Chemistry - Atomic Structure Question with Solution | TestHub

For hydrogen like species, from Bohr's formulas we get,
$$$\mathrm{r}\propto \frac{{\mathrm{n}}^2}{\mathrm{Z}}$
where $\mathrm{n}$ is orbit number, $\mathrm{Z}$ is atomic number

$$Thus,

$\Rightarrow {\mathrm{r}}_{{\mathrm{He}}^{*}}={\mathrm{r}}_{\mathrm{H}}\times \frac{{\mathrm{n}}^2}{\mathrm{Z}}$

$\Rightarrow {\mathrm{r}}_{{\mathrm{He}}^{+}}=0.6\times \frac{{\left(3\right)}^2}{2}$

$$\begin{gathered} =2.7Å \\ =270 \times {10}^{-12} \mathrm{m} \end{gathered}$$

$\therefore {\mathrm{r}}_{{\mathrm{He}}^{+}}=270\mathrm{pm}$