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Q.
The stable nucleus that has a radius half that of $ F{{e}^{56}} $ is :
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Solution:
Key Idea: The radius of nucleus is proportional to cube root of atomic mass. The nuclear radius is $ R\propto {{A}^{1/3}} $ ?...(i) or $ A\propto {{R}^{3}} $ or $ \frac{{{A}_{1}}}{{{A}_{2}}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}} $ Here, $ {{R}_{1}}=R,{{R}_{2}}=\frac{R}{2},A=56 $ $ \therefore $ $ \frac{56}{{{A}^{2}}}={{\left( \frac{R}{R/2} \right)}^{3}}={{2}^{3}}=8 $ or $ {{A}_{2}}=\frac{56}{8}=7 $ Thus, stable nucleus will be $ L{{i}^{7}} $ .