Question

Verify that Eq. 39-44, the radial probability density for the ground state of the hydrogen atom, is normalized. That is, verify that the following is true:${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{P}\left(r\right)\mathbf{dr}\mathbf{=}\mathbf{1}$

Short answer

It is proved that ${\int }_0^{\infty }P\left(r\right)dr=1$.

Step-by-step solution

Radial probability density:

The expression of radial probability density is given by,

$\mathbf{P}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{4}}{{\mathbf{a}}^{\mathbf{3}}}{\mathbf{r}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\frac{\mathbf{r}}{\mathbf{a}}}$ ….. (1)

Here, $\mathbf{a}\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{292}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{}\mathbf{m}$is the Bohr radius.

Prove that∫0∞P(r)dr=1:

Integrate the radial probability density from zero to infinity.

$$\begin{gathered} {\int }_0^{\infty }P\left(r\right)dr={\int }_0^{\infty }\frac{4}{a^3}{r}^2{e}^{-2\frac{r}{a}}dr \\ =\frac{4}{a^3}{\int }_0^{\infty }\left(r^2{e}^{-2\frac{r}{a}}\right)dr \\ =\frac{4}{a^3}{\left(-r^2\left(\frac{a}{2}\right)e^{-2\frac{r}{a}}-\frac{a^{2}r}{2}{e}^{-2\frac{r}{a}}-\frac{a^3}{4}{e}^{-2\frac{r}{a}}\right)}_0^{\infty } \\ =\frac{4}{a^3}{\left(-\frac{r^2}{2}{e}^{-2\frac{r}{a}}-\frac{ar}{2}{e}^{-2\frac{r}{a}}-\frac{a^2}{4}{e}^{-2\frac{r}{a}}\right)}_0^{\infty } \end{gathered}$$

Simplify further:

$$\begin{gathered} {\int }_0^{\infty }P\left(r\right)dr=\frac{4}{a^2}\left(0-0-{\left|\frac{a^2}{4}{e}^{-\frac{2r}{a}}\right|}_0^{\infty }\right) \\ =-\frac{4}{a^2}\left(\frac{a^2}{4}{e}^{-\frac{2\infty }{a}}-e^0\right) \\ =1 \end{gathered}$$

Hence, it is proved that .${\int }_0^{\infty }P\left(r\right)dr=1$