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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 as |
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Answer» 9.6 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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