Question

If the radius of $3$rd orbit of hydrogen atom is '$x$' then the radius of $4$th orbit of ${Li}^{2+}$ ion would be:

• A. $\frac{27}{16}x$
• B. $\frac{16}{27}x$
• C. $\frac{9}{16}x$
• D. $\frac{16}{9}x$

Answer 1

Answer: (B)

$\frac{16}{27}x$

$r=\frac{n^2}{Z}\times 0.529\mathring{A}$
$x=\frac{9}{1}\times 0.529\mathring{A}$ ...(i)
$r_4\left({Li}^{2+}\right)=\frac{16}{3}\times 0.529\mathring{A}$ ...(ii)
Dividing equations (ii) by (i)
$\frac{r_4\left({Li}^{2+}\right)}{x}=\frac{16}{27}$
$r_4\left({Li}^{2+}\right)=\frac{16\times x}{27}$.