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Question
What is the nuclear radius of 125Fe, if that of 27Al is 3.6 fermi?
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Solution
RAl = 3.6 fermi
RFe = ?
Radius of the Nucleus R = R0 A1/3
`"R"_("Fe")/"R"_("Al") = ("A"_("Fe")/"A"_("Al"))^(1/3)`
= `(125/27)^(1/3)`
RFe = `5/3` RAl
= `5/3 xx 3.6`
= 6 fermi
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