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what is the nuclear radius ` of .^(125)` Fe , if that of `"".^(27) Al `is 6.4 fermi? |
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Answer» Radius of the nucleus is , `R=R_(0)A^(1//3)` `(R_(Fe))/(R_(Al))=((A_(Fe))/(A_(Al)))^(1//3)=((125)/(27))^(1//3)` `Radius ,R_(Fe) =(5)/(3) R_(Al)=(5)/(3)xx6.4=10Fm` |
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