What is the distance of closest approach to the nucleus for an `alpha`-particle of energy 5 MeV which undergoes scattering in the Gieger-Marsden experiment.

Options

A. `3.2xx10^(-16)` m
B. `3.8xx10^(-14)` m
C. `4.6xx10^(-15)` m
D. `3.2xx10^(-15)` m

Answer

To find the distance of closest approach \( d \) for an alpha particle of energy 5 MeV, we can use the principle of conservation of energy. The kinetic energy of the alpha particle is converted into potential energy when it is at the closest approach to the nucleus.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Energy of the alpha particle \( E = 5 \, \text{MeV} \)
- Charge of the alpha particle \( Q_1 = 2e \) (since an alpha particle consists of 2 protons)
- Charge of the nucleus (gold) \( Q_2 = Ze \) where \( Z = 79 \) for gold.
- Coulomb's constant \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

2. **Convert Energy from MeV to Joules:**
\[ 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]
Therefore,
\[ E = 5 \, \text{MeV} = 5 \times 1.6 \times 10^{-13} \, \text{J} = 8 \times 10^{-13} \, \text{J} \]

3. **Use the Conservation of Energy:**
At the distance of closest approach, the kinetic energy of the alpha particle is equal to the potential energy due to electrostatic force:
\[ KE = PE \]
\[ \frac{1}{2}mv^2 = \frac{k \cdot Q_1 \cdot Q_2}{d} \]
Here, we can express the potential energy as:
\[ PE = \frac{k \cdot (2e) \cdot (79e)}{d} \]

4. **Set Up the Equation:**
\[ 8 \times 10^{-13} = \frac{(9 \times 10^9) \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (79 \cdot 1.6 \times 10^{-19})}{d} \]

5. **Calculate the Right Side:**
\[ PE = \frac{(9 \times 10^9) \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (79 \cdot 1.6 \times 10^{-19})}{d} \]
\[ = \frac{(9 \times 10^9) \cdot (2.56 \times 10^{-37})}{d} \]

6. **Rearranging for \( d \):**
\[ d = \frac{(9 \times 10^9) \cdot (2.56 \times 10^{-37})}{8 \times 10^{-13}} \]

7. **Calculate \( d \):**
\[ d = \frac{2.304 \times 10^{-27}}{8 \times 10^{-13}} = 2.88 \times 10^{-15} \, \text{m} \]

8. **Final Result:**
The distance of closest approach \( d \) is approximately \( 2.88 \times 10^{-15} \, \text{m} \), which is of the order of \( 10^{-14} \, \text{m} \).

### Conclusion:
The distance of closest approach for an alpha particle of energy 5 MeV is approximately \( 2.88 \times 10^{-15} \, \text{m} \).