1.

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

Answer»

9.6 FM
12 fm
4.8 fm
6 fm

SOLUTION :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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