Question

Schrodinger’s equation for states of the hydrogen atom for which the orbital quantum number l is zero is

$\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{r}}\left(r^2\frac{d\psi }{dr}\right)\mathbf{+}\frac{\mathbf{8}\mathbf{\tau }{\mathbf{\tau }}^{\mathbf{2}}\mathbf{m}}{{\mathbf{r}}^{\mathbf{2}}}\left[E-U\left(r\right)\right]\mathit{\psi }\mathbf{=}\mathbf{0}$

Verify that Eq. 39-39, which describes the ground state of the hydrogen atom, is a solution of this equation?

Short answer

The ground state of the hydrogen atom is a solution to this equation.

Step-by-step solution

Identification of the given data

The given data is listed below as-

The orbital quantum number,$l=0$

Formula for the wave function

Theformula for the wave functionis given by-

$\mathit{\psi }\mathbf{=}\frac{\mathbf{1}}{\mathbf{\tau }\mathbf{\tau }{\mathbf{a}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}{\mathit{e}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{a}}$

Here, a is the Bohr radius.

To show that the ground state of the hydrogen atom is a solution to the given equation.

(a)

Let the wave function is given by-

$\psi =\frac{1}{{\mathrm{\tau \tau a}}^{3/2}}{e}^{-\mathrm{r}/\mathrm{a}}$

Now, the goal is to show that the result is zero when the wave function is substituted into the right side of Schrodinger’s equation.

The derivative of the wave function is:

$\frac{d\psi }{dr}=-\frac{1}{{\mathrm{\pi a}}^{5/2}}{e}^{-r/a}$

So,$r^2\frac{d\psi }{dr}=-\frac{r^2}{\sqrt{\mathrm{\pi }}{a}^{5/2}}{e}^{-r/a}$

Also, $\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\psi }{dr}\right)=\frac{1}{\sqrt{\mathrm{\pi }}{a}^{5/2}}\left[\frac{-2}{r}+\frac{1}{a}\right]e^{-r/a}=\frac{1}{2}\left[-\frac{2}{r}+\frac{1}{a}\right]\psi$ .

Now, the ground state energy is given by-

$E=-\frac{m{e}^4}{8{\epsilon }_0^2{h}^2}$

And the Bohr radius is given by-

$a=\frac{h^2{\epsilon }_0}{{\mathrm{\pi me}}^2}$ …… (1)

S0, $E=-\frac{e^2}{8{\mathrm{\pi \epsilon }}_0\mathrm{a}}$

Now, the potential energy is given by

$E=-\frac{e^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}$

So,

$$\begin{gathered} \frac{8{\pi }^{2}m}{h^2}[E-U]\psi =\frac{8{\pi }^{2}m}{h^2}\left[-\frac{e^2}{8\pi {\epsilon }_{0}a}+\frac{e^2}{4\pi {\epsilon }_{0}r}\right]\psi \\ =\frac{8{\pi }^{2}m}{h^2}\frac{e^2}{8\pi {\epsilon }_0}\left[-\frac{1}{a}+\frac{2}{r}\right] \\ =\frac{\pi m{e}^2}{h^2{\epsilon }_0}\left[-\frac{1}{a}+\frac{2}{r}\right]\psi \end{gathered}$$

Substitute value of a from equation (1) into above equation.

$\frac{8{\pi }^{2}m}{h^2}[E-U]\psi =\frac{8{\pi }^{2}m}{h^2}\left[-\frac{e^2}{8\pi {\epsilon }_{0}a}+\frac{e^2}{4\pi {\epsilon }_{0}r}\right]\psi$

Hence, the proposed function $\psi$satisfies the given equation.

Thus, the ground state of the hydrogen atom is a solution of this equation.