If the surface areas of two nuclei are in the | vedclass

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(A) The surface area $S$ of a nucleus is given by $S = 4\pi R^2$, where $R$ is the radius of the nucleus.Given that the radius $R$ is proportional to

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$27 : 343$

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(A) The surface area $S$ of a nucleus is given by $S = 4\pi R^2$, where $R$ is the radius of the nucleus.Given that the radius $R$ is proportional to the cube root of the mass number $A$, i.e., $R = R_0 A^{1/3}$.Therefore, the surface area $S \propto R^2 \propto (A^{1/3})^2 = A^{2/3}$.Given the ratio of surface areas $S_1/S_2 = 9/49$.Since $S_1/S_2 = (A_1/A_2)^{2/3}$, we have $(A_1/A_2)^{2/3} = 9/49$.Taking the power of $3/2$ on both sides: $A_1/A_2 = (9/49)^{3/2} = ( (3/7)^2 )^{3/2} = (3/7)^3$.Calculating the value: $A_1/A_2 = 27/343$.Thus, the ratio of their mass numbers is $27 : 343$.

If the surface areas of two nuclei are in the ratio $9 : 49$, then the ratio of their mass numbers is

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