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What is the nuclear radius of `Fe^(125)`, if that of `Al^(27)` is 3.6 fermi.

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What is the nuclear radius of `Fe^(125)`, if that of `Al^(27)` is 3.6 fermi.
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`R=R_0A^(1//3)`
`R_"Fe"/R_"Al"=(A_"Fe"/A_"Al")^(1//3) = (125/27)^(1//3)`
`R_"Fe"=5/3 R_"Al"=5/3xx3.6`=6.0 fermi.
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