Question

If radius of second Bohr orbit of the $H{e}^{+}$ion is $105.8$ $pm$, what is the radius of third Bohr orbit of $L{i}^{2+}$ ion?

• A. $158.7\mathring{A}$
• B. $158.7pm$
• C. $15.87pm$
• D. $1.587pm$

Answer 1

Answer: (B)

$\frac{{\left(r_3\right)}_{L^{+2}}}{{\left(r_2\right)}_{H{e}^{+}}}=\frac{n_1^2}{n_2^2}\times \frac{Z_2}{Z_1}$
$\frac{{\left(r_3\right)}_{L^{+2}}}{105.8pm}=\frac{3\times 3}{2\times 2}\times \frac{2}{3}$
${\left(r_3\right)}_{L{i}^{+2}}=158.7pm$