A heavy nucleus of mass number 64 hypothetically breaks into 8 identical lighter nuclei, each of mass number 8 . The ratio of the total surface area of all the 8 lighter nuclei to the surface area of the original nucleus is:

Step-by-step solution

The radius of a nucleus is given by R = R₀ A^ 1/3 . The surface area of a nucleus is S = 4 R^2 . Substituting the expression for R , we get S A^ 2/3 . For the original nucleus ( A = 64 ), the surface area is: S₀ (64)^ 2/3 = 16 For one lighter nucleus ( A' = 8 ), the surface area is: S₁ (8)^ 2/3 = 4 The total surface area of all 8 lighter nuclei is: S_ total = 8 S₁ 8 4 = 32 The required ratio is: S_ total S₀ = 32 16 = 2 1 Thus, the ratio is 2:1 . Answer: 2:1