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\({\rm{900}}\)?

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.

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

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

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

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