An alpha particle having KE equal to `8.7MeV` is projected towards the nucleus of copper with Z=29. Calculate its distance of closest approach.

Explanation

To find the distance of closest approach of an alpha particle to a copper nucleus, we can use the formula:

\[ r_0 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e \cdot 2e}{K} \]

Where:
- \( r_0 \) is the distance of closest approach,
- \( Z \) is the atomic number of the copper nucleus (which is 29),
- \( e \) is the charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \( K \) is the kinetic energy of the alpha particle,
- \( \epsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\)).

### Step 1: Convert Kinetic Energy from MeV to Joules
The kinetic energy of the alpha particle is given as \(8.7 \, \text{MeV}\). To convert this to joules, we use the conversion factor \(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\):

\[ K = 8.7 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 1.392 \times 10^{-12} \, \text{J} \]

### Step 2: Substitute Values into the Formula
Now we substitute the known values into the formula for \( r_0 \):

\[ r_0 = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{(29)(1.6 \times 10^{-19}) \cdot 2(1.6 \times 10^{-19})}{1.392 \times 10^{-12}} \]

### Step 3: Calculate the Numerator
First, calculate the numerator:

\[ \text{Numerator} = 29 \cdot 1.6 \times 10^{-19} \cdot 2 \cdot 1.6 \times 10^{-19} = 29 \cdot 2.56 \times 10^{-38} = 7.424 \times 10^{-38} \]

### Step 4: Calculate the Denominator
Now calculate the denominator:

\[ \text{Denominator} = 4 \pi (8.85 \times 10^{-12}) \cdot 1.392 \times 10^{-12} \]

Calculating \(4 \pi \approx 12.566\):

\[ \text{Denominator} \approx 12.566 \cdot 8.85 \times 10^{-12} \cdot 1.392 \times 10^{-12} \approx 1.1 \times 10^{-22} \]

### Step 5: Final Calculation of \( r_0 \)
Now we can calculate \( r_0 \):

\[ r_0 = \frac{7.424 \times 10^{-38}}{1.1 \times 10^{-22}} \approx 6.75 \times 10^{-15} \, \text{m} \]

### Step 6: Conclusion
Thus, the distance of closest approach \( r_0 \) is approximately:

\[ r_0 \approx 6.75 \times 10^{-15} \, \text{m} \]