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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
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Across a metallic conductor of non-uniform cross-section, a constant potential
- Planck’s constant has same dimensions as those of
- The energy released by the fission of one uranium atom is 200 MeV.
- A ball is dropped from a height of 20 cm. Ball rebounds to a height of 10cm
- A glass flask weighing 390 g having internal volume 500 cc just floats
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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