If radius of the ₁₃Al²⁷ nucleus is taken to be Rᴀı, then the radius of ₅₃Te¹²⁵ nucleus

If radius of the ₁₃Al²⁷ nucleus is taken to be Rᴀı, then the radius of ₅₃Te¹²⁵ nucleus is nearly

(a) (53/13)¹⁄₃ Rᴀı
(b) 5/3Rᴀı
(c) 3/5Rᴀı
(d) (13/53)¹⁄₃ Rᴀı

5/3Rᴀı

Radius of the nucleus is given by,
R = R₀A¹/³ ⇒ R ∝ A¹/³
.·. Rᴀ₁ / Rᴛₑ = [Aᴀ₁ / Aᴛₑ]¹/³ = [27 / 125]¹/³ = 3 / 5
Rᴛₑ = (5 / 3) Rᴀ₁

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