When a silver foil (Z = 47) was used in an `alpha ` ray scattering experiment , the number of `alpha`-particles scattered at `30^(@)` was found to be 200 per minute. If the silver foil is replaced by alumninium (z = 13) foil of same thickness, the number of `alpha`-particles scattered per minute at `30^(@)` is nearly equal to

Options

A. 15
B. 30
C. 10
D. 20

Answer

To solve the problem, we need to determine the number of alpha particles scattered per minute when the silver foil is replaced by aluminum foil of the same thickness. We will use the relationship between the number of scattered alpha particles and the atomic number of the material.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Atomic number of silver (Z₁) = 47
- Number of alpha particles scattered with silver (N₁) = 200 particles/minute
- Atomic number of aluminum (Z₂) = 13

2. **Understand the Relationship**:
- The number of alpha particles scattered (N) is directly proportional to the square of the atomic number (Z) of the material:
\[ N \propto Z^2 \]
- Therefore, we can express this relationship for both materials:
\[ \frac{N_1}{N_2} = \frac{Z_1^2}{Z_2^2} \]

3. **Set Up the Equation**:
- Substitute the known values into the equation:
\[ \frac{200}{N_2} = \frac{47^2}{13^2} \]

4. **Calculate the Squares**:
- Calculate \(47^2\) and \(13^2\):
\[ 47^2 = 2209 \]
\[ 13^2 = 169 \]

5. **Substitute and Solve for N₂**:
- Now substitute these values back into the equation:
\[ \frac{200}{N_2} = \frac{2209}{169} \]
- Cross-multiply to solve for \(N_2\):
\[ 200 \cdot 169 = 2209 \cdot N_2 \]
\[ 33800 = 2209 \cdot N_2 \]
\[ N_2 = \frac{33800}{2209} \]

6. **Calculate N₂**:
- Performing the division:
\[ N_2 \approx 15.3 \]
- Since we are looking for the number of alpha particles, we can round this to:
\[ N_2 \approx 15 \text{ particles/minute} \]

### Final Answer:
The number of alpha particles scattered per minute at \(30^\circ\) when using aluminum foil is approximately **15 particles/minute**.