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

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

In This post you will get answer of Q.
Radius of the $_4Be^8$ is $2.6$ fermi. Radius of $_{13}AI^{27}$ nucleus in units of fermi is

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} $

$R_{Al} = frac{3}{2}times2.6 = 39 fermi$

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