In experiment of Rutherford scattering, the number of `alpha` - particles scattered at `60^@` is `4.5 xx 10^6`. The number of a - particles scattered at `120^@` will be

Options

A. `0.4 xx 10^6`
B. `0.5 xx 10^6`
C. `0.4 xx 10^5`
D. `0.5 xx 10^5`

Explanation

To solve the problem of determining the number of alpha particles scattered at an angle of \(120^\circ\) given the number of alpha particles scattered at \(60^\circ\), we can use the relationship derived from Rutherford's scattering theory.

### Step-by-Step Solution:

1. **Identify Given Values**:
- The number of alpha particles scattered at \(60^\circ\) (denoted as \(n_1\)) is \(4.5 \times 10^6\).
- The angle for which we want to find the number of scattered particles is \(120^\circ\) (denoted as \(n_2\)).

2. **Understanding the Relationship**:
- The number of alpha particles scattered at an angle \(\theta\) is inversely proportional to \(\sin^4(\theta/2)\).
- This can be expressed mathematically as:
\[ \frac{n_1}{n_2} = \frac{\sin^4(\theta_2/2)}{\sin^4(\theta_1/2)} \]

3. **Substituting the Angles**:
- For \(\theta_1 = 60^\circ\) and \(\theta_2 = 120^\circ\):
- \(\theta_1/2 = 30^\circ\)
- \(\theta_2/2 = 60^\circ\)
- Therefore, we can rewrite the equation as:
\[ \frac{n_1}{n_2} = \frac{\sin^4(60^\circ)}{\sin^4(30^\circ)} \]

4. **Calculating the Sine Values**:
- We know:
- \(\sin(30^\circ) = \frac{1}{2}\)
- \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)
- Now substituting these values into the equation:
\[ \frac{n_1}{n_2} = \frac{\left(\frac{\sqrt{3}}{2}\right)^4}{\left(\frac{1}{2}\right)^4} \]

5. **Simplifying the Equation**:
- Simplifying the right-hand side:
\[ \frac{n_1}{n_2} = \frac{\frac{3}{4}}{\frac{1}{16}} = \frac{3}{4} \times 16 = 12 \]
- Thus, we have:
\[ n_2 = \frac{n_1}{12} \]

6. **Calculating \(n_2\)**:
- Now substituting \(n_1 = 4.5 \times 10^6\):
\[ n_2 = \frac{4.5 \times 10^6}{12} = 0.375 \times 10^6 = 3.75 \times 10^5 \]

### Final Answer:
The number of alpha particles scattered at \(120^\circ\) is \(3.75 \times 10^5\).