P is the probability of finding the 1 s electron of the hydrogen atom in a spherical shell of infinitesimal…

Solution

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. :

Asked in: JEE Advanced 2016 (Paper 1)

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