A) 9.6 fm
B) 12.0 fm
C) 4.8 fm
D) 6.0 fm.
Correct Answer: D
Solution :
[d] It has been known that a nucleus of mass number A has radius \[R={{R}_{0}}{{A}^{1/3}},\] where \[{{R}_{0}}=1.2\times {{10}^{-15}}m\] and A = mass number In case of \[_{13}^{27}A\ell ,\] let nuclear radius be \[{{R}_{1}}\] and for \[_{32}^{125}Te,\] nuclear radius be \[{{R}_{2}}\] For \[_{13}^{27}Al,\,\,\,{{R}_{1}}={{R}_{0}}{{(27)}^{1/3}}=3{{R}_{0}}\] For \[_{32}^{125}Te,\,\,{{R}_{2}}={{R}_{0}}{{(125)}^{1/3}}=5{{R}_{0}}\] \[\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{5{{R}_{0}}}{3{{R}_{0}}}=\frac{5}{3}{{R}_{1}}=\frac{5}{3}\times 3.6=6fm\]
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