Chemistry - Atomic Structure Question with Solution | TestHub

${\text{r}}_{\text{n}-\text{1}}\text{+}{\text{r}}_{\text{n}}\text{=}{\text{r}}_{\text{n}\text{+}\text{1}}$
and${\text{r}}_{\text{n}}\text{=}{\text{n}}^{\text{2}}{\text{r}}_{\text{1}}$
$\text{i}\text{.e}\text{.,}{\text{(n}-\text{1)}}^{\text{2}}{\text{r}}_{\text{1}}\text{+}{\text{n}}^{\text{2}}{\text{r}}_{\text{1}}\text{=}{\text{(n}\text{+}\text{1)}}^{\text{2}}{\text{r}}_{\text{1}}$
$\Rightarrow {\text{(n}-\text{1)}}^{\text{2}}\text{+}{\text{n}}^{\text{2}}\text{=}{\text{(n}\text{+}\text{1)}}^{\text{2}}$
$\Rightarrow {\text{n}}^{\text{2}}-\text{2n}\text{+}\text{1}\text{+}{\text{n}}^{\text{2}}$
$\text{=}{\text{n}}^{\text{2}}\text{+}\text{2n}\text{+}\text{1}$
$\Rightarrow {\text{n}}^{\text{2}}-\text{4n}\text{=}\text{0}$
$\therefore \text{n}\text{=}\text{4}$