The nucleus is approximately spherical in shape. Then the surface area of nucleus haviing mass number A varies as.

Options

A. `A^(2//3)`
B. `A^(4//3)`
C. `A^(1//3)`
D. `A^(5//3)`

Explanation

Volume of nucleus is proportional to mass number
`4/3pi R^3 prop A , R =R_0 A^(1//3)`
So, `4piR^2=R_0A^(2//3) rArr 4piR^2 prop A^(2//3)`
Surface area is proportional to `"(mass number)"^(2//3)`