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The radius of $_{13}^{27} Al$ is $3.6$ Fermi. Find the radius of $_{52}^{125} Te$
$6$ Fermi
$8$ Fermi
$4$ Fermi
$5$ Fermi

A

8

Fermi

B

5

Fermi

C

6

Fermi

D

4

Fermi

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Solution

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$\frac{R_{1}}{R_{2}}$ = $\frac{A_{1}^{1 / 3}}{A_{2}^{1 / 3}}$
$R_{2}$ = ${(\frac{125}{27})}^{1 / 3} \times 3.6$ $= 6$ Fermi

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