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Step-by-Step Solution
Step 1: Understand the Concept of Distance of Closest Approach
When an alpha particle of charge $+2e$ (where $e$ is the elementary charge) is directed toward a heavy nucleus of charge $Ze$, the distance of closest approach is derived by equating the initial kinetic energy of the alpha particle to the electrostatic potential energy due to repulsion between the alpha particle and the nucleus. This distance is sometimes referred to as $r_0$ and can be given by:
$$r_0 = \frac{2Ze^2}{4 \pi \varepsilon_0 \, E}$$
Here,
$Z$: Atomic number of the target nucleus (uranium, $Z=92$).
$e$: Elementary charge ($1.6 \times 10^{-19}\,\text{C}$).
$\varepsilon_0$: Permittivity of free space ($8.854 \times 10^{-12}\,\text{C}^2 \text{N}^{-1} \text{m}^{-2}$).
$E$: Kinetic energy of the incoming alpha particle.
Step 2: Convert the Kinetic Energy from MeV to Joules
The alpha particle is given to have an energy of $5\,\text{MeV}$. Recall that:
$$1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}, \quad 1\,\text{MeV} = 10^6\,\text{eV}.$$
Therefore,
$$5\,\text{MeV} = 5 \times 10^6 \times 1.6 \times 10^{-19}\,\text{J} = 8 \times 10^{-13}\,\text{J}.$$
Step 3: Substitute the Known Values into the Formula
Using $Z=92$, $e = 1.6 \times 10^{-19}\,\text{C}$, and $\varepsilon_0 = 8.854 \times 10^{-12}\,\text{C}^2 \text{N}^{-1} \text{m}^{-2}$, and $E=8 \times 10^{-13}\,\text{J}$, we write:
$$r_0
= \frac{2 \times 92 \times (1.6 \times 10^{-19})^2}{4 \pi \times 8.854 \times 10^{-12} \times 8 \times 10^{-13}}.$$
Step 4: Numerical Evaluation
Performing the calculation (often grouped by first handling the constants and then powers of 10):
$$r_0 \approx 5.2 \times 10^{-14}\,\text{m}.$$
Step 5: Convert the Result to Centimeters
We know that $1\,\text{m} = 100\,\text{cm}$, so:
$$5.2 \times 10^{-14}\,\text{m} = 5.2 \times 10^{-14} \times 100\,\text{cm} = 5.2 \times 10^{-12}\,\text{cm}.$$
This is on the order of $10^{-12}\,\text{cm}$.
Step 6: Final Answer
Thus, the distance of closest approach of a $5\,\text{MeV}$ alpha particle scattered through $180^\circ$ by a uranium nucleus is of the order of $10^{-12}\,\text{cm}$.
