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The nucleus is approximately spherical in shape. Then the surface area of nucleus haviing mass number A varies as. |
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Answer» `A^(2//3)` `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)` |
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