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  4. Radius of the 4Be8 is 2.6 fermi. Radius of 13AI27 nucleus in units of fermi is

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Q. Radius of the $_4Be^8$ is $2.6$ fermi. Radius of $_{13}AI^{27}$ nucleus in units of fermi is

COMEDKCOMEDK 2010Nuclei

Solution:

Radius of a nucleus is given by $R = R_0A^{1/3}$
$ \therefore \, \frac{R_{Be}}{R_{Al}} = \left(\frac{A_{Be}}{A_{Al}}\right)^{1 3} $
Here $ R_{Be} = 2.6 fermi , A_{Be} = 8, A_{Al} = 27 R_{Al} = ? $
So, $ \frac{2.6}{R_{Al}} = \left(\frac{8}{27}\right)^{1/ 3 }= \frac{2}{3} $
$R_{Al} = \frac{3}{2}\times2.6 = 39 fermi$