Chemistry - Atomic Structure Question with Solution | TestHub

${\psi }^2(r)=\frac{1}{81\times 6}{\left(\frac{\mathrm{Z}}{{\mathrm{a}}_0}\right)}^3{\left(4\sigma -{\sigma }^2\right)}^2{\mathrm{e}}^{-\sigma }$

$\therefore \ell =1$ and $\mathrm{n}-\ell -1=1$

$\therefore \mathrm{n}=3,\mathrm{z}=3$

For ${\psi }^2(r)$ to be maximum / minimum

$\frac{\mathrm{d}{\psi }^2(\mathrm{r})}{\mathrm{d}.\sigma }=0$

${\mathrm{e}}^{-\sigma }2\left(4\sigma -{\sigma }^2\right)(4-2\sigma )+{\left(4\sigma -{\sigma }^2\right)}^2{\mathrm{e}}^{-\sigma }(-1)=0$

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 $4-\sigma =0$

or, $\sigma =4$

or, $\frac{2\times 3\mathrm{r}}{3{\mathrm{a}}_0}=4\Rightarrow \frac{\mathrm{r}}{{\mathrm{a}}_0}=2$

when ${\sigma }^2-8\sigma +8=0$

or, $\sigma =\frac{8\pm \sqrt{64-32}}{2}$

or, $\sigma =4\pm 2\sqrt{2}$

or, $\frac{2\mathrm{r}}{{\mathrm{a}}_0}=4+2\sqrt{2}$

or, $\frac{\mathrm{r}}{{\mathrm{a}}_0}=2\pm \sqrt{2}$

∴ 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 $\frac{\mathrm{r}}{{\mathrm{a}}_0}=2+\sqrt{2}$.