Question

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

• A. 3.9
• B. 3.6
• C. 1.3
• D. 7.8

Answer 1

Answer: (A)

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, $\frac{2.6}{R_{Al}} = \left(\frac{8}{27}\right)^{1/ 3 }= \frac{2}{3}$
$R_{Al} = \frac{3}{2}\times2.6 = 39 fermi$