An alpha particle of energy `5 MeV` is scattered through `180^(@)` by a found uramiam nucleus . The distance of appreach is of the order of

Options

A. `1Å`
B. `10^(-10) cm`
C. `10^(-12) cm`
D. `10^(-15) cm`

Explanation

To solve the problem of finding the distance of closest approach for an alpha particle scattering off a uranium nucleus, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have an alpha particle with energy \(E = 5 \, \text{MeV}\) scattering through \(180^\circ\) by a uranium nucleus. We need to find the distance of closest approach.

2. **Convert Energy to Joules**:
The energy of the alpha particle in joules can be calculated using the conversion \(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\).
\[ E = 5 \, \text{MeV} = 5 \times 1.6 \times 10^{-13} \, \text{J} = 8 \times 10^{-13} \, \text{J} \]

3. **Use the Concept of Energy Conservation**:
At the distance of closest approach, the kinetic energy of the alpha particle is converted into electrostatic potential energy. The potential energy \(U\) between two charges is given by:
\[ U = \frac{k \cdot |Q_1| \cdot |Q_2|}{r} \]
where \(k\) is Coulomb's constant (\(k \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2\)), \(Q_1\) is the charge of the alpha particle, \(Q_2\) is the charge of the uranium nucleus, and \(r\) is the distance of closest approach.

4. **Identify the Charges**:
- The charge of the alpha particle \(Q_1 = 2e\) (since it consists of 2 protons).
- The charge of the uranium nucleus \(Q_2 = 92e\) (since uranium has 92 protons).
- The elementary charge \(e \approx 1.6 \times 10^{-19} \, \text{C}\).

5. **Set Up the Equation**:
At the closest approach:
\[ \frac{k \cdot (2e) \cdot (92e)}{r} = E \]
Substituting the values:
\[ \frac{9 \times 10^9 \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (92 \cdot 1.6 \times 10^{-19})}{r} = 8 \times 10^{-13} \]

6. **Rearranging to Solve for \(r\)**:
\[ r = \frac{9 \times 10^9 \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (92 \cdot 1.6 \times 10^{-19})}{8 \times 10^{-13}} \]

7. **Calculate \(r\)**:
First, calculate the numerator:
\[ 9 \times 10^9 \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (92 \cdot 1.6 \times 10^{-19}) = 9 \times 10^9 \cdot 2 \cdot 92 \cdot (1.6 \times 10^{-19})^2 \]
\[ = 9 \times 10^9 \cdot 184 \cdot 2.56 \times 10^{-38} = 9 \times 10^9 \cdot 4.6944 \times 10^{-36} \approx 4.22496 \times 10^{-26} \]
Now substituting back:
\[ r \approx \frac{4.22496 \times 10^{-26}}{8 \times 10^{-13}} \approx 5.28 \times 10^{-14} \, \text{m} = 5.28 \times 10^{-12} \, \text{cm} \]

8. **Final Result**:
The distance of closest approach is approximately \(5.28 \times 10^{-12} \, \text{cm}\), which is of the order of \(10^{-12} \, \text{cm}\).