A) \[6\times {{10}^{-15}}m\]
B) \[3\times {{10}^{-15}}m\]
C) \[4.5\times {{10}^{-15}}m\]
D) none of these
Correct Answer: A
Solution :
Radius of nucleus is proportional the cube root of atomic number A Hence, \[R\propto {{A}^{1/3}}\] Hence, \[\frac{{{R}_{2}}}{{{R}_{1}}}={{\left( \frac{{{A}_{2}}}{{{A}_{1}}} \right)}^{1/3}}\] ...(i) Here, \[{{R}_{1}}=3\times {{10}^{-15}}m,\,\,{{A}_{1}}16,\,{{A}_{2}}=128\] \[\frac{{{R}_{2}}}{{{R}_{1}}}={{\left( \frac{128}{16} \right)}^{1/3}}\] \[{{R}_{2}}={{\left( \frac{128}{16} \right)}^{1/3}}\times {{R}_{1}}=2{{R}_{1}}\] \[=2\times 3\times {{10}^{-15}}=6\times {{10}^{-15}}m\]
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