If the nucleus ²⁷₁₃Al has nuclear radius of about 3.6 fm, then ¹²⁵₃₂Te would have its radius approximately as

Options

(a) 9.6 fm
(b) 12.0 fm
(c) 4.8 fm
(d) 6.0 fm

Correct Answer:

6.0 fm

Explanation:

It has been known that a nucleus of mass number A has radius
R = R₀A¹/³, Where R₀ = 1.2 x 10⁻¹⁵ m, and A = mass number
In case of ²⁷₁₃Al, let nuclear radius be R₁ and for ¹²⁵₃₂Te, nuclear radius be R₂
For ²⁷₁₃Al, R₁ = R₀ (27)¹/³ = 3R₀
For ¹²⁵₃₂Te, R₂ = R₀ (125)¹/³ = 5R₀
R₂ / R₁ = 5R₀ / 3R₀ = 5/3 R₁ = 5/3 x 3.6 = 6 fm.

Alternative Explanation:

Nuclear radii R = R₀ (A)¹/³

Where A is the mass number.

.·. Rтₑ / Rᴀ₁ = [Aтₑ / Aᴀ₁]¹/³ = (125 / 27)¹/³ = 5 / 3

(or), Rтₑ = (5 / 3) × Rᴀ₁ = (5 / 3) × 3.6 = 6 fm