Question

The radial probability density for the ground state of the hydrogen atom is a maximum when r = a , where is the Bohr radius. Show that the average value of r, defined as

${\mathit{r}}_{\mathbf{a}\mathbf{v}\mathbf{g}}\mathbf{=}\mathbf{\int }\mathit{P}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathit{r}\mathbf{}\mathit{d}\mathit{r}$,

has the value 1.5a. In this expression for ${\mathit{r}}_{\mathbf{a}\mathbf{v}\mathbf{g}}$ , each value of (P)r is weighted with the value of r at which it occurs. Note that the average value of is greater than the value of r for which (P)r is a maximum.

Short answer

$r_{avg}=1.5a$

Step-by-step solution

Identification of the given data

The given data is listed below as-

The radial probability density is maximum when r = a

The radial probability function

Theradial probability functionfor the ground state is given by-

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

Here, r is the radius of Hydrogen atom.

To Show that the average value of r, defined as ravg=∫P(r)r dr , has the value 1.5 a   

The average value of is defined as:

$r_{avg}={\int }_0^{\infty }rP\left(r\right)dr$……(1)

Now, $P\left(r\right)=\left(\frac{4r^2}{a^3}\right)e^{-2/a}$

Substitute the value of P(r) in equation (1)

$r_{avg}={\int }_0^{\infty }\left(\frac{4r^3}{a^3}\right)e^{-2r/a}dr$

Now, let $x=\frac{2r}{a}$

$dx=\frac{2dr}{a}$

$r_{avg}={\int }_0^{\infty }\left(\frac{{\left(ax\right)}^3}{4a^3}\right)e^{-x}\left(adx\right)$

Solving the above equation,

$r_{avg}=\frac{a}{4}{\int }_0^{\infty }{x}^3{e}^{-x}dx$

Now, $r_{avg}$will become $r_{avg}.=\frac{a(3!)}{4}$

Therefore, it is shown that $r_{avg}=1.5a$.