If in Rutherford's experiment, the number of particles scattered at `90^(@)` angle are `28` per min, then number of scattered particles at an angle `60^(@)` and `120^(@)` will be

Options

A. `112//min, 12.5//min`
B. `100//min, 200//min`
C. `50//min, 12.5//min`
D. `117//min, 25//min`

Explanation

To solve the problem of finding the number of particles scattered at angles of \(60^\circ\) and \(120^\circ\) given that \(28\) particles are scattered at \(90^\circ\), we can use the relationship derived from Rutherford's scattering formula. The number of scattered particles is proportional to \( \sin^4(\theta/2) \).

### Step-by-Step Solution:

1. **Understand the relationship**:
The number of particles scattered at an angle \( \theta \) is given by:
\[ n \propto \sin^4\left(\frac{\theta}{2}\right) \]

2. **Set up the ratio for \(60^\circ\)**:
Let \( n_1 = 28 \) (number of particles scattered at \(90^\circ\)) and \( n_2 \) (number of particles scattered at \(60^\circ\)).
\[ \frac{n_1}{n_2} = \frac{\sin^4\left(\frac{60^\circ}{2}\right)}{\sin^4\left(\frac{90^\circ}{2}\right)} \]
Simplifying the angles:
\[ \frac{n_1}{n_2} = \frac{\sin^4(30^\circ)}{\sin^4(45^\circ)} \]

3. **Calculate the sine values**:
\[ \sin(30^\circ) = \frac{1}{2}, \quad \sin(45^\circ) = \frac{\sqrt{2}}{2} \]
Therefore:
\[ \sin^4(30^\circ) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}, \quad \sin^4(45^\circ) = \left(\frac{\sqrt{2}}{2}\right)^4 = \frac{4}{16} = \frac{1}{4} \]

4. **Substitute the sine values into the ratio**:
\[ \frac{28}{n_2} = \frac{\frac{1}{16}}{\frac{1}{4}} = \frac{1}{16} \times \frac{4}{1} = \frac{4}{16} = \frac{1}{4} \]

5. **Solve for \( n_2 \)**:
\[ n_2 = 28 \times 4 = 112 \text{ particles per minute} \]

6. **Set up the ratio for \(120^\circ\)**:
Let \( n_3 \) be the number of particles scattered at \(120^\circ\).
\[ \frac{n_1}{n_3} = \frac{\sin^4\left(\frac{120^\circ}{2}\right)}{\sin^4\left(\frac{90^\circ}{2}\right)} \]
Simplifying the angles:
\[ \frac{n_1}{n_3} = \frac{\sin^4(60^\circ)}{\sin^4(45^\circ)} \]

7. **Calculate the sine values**:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Therefore:
\[ \sin^4(60^\circ) = \left(\frac{\sqrt{3}}{2}\right)^4 = \frac{9}{16} \]

8. **Substitute the sine values into the ratio**:
\[ \frac{28}{n_3} = \frac{\frac{9}{16}}{\frac{1}{4}} = \frac{9}{16} \times 4 = \frac{9}{4} \]

9. **Solve for \( n_3 \)**:
\[ n_3 = 28 \times \frac{4}{9} = \frac{112}{9} \approx 12.44 \text{ particles per minute} \]

### Final Answers:
- Number of particles scattered at \(60^\circ\): \(112\) particles per minute.
- Number of particles scattered at \(120^\circ\): \(12.44\) particles per minute.