In Rutherford's `alpha` particle scattering experiment with their gold foil 8100 scintillations per minute are observed at an angle of `60^(@)` . The number of seintillations per minute at `120^(@)` will be

Options

A. 100
B. 2025
C. 32400
D. 900

Answer

To solve the problem of determining the number of scintillations per minute at an angle of 120 degrees in Rutherford's alpha particle scattering experiment, we can follow these steps:

### Step 1: Understand the Given Information
We know that:
- At an angle of \( \theta_1 = 60^\circ \), the number of scintillations per minute, \( n_1 = 8100 \).
- We need to find the number of scintillations per minute, \( n_2 \), at an angle of \( \theta_2 = 120^\circ \).

### Step 2: Use the Relation for Scattering
The relationship between the number of alpha particles scattered and the angle is given by the formula:
\[ n \propto \frac{1}{\sin^4(\theta/2)} \]
This means that the number of scintillations is inversely proportional to the sine of half the angle raised to the fourth power.

### Step 3: Set Up the Equations
For \( n_1 \) at \( \theta_1 = 60^\circ \):
\[ n_1 = k \cdot \frac{1}{\sin^4(30^\circ)} \]
For \( n_2 \) at \( \theta_2 = 120^\circ \):
\[ n_2 = k \cdot \frac{1}{\sin^4(60^\circ)} \]
Where \( k \) is a constant of proportionality.

### Step 4: Relate \( n_1 \) and \( n_2 \)
Dividing the two equations:
\[ \frac{n_1}{n_2} = \frac{\sin^4(60^\circ)}{\sin^4(30^\circ)} \]
This allows us to express \( n_2 \) in terms of \( n_1 \):
\[ n_2 = n_1 \cdot \frac{\sin^4(30^\circ)}{\sin^4(60^\circ)} \]

### Step 5: Calculate the Sine Values
We know:
- \( \sin(30^\circ) = \frac{1}{2} \)
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)

Now, substituting these values:
\[ \sin^4(30^\circ) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]
\[ \sin^4(60^\circ) = \left(\frac{\sqrt{3}}{2}\right)^4 = \frac{3^2}{16} = \frac{9}{16} \]

### Step 6: Substitute Back to Find \( n_2 \)
Now substituting these into the equation for \( n_2 \):
\[ n_2 = 8100 \cdot \frac{\frac{1}{16}}{\frac{9}{16}} = 8100 \cdot \frac{1}{9} \]
\[ n_2 = \frac{8100}{9} = 900 \]

### Final Answer
The number of scintillations per minute at an angle of \( 120^\circ \) is \( n_2 = 900 \).