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The radius of ${ }_{29} \mathrm{Cu}^{64}$ nucleus in Fermi is (given $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ )
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The correct answer is:
$4.8$
$\begin{aligned} \mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3} &=1.2 \times 10^{-15} \times(64)^{1 / 3} \\ &=4.8 \times 10^{-15} \mathrm{~m} \end{aligned}$ or $\quad \mathrm{R}=4.8$ fermi
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