Rankers Physics

Nucleus: Practice Problem & Solution

Surface area of \[Al^{27}\] is S0, then surface area of \[Zn^{64}\] is :
\[\frac{16}{9}S_{0}\]
\[\frac{4}{3}S_{0}\]
\[\frac{64}{27}S_{0}\]
\[\frac{3}{2}S_{0}\]

No solution provided for this question.

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