Chemistry - Atomic Structure Question with Solution | TestHub

$\begin{array}{rlrl}& {\psi }^2(\mathrm{r})=\frac{1}{81\times 6}{\left(\frac{\mathrm{Z}}{{\mathrm{a}}_0}\right)}^3{\left(4\sigma -{\sigma }^2\right)}^2{\mathrm{e}}^{-\sigma }& \therefore \ell =1\text{ and }\mathrm{n}-\ell -1=1& \therefore \mathrm{n}=3,\mathrm{z}=3\end{array}$ For ${\psi }^2(\mathrm{r})$ to be maximum / minimum $\begin{array}{rlr}& \frac{\mathrm{d}{\psi }^2(\mathrm{r})}{\mathrm{d}\cdot \sigma }=0& {\mathrm{e}}^{-\sigma }2\left(4\sigma -{\sigma }^2\right)(4-2\sigma )+{\left(4\sigma -{\sigma }^2\right)}^2{\mathrm{e}}^{-\sigma }(-1)=0\end{array}$ or $\left(4\sigma -{\sigma }^2\right)\left(8-4\sigma -4\sigma +{\sigma }^2\right)=0$ when $4\sigma -{\sigma }^2=0\Rightarrow \sigma =0\Rightarrow \mathrm{r}=0$ Also $4-\sigma =0$ $\begin{array}{rlr}& \text{ or, }\sigma =4& \text{ or, }\frac{2\times 3r}{3a_0}=4\Rightarrow \frac{r}{a_0}=2\end{array}$ when ${\sigma }^2-8\sigma +8=0$ $\begin{array}{rlrlr}& \text{ or, }\sigma =\frac{8\pm \sqrt{64-32}}{2}& \text{ or, }\sigma =4\pm 2\sqrt{2}& \text{ or, }\frac{2r}{a_0}=4+2\sqrt{2}& \text{ or, }\frac{r}{a_0}=2\pm \sqrt{2}\end{array}$ ∴ Node is at $\frac{\mathrm{r}}{{\mathrm{a}}_0}=2$ and two maxima is reached at $\frac{\mathrm{r}}{{\mathrm{a}}_0}=2-\sqrt{2}$ and $\frac{\mathrm{r}}{{\mathrm{a}}_0}=2+\sqrt{2}$.