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If radius of ${ }_{13}^{27} \mathrm{Al}$ nucleus is estimated to be $3.6$ Fermi then the radius ${ }_{52}^{125} \mathrm{Te}$ nucleus be nearly
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6 fermi
6 fermi
$\frac{\mathrm{R}}{3.6}=\left(\frac{125}{27}\right)^{\frac{1}{3}} \Rightarrow \mathrm{R}=6$ fermi
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