An `alpha-` paticle approaches the target nucleus of copper (Z = 29) in such a way that the value of impact parameter is zero. The distance of closest approach will be

Options

A. `(2pi epsilon_(0)(K.E.)_(alpha))/(29e^(2))`
B. `(29e^(2))/(2pi epsilon_(0)(K.E.)_(alpha))`
C. `(4pi epsilon_(0)(K.E.)_(alpha))/(29e^(2))`
D. `(K.E.)_(alpha)`

Answer

To find the distance of closest approach of an alpha particle to a copper nucleus, we can use the concept of energy conservation. The kinetic energy of the alpha particle will be converted into potential energy at the point of closest approach.

### Step-by-Step Solution:

1. **Identify the Charges**:
- The copper nucleus has an atomic number \( Z = 29 \), which means it has a charge of \( +29e \) (where \( e \) is the elementary charge).
- An alpha particle, being a helium nucleus, has a charge of \( +2e \).

2. **Write the Kinetic Energy**:
- The kinetic energy (KE) of the alpha particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the alpha particle and \( v \) is its velocity.

3. **Write the Potential Energy at Closest Approach**:
- The potential energy (PE) at the distance of closest approach \( d \) is given by:
\[ PE = \frac{k \cdot (29e) \cdot (2e)}{d} \]
where \( k = \frac{1}{4\pi \epsilon_0} \) is Coulomb's constant.

4. **Set Kinetic Energy Equal to Potential Energy**:
- At the distance of closest approach, all the kinetic energy is converted into potential energy:
\[ \frac{1}{2} mv^2 = \frac{k \cdot (29e) \cdot (2e)}{d} \]

5. **Rearranging for Distance \( d \)**:
- Rearranging the equation gives:
\[ d = \frac{2k \cdot (29e) \cdot (2e)}{mv^2} \]
- Simplifying this, we have:
\[ d = \frac{58ke^2}{mv^2} \]

6. **Substituting the Value of \( k \)**:
- Substitute \( k = \frac{1}{4\pi \epsilon_0} \):
\[ d = \frac{58 \cdot \frac{1}{4\pi \epsilon_0} \cdot e^2}{mv^2} \]
- This gives us the final formula for the distance of closest approach.

### Final Answer:
The distance of closest approach \( d \) can be expressed as:
\[ d = \frac{58e^2}{4\pi \epsilon_0 \cdot KE} \]
where \( KE \) is the kinetic energy of the alpha particle.