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The radius of `._3Al^(27)` nucleus is 5 fermi. Find the radius of `._52Te^(125)` nucleus. |
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Answer» Here, `A_1=27, R_1=6 fermi` `A_2=125,R_2=?` As `(R_2)/(R_1)=((A_2)/(A_1))^(1//3)=(125/27)^(1//3)=5/3` `R_2=5/3R_1=5/3xx6=10fermi` |
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