Eduprobe
Question:
If radius of the 2713Al nucleus is taken to be RAl, then the radius of 12552Te nucleus is nearly:
53RAl
(1353)13RAl
(5313)13RAl
35RAl
Solution:
R = R₀A⅓
∴ RAl/RTe = (27/125)⅓ = 3/5
RTe = 5/3RAl = 5/3RAl
53RAl
(1353)13RAl
(5313)13RAl
35RAl
R = R₀A⅓
∴ RAl/RTe = (27/125)⅓ = 3/5
RTe = 5/3RAl = 5/3RAl