Determine the value of r1 and E for given wavefunction of hydrogen

[dark_matter_is_neat]

Homework Statement

An electron in the hydrogen atom in the ground state is described by the wavefunction: $\Psi(x,y,z) = Ae^{-\frac{r}{r_{1}}}$
where $r = \sqrt{x^{2}+y^{2}+z^{2}}$ and A and $r_{1}$ are constants.
Use the Schrodinger equation to find $r_{1}$ and the energy eigenvalue E in terms of the electron mass and charge.

Relevant Equations

$-\frac{\hbar^{2}}{2m} \nabla^{2} \Psi + V \Psi = E \Psi$

In this case, ignoring derivatives that go to zero, (denoting the charge of the electron as q to avoid confusion) $-\frac{\hbar^{2}}{2m} \frac{1}{r} \frac{\partial^{2}}{\partial r^{2}} (rAe^{-\frac{r}{r_{1}}}) - \frac{q^{2}}{4 \pi \epsilon_{0} r} Ae^{-\frac{r}{r_{1}}} = E A e^{-\frac{r}{r_{1}}}$.
So going through the derivatives:
$(\frac{\hbar^{2}}{mrr_{1}} - \frac{\hbar^{2}}{2mr_{1}^{2}} - \frac{q^{2}}{4 \pi \epsilon_{0} r}) A e^{-\frac{r}{r_{1}}} = E A e^{-\frac{r}{r_{1}}}$.
I can cancel $A e^{-\frac{r}{r_{1}}}$ on each side to get $E = \frac{\hbar^{2}}{mrr_{1}} - \frac{\hbar^{2}}{2mr_{1}^{2}} - \frac{q^{2}}{4 \pi \epsilon_{0} r}$, which isn't good since it contains two unknowns $r_{1}$ and E, and it contains r. I'm not sure how to get two separate equations for $r_{1}$ and E from just the Schrodinger equation and I'm not sure how to get rid of r, since neither expression should depend r.

 

[kuruman]

E is assumed known. The problem tells you that the hydrogen atom is in the ground state. So what's E for the hydrogen ground state?

As far as getting rid of $r$ is concerned, don't forget that the equation you get, after you substitute the solution into the Schrodinger equation, must hold for any value of $r$.

 

[vela]

I can cancel $A e^{-\frac{r}{r_{1}}}$ on each side to get $E = \frac{\hbar^{2}}{mrr_{1}} - \frac{\hbar^{2}}{2mr_{1}^{2}} - \frac{q^{2}}{4 \pi \epsilon_{0} r}$, which isn't good since it contains two unknowns $r_{1}$ and E, and it contains r. I'm not sure how to get two separate equations for $r_{1}$ and E from just the Schrodinger equation and I'm not sure how to get rid of r, since neither expression should depend r.

You determine $r_1$ so that the $r$ dependence disappears. In other words, what value does $r_1$ have to take so that the terms with the $r$ dependence cancel out?

 

[dark_matter_is_neat]

E is assumed known. The problem tells you that the hydrogen atom is in the ground state. So what's E for the hydrogen ground state?

As far as getting rid of $r$ is concerned, don't forget that the equation you get, after you substitute the solution into the Schrodinger equation, must hold for any value of $r$.

The problem asks for you to solve for the energy eigenvalue in terms of the electron mass and charge, so I don't think it is supposed to be assumed as being known.

E is -13.6 eV = $-\frac{me^{4}}{2 \hbar^{2} (4 \pi \epsilon_{0})^{2}}$ for the hydrogen atom ground state, so I could put in $r_{1}$ for r and then solve for it in terms of E.

 

[kuruman]

The problem asks for you to solve for the energy eigenvalue in terms of the electron mass and charge, so I don't think it is supposed to be assumed as being known.

E is -13.6 eV = $-\frac{me^{4}}{2 \hbar^{2}}$ for the hydrogen atom ground state, so I could put in $r_{1}$ for r and then solve for it in terms of E.

Yes, you are correct. You can get both the energy and $r_1$ following @vela's suggestion, or mine. To put it simply, if you substitute the wavefunction into the Schrodinger equation, you can move everything to the left side and bring it to the form $$A +f(r)=0$$ where $A$ is some constant and $f(r)$ is a function of $r$. This must be true for any value of $r$ because that's what "solution" means. What does that imply for $A$ and $f(r)$?

 

[dark_matter_is_neat]

Okay, so solving for $r_{1}$ so that the r dependence cancels, $r_{1} = \frac{4 \pi \epsilon_{0} \hbar^{2}}{me^{2}}$ which is as expected, is the Bohr radius.

With an expression for r_{1}, getting the energy is trivial, since I can just substitute in $r_{1}$. So $E = -\frac{me^{4}}{2(4 \pi \epsilon_{0})^{2} \hbar^{2}}$ as required.