The radius of { }_{29} Cu^{64} nucleus in Fer | vedclass

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(B) The radius of a nucleus is given by the formula $R = R_{0} A^{1/3}$,where $R_{0} = 1.2 	imes 10^{-15} \ m$ and $A$ is the mass number.For ${ }_{2

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$4.8$

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(B) The radius of a nucleus is given by the formula $R = R_{0} A^{1/3}$,where $R_{0} = 1.2 	imes 10^{-15} \ m$ and $A$ is the mass number.For ${ }_{29} Cu^{64}$,the mass number $A = 64$.Substituting the values into the formula:$R = 1.2 	imes 10^{-15} 	imes (64)^{1/3} \ m$Since $(64)^{1/3} = 4$,we have:$R = 1.2 	imes 4 	imes 10^{-15} \ m$$R = 4.8 	imes 10^{-15} \ m$Since $1 \ 	ext{Fermi} = 10^{-15} \ m$,the radius is $4.8 \ 	ext{Fermi}$.

The radius of ${ }_{29} Cu^{64}$ nucleus in Fermi is (given $R_{0}=1.2 \times 10^{-15} \ m$)

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