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If the nucleus ${ }_{13}^{27} \mathrm{Al}$ has a nuclear radius of about $3.6 \mathrm{fm}$, then ${ }_{32}^{125} \mathrm{Te}$ would have its radius approximately as:
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The correct answer is:
$6.0 \mathrm{fm}$.
Nuclear radii \(R=\left(R_0\right) A^{1 / 3}\)
where, \(A\) is the mass number.
\(\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}}=\left(\frac{A_{\mathrm{Te}}}{A_{\mathrm{Al}}}\right)^{1 / 3}=\left(\frac{125}{27}\right)^{1 / 3}=\left(\frac{5}{3}\right)\)
or, \(\quad R_{\mathrm{Te}}=\frac{5}{3} \times R_{\mathrm{Al}}=\frac{5}{3} \times 3.6=6 \mathrm{fm}\).
(Given \(R_{\mathrm{Al}}=3.6 \mathrm{fm}\))
where, \(A\) is the mass number.
\(\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}}=\left(\frac{A_{\mathrm{Te}}}{A_{\mathrm{Al}}}\right)^{1 / 3}=\left(\frac{125}{27}\right)^{1 / 3}=\left(\frac{5}{3}\right)\)
or, \(\quad R_{\mathrm{Te}}=\frac{5}{3} \times R_{\mathrm{Al}}=\frac{5}{3} \times 3.6=6 \mathrm{fm}\).
(Given \(R_{\mathrm{Al}}=3.6 \mathrm{fm}\))
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