The radius of the oxygen nucleus $({}_{8}^{16}O)$ is 2.8 × 10^–15m. Find the radius of the lead nucleus $({}_{82}^{205}{Pb})$ .

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6.55 × 10^–13m

8.55 × 10^–15m

6.55 × 10^–15m

4.55 × 10^–12m

answer is B.

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Detailed Solution

From $R = R_{0} A^{1 / 3} \Rightarrow \frac{R_{2}}{R_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3} = {(\frac{205}{16})}^{1 / 3}$
$[{(\frac{205}{16})}^{1 / 3} = 2 .33]$

$R_{2} = 2 .8 \times 10^{- 15} \times {(\frac{205}{16})}^{1 / 3} = 6 .55 \times 10^{- 15} m$

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