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

Question

Radius of the 4Be8 is 2.6 fermi. Radius of 13AI27 nucleus in units of fermi is (A) 3.9 (B) 3.6 (C) 1.3 (D) 7.8

Tardigrade Admin · Accepted Answer

Radius of a nucleus is given by R = R0A1/3 ∴ (RBe/RAl) = ((ABe/AAl))1 3 Here RBe = 2.6 fermi , ABe = 8, AAl = 27 RAl = ? So, (2.6/RAl) = ((8/27))1/ 3 = (2/3) RAl = (3/2)×2.6 = 39 fermi

Q. Radius of the $_{4} B e^{8}$ is $2.6$ fermi. Radius of $_{13} A I^{27}$ nucleus in units of fermi is

3.9

3.6

1.3

7.8

Radius of a nucleus is given by $R = R_{0} A^{1/3}$
$∴ \frac{R _{B e}}{R _{A l}} = (\frac{A _{B e}}{A _{A l}})^{13}$
Here $R_{B e} = 2.6 f er mi, A_{B e} = 8, A_{A l} = 27 R_{A l} = ?$
So, $\frac{2.6}{R _{A l}} = (\frac{8}{27})^{1/3} = \frac{2}{3}$
$R_{A l} = \frac{3}{2} \times 2.6 = 39 f er mi$

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

3.9

3.6

1.3

7.8

Radius of a nucleus is given by R=R0​A1/3 ∴RAl​RBe​​=(AAl​ABe​​)13 Here RBe​=2.6fermi,ABe​=8,AAl​=27RAl​=? So, RAl​2.6​=(278​)1/3=32​ RAl​=23​×2.6=39fermi

Q. Radius of the $_{4} B e^{8}$ is $2.6$ fermi. Radius of $_{13} A I^{27}$ nucleus in units of fermi is

Radius of a nucleus is given by $R = R_{0} A^{1/3}$
$∴ \frac{R _{B e}}{R _{A l}} = (\frac{A _{B e}}{A _{A l}})^{13}$
Here $R_{B e} = 2.6 f er mi, A_{B e} = 8, A_{A l} = 27 R_{A l} = ?$
So, $\frac{2.6}{R _{A l}} = (\frac{8}{27})^{1/3} = \frac{2}{3}$
$R_{A l} = \frac{3}{2} \times 2.6 = 39 f er mi$