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The radius of .3Al^(27) nucleus is 5 fer...

The radius of `._3Al^(27)` nucleus is 5 fermi. Find the radius of `._52Te^(125)` nucleus.

Text Solution

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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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