A) \[L{{i}^{7}}\]
B) \[N{{a}^{21}}\]
C) \[{{S}^{16}}\]
D) \[C{{a}^{40}}\]
Correct Answer: A
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 \[L{{i}^{7}}\].
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