If the nucleus 2713Al has a nuclear radius of about 3.6 fm, then 12552Te would have its radius approximately as
- 9.6 fm
- 12.0 fm
- 4.8 fm
- 6.0 fm
Answer
(4) 6.0 fm
Solution
As per NCERT, the radius of a nucleus of mass number A is R = R0 A1/3
⇒ RTe RAl = ( 12527 )1/3 = 53
⇒ RTe = 53 3.6 fm
∴ RTe = 6.0 fm
Answer: (4)