An `alpha`-particle of kinetic energy 7.68 MeV is projected towards the nucleus of copper (Z=29). Calculate its distance of nearest approach.

Answer

To calculate the distance of nearest approach for an alpha particle projected towards the nucleus of copper, we can use the formula for the distance of closest approach \( r_0 \):

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

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

### Step 1: Convert the kinetic energy from MeV to Joules
The kinetic energy \( K \) of the alpha particle is given as \( 7.68 \, \text{MeV} \).

\[ K = 7.68 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 1.2288 \times 10^{-12} \, \text{J} \]

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

\[ r_0 = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{(29)(2)(1.6 \times 10^{-19})^2}{1.2288 \times 10^{-12}} \]

### Step 3: Calculate the numerator
Calculate \( (29)(2)(1.6 \times 10^{-19})^2 \):

\[ = 58 \cdot (2.56 \times 10^{-38}) = 1.4848 \times 10^{-36} \]

### Step 4: Calculate the denominator
Calculate \( 4 \pi (8.85 \times 10^{-12}) \):

\[ = 4 \cdot 3.14 \cdot 8.85 \times 10^{-12} \approx 1.112 \times 10^{-10} \]

### Step 5: Substitute and simplify
Now substitute back into the equation for \( r_0 \):

\[ r_0 = \frac{1.4848 \times 10^{-36}}{1.112 \times 10^{-10} \cdot 1.2288 \times 10^{-12}} \]

Calculating the denominator:

\[ 1.112 \times 10^{-10} \cdot 1.2288 \times 10^{-12} \approx 1.366 \times 10^{-22} \]

Now, substituting:

\[ r_0 = \frac{1.4848 \times 10^{-36}}{1.366 \times 10^{-22}} \approx 1.087 \times 10^{-14} \, \text{m} \]

### Final Answer
Thus, the distance of nearest approach \( r_0 \) is approximately:

\[ r_0 \approx 1.087 \times 10^{-14} \, \text{m} \, \text{or} \, 10.87 \, \text{fm} \]