For an electron in the$\left(n,l,m_{\mathcal{l}}\right)\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$ state in a hydrogen atom, (a) write the solution of the time-independent Schrodinger equation,

(b) verify explicitly that it is a solution with the expected angular momentum and energy.

For electron the state is given as: (n,l,m) = (2,0,0).

For an electron in the $\left(n,l,m_{\mathcal{l}}\right)\mathbf{=}\left(2,0,0\right)$state in a hydrogen atom, the solution of the time-independent Schrodinger equation will be given by,

${\mathbf{R}}_{\mathbf{2}\mathbf{,}\mathbf{0}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}{\mathbf{⊝}}_{\mathbf{0}\mathbf{,}\mathbf{0}}\left(\theta \right){\mathbf{\Phi }}_{\mathbf{0}}\left(\varphi \right)\mathbf{=}\left(\frac{2}{{\left(2a_0\right)}^{3/2}}\left(1-\frac{r}{2a_0}\right)e^{-r/2a_0}\right)\left(\sqrt{\frac{1}{4\pi }}\right)$

Where, r = radius

${\mathit{a}}_{\mathbf{0}}$ = radius of the hydrogen atom

$\mathit{\theta }$= colatitude

$\mathit{\varphi }$ = azimuth

Solving the equation further as:

${\mathrm{R}}_{2,0}\left(\mathrm{r}\right){⊝}_{0,0}\left(\mathrm{\theta }\right){\mathrm{\Phi }}_0\left(\mathrm{\varphi }\right)=\left(\frac{2}{{\left(2{\mathrm{a}}_0\right)}^{3/2}}(1-\frac{\mathrm{r}}{2{\mathrm{a}}_0}){\mathrm{e}}^{-\mathrm{r}/2{\mathrm{a}}_0}\right)\left(\sqrt{\frac{1}{4\mathrm{\pi }}}\right)$

Thus the required equation is $\frac{1}{8{a_0}^{5/2}\sqrt{\mathrm{\pi }}}r{e}^{-r/2a_0}\sin \theta e^{+i\varphi }$.

The equation in Step (1) obeys the polar equation, thus $\mathcal{l}\mathbf{=}\mathbf{0}$.

According to the hydrogen radial equation,

$\frac{\mathbf{-}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{d}}{\mathbf{dr}}\left(r^2\frac{d}{\mathrm{dr}}\right)\mathbf{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}\mathbf{l}\left(\mathcal{l}+1\right)}{\mathbf{2}{\mathbf{mr}}^{\mathbf{2}}}\mathbf{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{ER}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Where, $\mathcal{l}$ = azimuthal quantum number

h = Plank’s constant

${\mathit{\epsilon }}_{\mathbf{0}}$= permittivity

e = charge on an electron

The first term of the eq. (1) is :

$$\begin{gathered} =\frac{-h^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\left(\left(1-\frac{r}{2a_0}\right)e^{-r/2a_0}\right)\right) \\ =\frac{-h^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(\left(\left(-\frac{r^2}{a_0}+\frac{r^3}{4a_0^2}\right)e^{-r/2a_0}\right)\right) \\ =\frac{-h^2}{2m}\left(-\frac{2}{a_{0}r}+\frac{5}{4a_0^2}-\frac{r}{8a_0^3}\right)e^{-r/2a_0} \end{gathered}$$

Now, if$\mathcal{l}\mathbf{=}\mathbf{0}$,

The second term will also be zero.

Hence, the radial equation now becomes,

$$\begin{gathered} \frac{-h^2}{2m}\left(-\frac{2}{a_{0}r}+\frac{5}{4a_0^2}-\frac{r}{8a_0^3}\right)e^{-r/2a_0}-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{r}\left(1-\frac{r}{2a_0}\right)e^{-r/2a_0}=E\left(1-\frac{r}{2a_0}\right)e^{-r/2a_0} \\ \frac{-h^2}{2m}\left(-\frac{2}{a_{0}r}+\frac{5}{4a_0^2}-\frac{r}{8a_0^3}\right)-\frac{h^2}{m{a}_0}\frac{1}{r}\left(1-\frac{r}{2a_0}\right)=E\left(1-\frac{r}{2a_0}\right) \\ \left(-\frac{2}{a_{0}r}+\frac{5}{4a_0^2}-\frac{r}{8a_0^3}\right)+\left(\frac{2}{a_{0}r}-\frac{1}{a_0^2}\right)=\frac{-2mE}{h_0^2}\left(1-\frac{r}{2a_0}\right) \\ \frac{1}{4a_0^2}\left(1-\frac{r}{2a_0}\right)=\frac{-2mE}{h_0^2}\left(1-\frac{r}{2a_0}\right) \end{gathered}$$

Hence, the equation will hold if,