Radius of `._2^4 He` nucleus is `3` Fermi. The radius of `._82^206 Pb` nucleus will be.

Options

A. `5` Fermi
B. `6` Fermi
C. `11.16` Fermi
D. `8` Fermi

Explanation

To find the radius of the \( _{82}^{206}Pb \) nucleus given the radius of the \( _{2}^{4}He \) nucleus, we can use the formula that relates the radius of a nucleus to its mass number \( A \):

\[ R \propto A^{1/3} \]

### Step-by-Step Solution:

1. **Identify the Given Values**:
- The radius of the helium nucleus (\( R_1 \)) is given as 3 Fermi.
- The mass number of helium (\( A_1 \)) is 4.
- The mass number of lead (\( A_2 \)) is 206.

2. **Use the Proportionality Relation**:
We know that the radius of the nucleus is proportional to the cube root of its mass number. Therefore, we can write:
\[ \frac{R_2}{R_1} = \left(\frac{A_2}{A_1}\right)^{1/3} \]

3. **Substituting the Values**:
Substitute \( R_1 = 3 \) Fermi, \( A_1 = 4 \), and \( A_2 = 206 \) into the equation:
\[ \frac{R_2}{3} = \left(\frac{206}{4}\right)^{1/3} \]

4. **Calculate the Ratio of Mass Numbers**:
Calculate \( \frac{206}{4} \):
\[ \frac{206}{4} = 51.5 \]

5. **Calculate the Cube Root**:
Now, calculate the cube root of 51.5:
\[ R_2 = 3 \times (51.5)^{1/3} \]

6. **Finding the Cube Root**:
Using a calculator or estimation:
\[ (51.5)^{1/3} \approx 3.68 \quad (\text{approximately}) \]

7. **Final Calculation**:
Now calculate \( R_2 \):
\[ R_2 \approx 3 \times 3.68 \approx 11.04 \text{ Fermi} \]

### Conclusion:
The radius of the \( _{82}^{206}Pb \) nucleus is approximately 11.04 Fermi.