Atomic StructureHard
Question
If the radius of first orbit of H–atom is x Å, then the radius of the second orbit of Li2+ ion will be
Options
A.x Å
B.$\frac{4x}{3}$Å
C.$\frac{9x}{2}$ Å
D.4x Å
Solution
$r_{n,z} = 0.529 \times \frac{n^{2}}{z}\overset{0}{A}$
$\therefore\frac{r_{2,Li^{2 +}}}{r_{1,H}} = \frac{0.529 \times \frac{2^{2}}{3}}{0.529 \times \frac{1^{2}}{1}} \Rightarrow \frac{r_{2,Li^{2 +}}}{x\overset{0}{A}} = \frac{4}{3} $$$\Rightarrow r_{2,Li^{2 +}} = \frac{4}{3}x\overset{0}{A}$$
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