A) 6.0 fm
B) 9.6 fm
C) 12.0 fm
D) 4.8 fm
Correct Answer: A
Solution :
| If R is the radius of the nucleus, the corresponding volume \[\frac{4}{3}\pi {{R}^{3}}\] has been found to be proportional to A. This relationship is expressed in inverse form as \[R={{R}_{0}}{{A}^{1/3}}\] |
| The value of \[{{R}_{0}}\] is \[1.2\times {{10}^{-15}}\,m\] i.e., 1.2 fm |
| Therefore, \[\frac{{{R}_{Al}}}{{{R}_{Te}}}=\frac{{{R}_{0}}{{({{A}_{Al}})}^{1/3}}}{{{R}_{0}}{{({{A}_{Te}})}^{1/3}}}\] |
| \[\frac{{{R}_{Al}}}{{{R}_{Te}}}=\frac{{{({{A}_{Al}})}^{1/3}}}{{{({{A}_{Te}})}^{1/3}}}\] |
| \[=\frac{{{(27)}^{1/3}}}{{{(125)}^{1/3}}}=\frac{3}{5}\] |
| or \[{{R}_{Te}}=\frac{5}{3}\times {{R}_{Al}}=\frac{5}{3}\times 3.6\] |
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