A beam of beryllium nucleus (z = 4) of kinetic energy 5.3 MeV is headed towards the nucleus of gold atom (Z = 79). What is the distance of closest approach?

Options

A. `10.32xx10^(-14)m`
B. `8.58xx10^(-14)m`
C. `3.56xx10^(-14)m`
D. `1.25xx10^(-14)m`

Explanation

To find the distance of closest approach between a beryllium nucleus and a gold nucleus, we can use the concept of conservation of energy. The kinetic energy of the beryllium nucleus will be converted into electrostatic potential energy at the point of closest approach.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Charge of beryllium nucleus (Z1) = 4 (since beryllium has atomic number 4)
- Charge of gold nucleus (Z2) = 79 (since gold has atomic number 79)
- Kinetic energy (KE) of the beryllium nucleus = 5.3 MeV

2. **Convert Kinetic Energy to Joules:**
- 1 MeV = \(1.6 \times 10^{-13}\) Joules
- Therefore, KE = \(5.3 \, \text{MeV} = 5.3 \times 1.6 \times 10^{-13} \, \text{J} = 8.48 \times 10^{-13} \, \text{J}\)

3. **Calculate the Electrostatic Potential Energy at Closest Approach:**
- The potential energy (PE) between two point charges is given by:
\[ PE = \frac{k \cdot Z_1 \cdot Z_2 \cdot e^2}{r} \]
where:
- \(k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\) (Coulomb's constant)
- \(e = 1.6 \times 10^{-19} \, \text{C}\) (charge of an electron)
- \(r\) is the distance of closest approach.

4. **Set Kinetic Energy Equal to Potential Energy:**
- At the point of closest approach, the kinetic energy is converted to potential energy:
\[ KE = PE \]
\[ 8.48 \times 10^{-13} = \frac{(8.99 \times 10^9) \cdot (4) \cdot (79) \cdot (1.6 \times 10^{-19})^2}{r} \]

5. **Rearranging to Solve for r:**
- Rearranging the equation gives:
\[ r = \frac{(8.99 \times 10^9) \cdot (4) \cdot (79) \cdot (1.6 \times 10^{-19})^2}{8.48 \times 10^{-13}} \]

6. **Calculating the Values:**
- First calculate \( (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \)
- Then calculate:
\[ r = \frac{(8.99 \times 10^9) \cdot (4) \cdot (79) \cdot (2.56 \times 10^{-38})}{8.48 \times 10^{-13}} \]
- Calculate the numerator:
\[ = (8.99 \times 10^9) \cdot (4) \cdot (79) \cdot (2.56 \times 10^{-38}) = 8.99 \times 10^9 \cdot 316 \cdot 2.56 \times 10^{-38} \approx 7.16 \times 10^{-25} \]
- Now divide by \(8.48 \times 10^{-13}\):
\[ r \approx \frac{7.16 \times 10^{-25}}{8.48 \times 10^{-13}} \approx 8.45 \times 10^{-13} \, \text{m} \approx 8.45 \, \text{fm} \]

### Final Answer:
The distance of closest approach is approximately \(8.45 \, \text{fm}\).