaller than the actual current since the applied voltage in the Geiger tube accelerates the separated ions, which then create other ion pairs in subsequent collisions. What is the current if this last effect multiplies the number of ion pairs by\({\rm{900}}\)?

Short answer

1. The current is obtained as:\(1.0666 \times {10^{ - 8}}\,{\rm{A}}\).
2. The current if last effect multiplies the number of ion pairs is obtained as: \(9.5994 \times {10^{ - 6}}\,{\rm{A}}\).

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 current

a) The required energy to ionize the molecule is given as:

\({E_{ion}} = 30\,{\rm{eV}}\)

The energy of the radiation in the Geiger tube is given as:\({E_{rad}} = 1.0\,{\rm{MeV}}\).

The number of ion pairs, is then obtained as:

\(\begin{align}Number{\rm{ }}of{\rm{ }}ion{\rm{ }}pairs(N) &= \frac{{The{\rm{ }}energy{\rm{ }}of{\rm{ }}the{\rm{ }}radiation{\rm{ }}in{\rm{ }}the{\rm{ }}Geiger{\rm{ }}tube}}{{The{\rm{ }}energy{\rm{ }}for{\rm{ }}ionize{\rm{ }}molecule}}\\ &= \frac{{1.0 \times {{10}^6}\,eV}}{{30\,eV}}\\ &= 3.333 \times {10^4}\,pairs\end{align}\)

The following relation is used to evaluate the current as:

\(\begin{align}I &= \frac{Q}{{\Delta t}}\\ &= \frac{{2N{\rm{ }}e}}{{\Delta t}}\end{align}\)

Substitute all the value in the above equation

\(\begin{align}I &= \frac{{2 \times 3.333 \times {{10}^4}\,pairs \times 1.6 \times {{10}^{ - 19}}\,C}}{{1 \times {{10}^{ - 6}}\,s}}\\ &= 1.0666 \times {10^{ - 8}}\,A\end{align}\)

Therefore, the current is: \(1.0666 \times {10^{ - 8}}\,{\rm{A}}\).

Evaluating the current of the last effect if it multiplies the ion

b) The second case is then obtained as:

\(\begin{align}{I_n} &= nI\\ &= 900\,ions \times 1.0666 \times {10^{ - 8}}\,A\\ &= 9.5994 \times {10^{ - 6}}\,A\end{align}\)

Therefore, current if last effect multiplies the number of ion pairs is: \(9.5994 \times {10^{ - 6}}\,{\rm{A}}\).