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Q.

Radius of $_{2} H e^{4}$ nucleus is 3 Fermi. The radius of $_{16} S^{32}$ nucleus will be

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a

4 Fermi

b

5 Fermi

c

6 Fermi

d

8 Fermi

answer is A.

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Detailed Solution

$R \propto A^{1 / 3} \frac{R_{2}}{R_{1}} = {(\frac{A_{2}}{A_{1}})}^{1 / 3} = {(\frac{32}{4})}^{1 / 3} = 2 R_{2} = 2 R_{1} = 2 \times 3 = 6 F e r m i$

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