If the nucleus ₁₃²⁷Al has a nuclear radius of about 3.6 fm, them ₅₂¹²⁵Te would have its radius approximately as (a) 9.6 fm (b) 12.0 fm (c) 4.8 fm (d) 6.0 fm

Answer: (d). R ∝ A1/3, so RTe = 3.6 × (125/27)1/3. = 3.6 × 5/3 = 6.0 fm So the answer is (d) 6.0 fm.

If the nucleus ₁₃²⁷Al has a nuclear radius of about 3.6 fm, them… | NEET 2007 (AIPMT)

Q30NEET 2007 · AIPMTEasy

Nucleus: Composition, Size & Nuclear Force

If the nucleus ${}_{13}^{27}Al$ has a nuclear radius of about 3.6 fm, them ${}_{52}^{125}Te$ would have its radius approximately as

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Answer: (d)

$R \propto A^{1/3}$ , so $R_{Te} = 3.6 \times \left(\frac{125}{27}\right)^{1/3}$ .
$= 3.6 \times \frac{5}{3} = 6.0$ fm
So the answer is (d) 6.0 fm.