Normalization of the wave function for the electron in a hydrogen atom

[jjson775]

Homework Statement

The ground state wave function for the electron in a hydrogen atom is
Psi 1s = (1/(pi x a0^3)) x e^-r/a0
where r is. the radial coordinate of the electron and a0 is the Bohr radius. Show that the wave function as given is normalized.

Relevant Equations

First equation below

 

[hutchphd]

This is not a one dimensional atom. You must integrate over three dimensional space (use spherical coordinates).

 

[Steve4Physics]

There are 2 identical columns of equations. I guess this is an accidental mistake.

The 'first equation' is for a 1-dimensional (x-axis) wave, which is *not* what you need. You need the equivalent equation in 3D (spherical polar coordinates).

What is the purpose of the other equations?

What have you attempted so far?

 

[jjson775]

I think your reply is beyond the level I am studying. The textbook (Serway for Scientists and Engineers) takes advantage of spherical symmetry to determine the radial probability density to solve for location of the electron. My attempt is shown above beginning with the wave function given in the problem statement. Sorry about the duplication.

 

[Steve4Physics]

In 1D, \int |\phi^2| dx = 1 is wrong and should be \int |\phi|^2 dx = 1

In 3D, the normalisation requires \int |\phi|^2 dV = 1 where the volume integral is over all of 3D-space. This is simply saying the probability of finding the particle somewhere is 1.

Since the wave function is spherically symmetric (function of r ony) we can use spherical polar cootrdinates and write
\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty}|\phi|^2 r^2 sin(\phi) dr d\phi d\theta = 1
If that's unfamiliar, you need to spend a little time reading-up about volume integrals in spherical polar coordinates. But you should have covered it. There's no other way to do it that I know of.

Th triple integral looks scary but is actually not that hard.

 

[kuruman]

The probability that the electron be found within a shell of radius $r$ and thickness $dr$ is
$dP=|\phi(r)|^2dV$ where $dV=\text{Area of shell}\times\text{thickness}=4\pi r^2~dr$. Put it together and integrate with $r$ from zero to infinity.

 

[Steve4Physics]

The probability that the electron be found within a shell of radius $r$ and thickness $dr$ is
$dP=|\phi(r)|^2dV$ where $dV=\text{Area of shell}\times\text{thickness}=4\pi r^2~dr$. Put it together and integrate with $r$ from zero to infinity.

That's much neater than my triple integral (post#5).

 

[kuruman]

That's much neater than my triple integral (post#5).

Because of the spherical symmetry, the triple integral is really a one-dimensional radial integral.

 

[Charles Link]

It might interest the OP that there is a mistake in the evaluation of their integral: $\int e^{-2r/a_o} \, dr=-(a_o/2)e^{-2r/a_o}$. In any case, the previous comments of the others apply=you need to do a 3-D integral.

 

[jjson775]

It might interest the OP that there is a mistake in the evaluation of their integral: $\int e^{-2r/a_o} \, dr=-(a_o/2)e^{-2r/a_o}$. In any case, the previous comments of the others apply=you need to do a 3-D integral.

You are right about the mistake in the integration but kuruman is right that the problem can be solved with a one dimensional radial integral. After fixing the mistake the best I can get is 1/2πa0^2. Still not 1 like it is supposed to be.

 

[kuruman]

Please show your work. We cannot find what went wrong if you do not.

 

[jjson775]

 

[jjson775]

I fixed the integration mistake identified by Charles Link. See revised work above.

 

[Charles Link]

In the third line from the bottom, $e^{-2r/a_o}$ needs to be in the numerator, or you can have $e^{2r/a_o}$ in the denominator.

 

[hutchphd]

The integral is a 3D (volume) integral. It can be manipulated into a one dimensional integral because it does not depend upon angles.
What are the calculus prerequisites for the course? Do you know how to write an infinitesimal volume element in spherical coordinates?
More is required than changing dx to dr.

 

[Charles Link]

The integral the OP needs will be of the form $\int\limits_{0}^{+\infty} e^{-2r/a_o} r^2 \, dr$, which can be integrated by parts. It appears the OP has yet to see this type of integration in their coursework.

 

[kuruman]

It looks like you didn't read post #6 or if you did, you did not follow my suggestion. Please read it carefully. It gives you what you need to integrate.

On Edit:

The integral the OP needs will be of the form $\int e^{-2r/a_o} r^2 \, dr$, which can be integrated by parts. It appears the OP has yet to see this type of integration in their coursework.

You are probably right. This integral can also be found in a Table of Integrals.

 

[hutchphd]

I sympathize with the OP if volume integrals are foreign. But you need to understand explicitly how to do this or you will be in some ongoing difficulty. Integration in 3-D is a basic skill !

 

[jjson775]

It looks like you didn't read post #6 or if you did, you did not follow my suggestion. Please read it carefully. It gives you what you need to integrate.

On Edit:

You are probably right. This integral can also be found in a Table of Integrals.

I can solve the problem using Post #6 but have a conceptual issue. The problem statement gives the ground state wave function for the ELECTRON in a hydrogen atom. My textbook gives this same function for the hydrogen ATOM 1s state. Can someone please why they are the same?

 

[kuruman]

There might be a semantic difference between the two but most people use them interchangeably. I wouldn't worry about it.

 

[Steve4Physics]

I can solve the problem using Post #6 but have a conceptual issue. The problem statement gives the ground state wave function for the ELECTRON in a hydrogen atom. My textbook gives this same function for the hydrogen ATOM 1s state. Can someone please why they are the same?

Just different terms for the same thing. I prefer to call it the electron's wave function, but you'll see both terms used.

 

[hutchphd]

Also what are the units for the wavefunction?? How about your result? I don't think you are getting this...

 

[jjson775]

Also what are the units for the wavefunction?? How about your result? I don't think you are getting this...

Meters. I solved it using a table of integrals.

 

[Charles Link]

One comment (the very last line above)=you've made this error previously as well: When you take the lower limit in the integral, it gets an extra minus sign, which then gives it a "plus" sign.
Meanwhile, the units of the wave function are $meters^{-3/2}$.