Chemistry - Atomic Structure Question with Solution | TestHub

For$1\mathrm{s}$, the radial part of the wave function is:
${\mathrm{\Psi }}_{\left(\mathrm{r}\right)}=2{\left(\frac{1}{{\mathrm{a}}_0}\right)}^{\frac{3}{2}}{\mathrm{e}}^{-\frac{\mathrm{r}}{{\mathrm{a}}_0}}$
Probability of finding an${\mathrm{e}}^{–}$in a spherical shell of thickness,$\text{'}\mathrm{dr}\text{'}$at a distance$\text{'}\mathrm{r}\text{'}$from the nucleus:
$\mathrm{P}={\mathrm{\Psi }}_{\left(\mathrm{r}\right)}^2.4\mathrm{\pi }{\mathrm{r}}^2\mathrm{d}\mathrm{r}$
$=16\mathrm{\pi }{\mathrm{r}}^2{\left(\frac{1}{{\mathrm{a}}_0}\right)}^3{\mathrm{e}}^{\frac{-2\mathrm{r}}{{\mathrm{a}}_0}}\mathrm{ }\mathrm{d}\mathrm{r}$
So,$\mathrm{P}$is zero at$\mathrm{r}=0$and$\mathrm{r}=\mathrm{\infty }$.

So, the plot of radial probability function,$4\mathrm{\pi }{\mathrm{r}}^2\mathrm{d}\mathrm{r}.{\mathrm{\Psi }}_{\left(\mathrm{r}\right)}^2$V/s$\text{r}$is as shown above.