The mass number of He is 4 and that for sulph | vedclass

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(C) The radius $R$ of a nucleus with mass number $A$ is given by the relation $R = R_0 A^{1/3}$,where $R_0$ is a constant.Therefore,the ratio of the r

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$2$

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(C) The radius $R$ of a nucleus with mass number $A$ is given by the relation $R = R_0 A^{1/3}$,where $R_0$ is a constant.Therefore,the ratio of the radii of two nuclei is given by $\frac{R_S}{R_{He}} = \left( \frac{A_S}{A_{He}} \right)^{1/3}$.Given $A_S = 32$ and $A_{He} = 4$,we have:$\frac{R_S}{R_{He}} = \left( \frac{32}{4} \right)^{1/3} = (8)^{1/3} = 2$.Thus,the radius of the sulphur nucleus is $2$ times larger than that of the helium nucleus.

The mass number of $He$ is $4$ and that for sulphur is $32$. The radius of the sulphur nucleus is larger than that of the helium nucleus by a factor of:

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