The ratio of the volume of an Al^{27} nucleus | vedclass

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(D) The volume of a nucleus is given by $V = \frac{4}{3} \pi R^3$ and its surface area is $S = 4 \pi R^2$.The ratio of volume to surface area is $\fra

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$1.2 	imes 10^{-15} \ m$

Vedclass · Answer

(D) The volume of a nucleus is given by $V = \frac{4}{3} \pi R^3$ and its surface area is $S = 4 \pi R^2$.The ratio of volume to surface area is $\frac{V}{S} = \frac{\frac{4}{3} \pi R^3}{4 \pi R^2} = \frac{R}{3}$.The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number.For $Al^{27}$,$A = 27$. Thus,$R = R_0 (27)^{1/3} = 3 R_0$.Substituting this into the ratio: $\frac{V}{S} = \frac{3 R_0}{3} = R_0$.Given $R_0 = 1.2 	imes 10^{-15} \ m$,the ratio is $1.2 	imes 10^{-15} \ m$.

The ratio of the volume of an $Al^{27}$ nucleus to its surface area is (Given,$R_0 = 1.2 \times 10^{-15} \ m$):

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