Q2 PE - College Physics (Urone)

Question

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

Step-by-Step Solution

Answer

Maximum energy deposited is obtained as: \(0.12\,{\rm{MeV}}\).

Step 1: 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.

Step 2: Evaluating the minimum energy that was deposited

Number of ions is created as:

\(n = 4 \times {10^3}\,{\rm{ion}}\)

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

\({E_{rad}} = 30.0\,{\rm{eV}}\)

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

\({\rm{Number of ion pairs  =   }}\frac{{{\rm{The energy of the radiation in the Geiger tube}}}}{{{\rm{The energy for ionize molecule}}}}\)

Rearranging the values and we obtain:

\(\begin{align}E &= n \times {E_{ion}}\\ &= 4 \times {10^3}\,ion \times 30\,eV\\ &= 1.2 \times {10^5}\,eV\\ &= 0.12\,MeV\end{align}\)

Therefore, the minimum energy deposited was: \(0.12\,{\rm{MeV}}\).