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If the nuclear radius of ${ }^{27} \mathrm{Al}$ is $3.6 \mathrm{Fermi}$, the approximate nuclear radius of ${ }^{64} \mathrm{Cu}$ in Fermi is
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Verified Answer
The correct answer is:
4.8
Nuclear radius $r \propto A^{1 / 3}$, where $A$ is mass number
$\begin{aligned}
& r=r_0 A^{1 / 3} \\
& r=r_0(27)^{1 / 3}=3 r_0 \\
& r_0=\frac{3.6}{3}=1.2 \mathrm{fm}
\end{aligned}$
For ${ }^{64} \mathrm{Cu}$
$\begin{aligned}
r & =r_0 A^{1 / 3} \\
& =1.2 \mathrm{fm}(64)^{1 / 3} \\
& =4.8 \mathrm{fm}
\end{aligned}$
$\begin{aligned}
& r=r_0 A^{1 / 3} \\
& r=r_0(27)^{1 / 3}=3 r_0 \\
& r_0=\frac{3.6}{3}=1.2 \mathrm{fm}
\end{aligned}$
For ${ }^{64} \mathrm{Cu}$
$\begin{aligned}
r & =r_0 A^{1 / 3} \\
& =1.2 \mathrm{fm}(64)^{1 / 3} \\
& =4.8 \mathrm{fm}
\end{aligned}$
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