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

• dark_matter_is_neat
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.

Last edited:
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##.

dark_matter_is_neat
dark_matter_is_neat said:
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
kuruman said:
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.

dark_matter_is_neat said:
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.

vela

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