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 scatterd at an angle of`90^(@)` are

Options

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

Answer

To solve the problem regarding the Rutherford scattering experiment, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The number of alpha particles scattered at an angle θ is directly proportional to \( \frac{1}{\sin^4(\frac{\theta}{2})} \). Thus, we can express this relationship mathematically as:
\[ n \propto \frac{1}{\sin^4(\frac{\theta}{2})} \]
where \( n \) is the number of alpha particles scattered.

2. **Setting Up the Equation**:
We introduce a constant \( k \) to remove the proportionality:
\[ n = \frac{k}{\sin^4(\frac{\theta}{2})} \]

3. **Finding the Constant \( k \)**:
We know that when \( \theta = 60^\circ \), the number of alpha particles \( n = 12 \) per minute. We can substitute these values into the equation:
\[ 12 = \frac{k}{\sin^4(30^\circ)} \]
Since \( \sin(30^\circ) = \frac{1}{2} \), we have:
\[ \sin^4(30^\circ) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]
Substituting this back, we get:
\[ 12 = \frac{k}{\frac{1}{16}} \implies k = 12 \times \frac{1}{16} = 12 \times 16 = 192 \]

4. **Calculating for \( \theta = 90^\circ \)**:
Now, we need to find the number of alpha particles scattered at \( \theta = 90^\circ \):
\[ n = \frac{k}{\sin^4(45^\circ)} \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we have:
\[ \sin^4(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} \]
Substituting \( k = 192 \) into the equation:
\[ n = \frac{192}{\frac{1}{4}} = 192 \times 4 = 768 \]

5. **Final Answer**:
The number of alpha particles scattered at an angle of \( 90^\circ \) is \( 768 \) per minute.