The radius of the oxygen nucleus `(._8^16O)` is `2.8xx10^(-15)`m. Find the radius of the lead nucleus `(._82^205Pb)`.

Options

A. `8.55xx10^(-15)` m
B. `6.55xx10^(-15)` m
C. `6.55xx10^(-13)` m
D. `4.55xx10^(-12)` m

Answer

To find the radius of the lead nucleus \((^{205}_{82}Pb)\) given the radius of the oxygen nucleus \((^{16}_{8}O)\), we can use the relationship between the radius \(R\) of a nucleus and its atomic mass number \(A\). The relationship is given by:

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

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Radius of oxygen nucleus, \(R_O = 2.8 \times 10^{-15} \, \text{m}\)
- Atomic mass of oxygen, \(A_O = 16\)
- Atomic mass of lead, \(A_{Pb} = 205\)

2. **Set Up the Ratio**:
Using the proportionality relationship, we can write:
\[ \frac{R_O}{R_{Pb}} = \left(\frac{A_O}{A_{Pb}}\right)^{1/3} \]

3. **Substitute the Known Values**:
Substitute the known values into the equation:
\[ \frac{2.8 \times 10^{-15}}{R_{Pb}} = \left(\frac{16}{205}\right)^{1/3} \]

4. **Calculate the Right Side**:
First, calculate \(\frac{16}{205}\):
\[ \frac{16}{205} \approx 0.0780 \]
Now, take the cube root:
\[ \left(0.0780\right)^{1/3} \approx 0.0437 \]

5. **Rearrange to Find \(R_{Pb}\)**:
Rearranging the equation gives:
\[ R_{Pb} = \frac{2.8 \times 10^{-15}}{0.0437} \]

6. **Perform the Calculation**:
Now, calculate \(R_{Pb}\):
\[ R_{Pb} \approx \frac{2.8 \times 10^{-15}}{0.0437} \approx 6.41 \times 10^{-15} \, \text{m} \]

7. **Final Result**:
The radius of the lead nucleus is approximately:
\[ R_{Pb} \approx 6.41 \times 10^{-15} \, \text{m} \]

### Summary:
The radius of the lead nucleus \((^{205}_{82}Pb)\) is approximately \(6.41 \times 10^{-15} \, \text{m}\).