If radius of the ₁₃²⁷Al nucleus is estimated to be 3.6 fermi then the radius of ₅₂¹²⁵Te nucleus is estimated to be nearly (a) 8 fermi (b) 6 fermi (c) 5 fermi (d) 4 fermi

Answer: (b). R ∝ A1/3, so RTe = 3.6 × (125/27)1/3. = 3.6 × 5/3 = 6 fermi So the answer is (b) 6 fermi.

If radius of the ₁₃²⁷Al nucleus is estimated to be 3.6 fermi then the… | JEE Main 2005 (AIEEE)

Q33JEE Main 2005 · AIEEEEasy

Nucleus: Composition, Size & Nuclear Force

If radius of the ${}_{13}^{27}Al$ nucleus is estimated to be $3.6$ fermi then the radius of ${}_{52}^{125}Te$ nucleus is estimated to be nearly

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Answer: (b)

$R \propto A^{1/3}$ , so $R_{Te} = 3.6 \times \left(\frac{125}{27}\right)^{1/3}$ .
$= 3.6 \times \frac{5}{3} = 6$ fermi
So the answer is (b) 6 fermi.