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If the nucleus of `._13Al^27` has a nuclear radius of about 3.6 fm, then `._52Te^125` would have its radius approximately asA. 9.6 fmB. 12 fmC. 4.8 fmD. 6 fm |
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Answer» Correct Answer - d Here, `A_1`=27 , `A_2`=125 , `R_1`=3.6 fm As, `R_2/R_1=(A_2/A_1)^(1//3)=(125/27)^(1//3)=5/3` `therefore R_2=5/3R_1=5/3xx3.6=6` fm |
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