In the Rutherford scattering experiment the number of `alpha` particles scattered at anangle `theta = 60^(@)` is 12 per min. The number of `alpha` particles per minute when scattered at an angle of `90^(@)` are

Options

A. 160
B. 10
C. 6
D. 3

Answer

To solve the problem, we need to determine the number of alpha particles scattered at an angle of 90 degrees based on the number of alpha particles scattered at 60 degrees. We will use the relationship derived from Rutherford's scattering experiment.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
According to Rutherford's scattering experiment, the number of alpha particles scattered at an angle θ is directly proportional to \( \frac{Z^2}{\sin^4(\frac{\theta}{2})} \), where Z is the atomic number of the alpha particle.

2. **Setting Up the Equations**:
- For \( \theta = 60^\circ \):
\[ N_{60} \propto \frac{Z^2}{\sin^4(30^\circ)} \]
Given that \( N_{60} = 12 \) particles per minute.

- For \( \theta = 90^\circ \):
\[ N_{90} \propto \frac{Z^2}{\sin^4(45^\circ)} \]

3. **Dividing the Two Equations**:
To find the ratio of \( N_{60} \) to \( N_{90} \), we can divide the two equations:
\[ \frac{N_{60}}{N_{90}} = \frac{\frac{Z^2}{\sin^4(30^\circ)}}{\frac{Z^2}{\sin^4(45^\circ)}} \]
The \( Z^2 \) terms cancel out:
\[ \frac{N_{60}}{N_{90}} = \frac{\sin^4(45^\circ)}{\sin^4(30^\circ)} \]

4. **Calculating the Sine Values**:
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

Therefore:
\[ \sin^4(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} \]
\[ \sin^4(30^\circ) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

5. **Substituting the Sine Values**:
Now substituting these values back into the ratio:
\[ \frac{N_{60}}{N_{90}} = \frac{\frac{1}{4}}{\frac{1}{16}} = \frac{16}{4} = 4 \]

6. **Finding \( N_{90} \)**:
Now we can express \( N_{90} \) in terms of \( N_{60} \):
\[ N_{90} = \frac{N_{60}}{4} \]
Substituting \( N_{60} = 12 \):
\[ N_{90} = \frac{12}{4} = 3 \]

### Final Answer:
The number of alpha particles scattered at an angle of \( 90^\circ \) is **3 particles per minute**.

---