An `alpha` - particle after passing through a potential difference of V volts collides with a nucleus . If the atomic number of the nucleus is Z then the distance of closest approach of `alpha` – particle to the nucleus will be

Options

A. `14.4(Z)/(V)Å`
B. `14.4(Z)/(V)m`
C. `14.4(Z)/(V)cm`
D. all of these

Answer

To find the distance of closest approach of an alpha particle to a nucleus with atomic number Z after passing through a potential difference of V volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Energy Conversion**:
- When the alpha particle passes through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field. This kinetic energy \( KE \) can be expressed as:
\[ KE = 2eV \]
- Here, \( e \) is the charge of the alpha particle (which is \( 2e \) since it contains 2 protons).

2. **Setting Up the Potential Energy**:
- As the alpha particle approaches the nucleus, its kinetic energy is converted into electrostatic potential energy \( PE \) at the closest approach distance \( r_0 \). The potential energy between the alpha particle and the nucleus is given by:
\[ PE = \frac{1}{4\pi \epsilon_0} \cdot \frac{(Z \cdot e)(2e)}{r_0} \]
- Here, \( Z \) is the atomic number of the nucleus, and \( 2e \) is the charge of the alpha particle.

3. **Equating Kinetic and Potential Energy**:
- At the closest approach, the kinetic energy is equal to the potential energy:
\[ 2eV = \frac{1}{4\pi \epsilon_0} \cdot \frac{(Z \cdot e)(2e)}{r_0} \]

4. **Simplifying the Equation**:
- Cancel \( 2e \) from both sides:
\[ eV = \frac{Z \cdot e^2}{4\pi \epsilon_0 r_0} \]
- Further simplifying gives:
\[ r_0 = \frac{Z \cdot e^2}{4\pi \epsilon_0 V} \]

5. **Substituting Constants**:
- The value of \( \frac{e^2}{4\pi \epsilon_0} \) is known to be approximately \( 14.4 \, \text{MeV} \cdot \text{fm} \) (where 1 fm = \( 10^{-15} \) m).
- Thus, we can write:
\[ r_0 = \frac{14.4 \, Z}{V} \, \text{fm} \]

### Final Result:
The distance of closest approach \( r_0 \) of the alpha particle to the nucleus is given by:
\[ r_0 = \frac{14.4 \, Z}{V} \, \text{fm} \]