Question: 28. The energy of $30.0$ eV is required to ionize a molecule of the gas inside a Geiger tube, thereby producing an ion pair. Suppose a particle of ionizing radiation deposits $0.500$ MeV of energy in this Geiger tube. What maximum number of ion pairs can it create?
1. **State the problem:**
We want to find the maximum number of ion pairs created when a particle deposits $0.500$ MeV of energy in a Geiger tube, given that each ion pair requires $30.0$ eV.
2. **Convert units to be consistent:**
Energy deposited: $0.500$ MeV = $0.500 \times 10^6$ eV = $5.0 \times 10^5$ eV.
3. **Formula:**
The maximum number of ion pairs $N$ is given by dividing the total energy deposited by the energy required to create one ion pair:
$$N = \frac{\text{Energy deposited}}{\text{Energy per ion pair}}$$
4. **Calculate:**
$$N = \frac{5.0 \times 10^5 \text{ eV}}{30.0 \text{ eV}}$$
5. **Simplify:**
$$N = \frac{5.0 \times 10^5}{30.0} = \frac{5.0 \times 10^5}{\cancel{30.0}} \times \frac{1}{\cancel{30.0/30.0}} = 1.666\times 10^4$$
6. **Interpretation:**
The maximum number of ion pairs created is approximately $1.67 \times 10^4$.
**Final answer:**
$$\boxed{1.67 \times 10^4 \text{ ion pairs}}$$