The radius of a _2He^4 nucleus is 3 , fm. The | vedclass

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(C) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number.This implies that $R \propto A^{1/3}$.Given for $_2

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$11.16$

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(C) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number.This implies that $R \propto A^{1/3}$.Given for $_2He^4$,$R_1 = 3 \, fm$ and $A_1 = 4$.For $_{82}Pb^{206}$,$A_2 = 206$.Using the ratio: $\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} ight)^{1/3}$.$\frac{R_2}{3} = \left( \frac{206}{4} ight)^{1/3} = (51.5)^{1/3}$.Calculating the cube root: $(51.5)^{1/3} \approx 3.72$.Therefore,$R_2 = 3 	imes 3.72 = 11.16 \, fm$.Thus,the correct option is $C$.

The radius of a $_2He^4$ nucleus is $3 \, fm$. The radius of a $_{82}Pb^{206}$ nucleus will be .......... $fm$.

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