The nuclear radius of `._(8)^(16)O` is `3 xx 10^(-15) m`. What is the nuclear radius of `._(82)^(205)Pb`?

Answer

To find the nuclear radius of \( _{82}^{205}Pb \) given the nuclear radius of \( _{8}^{16}O \), we can use the relationship that the nuclear radius \( R \) is proportional to the cube root of the mass number \( A \):

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

### Step-by-Step Solution:

1. **Identify the mass numbers**:
- For oxygen \( _{8}^{16}O \), the mass number \( A_O = 16 \).
- For lead \( _{82}^{205}Pb \), the mass number \( A_{Pb} = 205 \).

2. **Write the proportionality relationship**:
\[ \frac{R_{Pb}}{R_O} = \left( \frac{A_{Pb}}{A_O} \right)^{1/3} \]

3. **Substitute the known values**:
\[ R_O = 3 \times 10^{-15} \text{ m} \]
\[ \frac{R_{Pb}}{3 \times 10^{-15}} = \left( \frac{205}{16} \right)^{1/3} \]

4. **Calculate the ratio of mass numbers**:
\[ \frac{205}{16} = 12.8125 \]

5. **Take the cube root**:
\[ \left( 12.8125 \right)^{1/3} \approx 2.32 \]

6. **Calculate \( R_{Pb} \)**:
\[ R_{Pb} = 2.32 \times (3 \times 10^{-15}) \text{ m} \]
\[ R_{Pb} \approx 6.96 \times 10^{-15} \text{ m} \]

7. **Convert to Fermi (1 Fermi = \( 10^{-15} \) m)**:
\[ R_{Pb} \approx 6.96 \text{ Fermi} \]

### Final Answer:
The nuclear radius of \( _{82}^{205}Pb \) is approximately \( 6.96 \times 10^{-15} \text{ m} \) or \( 6.96 \text{ Fermi} \).

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