An `alpha-` paticle approaches the target nucleus of copper (Z = 29) is such a way that the vlaue 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 conservation of energy. The alpha particle (which is a helium nucleus, \( \text{He}^{2+} \)) has an initial kinetic energy that will be converted into electric potential energy as it approaches the nucleus.

### Step-by-Step Solution:

1. **Identify the Charges**:
- The charge of the alpha particle (\( \text{He}^{2+} \)) is \( +2e \).
- The charge of the copper nucleus (with atomic number \( Z = 29 \)) is \( +29e \).

2. **Potential Energy Formula**:
The electric potential energy (\( U \)) between two point charges is given by:
\[ U = \frac{k \cdot Q_1 \cdot Q_2}{r} \]
where:
- \( k = \frac{1}{4\pi \epsilon_0} \) (Coulomb's constant),
- \( Q_1 = +2e \) (charge of the alpha particle),
- \( Q_2 = +29e \) (charge of the copper nucleus),
- \( r \) is the distance of closest approach.

3. **Kinetic Energy of the Alpha Particle**:
The initial 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.

4. **At the Distance of Closest Approach**:
At the distance of closest approach, all the kinetic energy will be converted into potential energy:
\[ KE = U \]
Therefore,
\[ \frac{1}{2} mv^2 = \frac{k \cdot (2e) \cdot (29e)}{r} \]

5. **Solving for \( r \)**:
Rearranging the equation gives:
\[ r = \frac{k \cdot (2e) \cdot (29e)}{\frac{1}{2} mv^2} \]
Simplifying further, we can express \( k \) as \( \frac{1}{4\pi \epsilon_0} \):
\[ r = \frac{(1/4\pi \epsilon_0) \cdot (2e) \cdot (29e)}{\frac{1}{2} mv^2} \]
This can be simplified to:
\[ r = \frac{(58e^2)}{(2 \cdot 4\pi \epsilon_0) \cdot \frac{1}{2} mv^2} \]
\[ r = \frac{29e^2}{2\pi \epsilon_0 \cdot mv^2} \]

6. **Final Expression**:
The distance of closest approach \( r \) can be expressed in terms of known constants and the kinetic energy of the alpha particle.

### Conclusion:
The distance of closest approach for the alpha particle to the copper nucleus can be calculated using the above formula, and the final answer will depend on the specific values of \( e \), \( \epsilon_0 \), and the mass \( m \) of the alpha particle.