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If the nucleus of .13Al^27 has a nuclear...

If the nucleus of `._13Al^27` has a nuclear radius of about 3.6 fm, then `._52Te^125` would have its radius approximately as

A

9.6 fm

B

12 fm

C

4.8 fm

D

6 fm

Text Solution

Verified by Experts

The correct Answer is:
D
Here, `A_1`=27 , `A_2`=125 , `R_1`=3.6 fm
As, `R_2/R_1=(A_2/A_1)^(1//3)=(125/27)^(1//3)=5/3`
`therefore R_2=5/3R_1=5/3xx3.6=6` fm
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