Hydrogen Atom Ground State Wavefunction Normalisation Solution

[White Ink]

Homework Statement

A hydrogen atom in the ground state can be described by the following wavefunction:

\Psi(r) = \frac{C}{\sqrt{4\pi}}e^{- \frac{r}{a_{0}}}

Normalise this wavefunction.

The Attempt at a Solution

I did this and got:

C = \sqrt{\frac{8\pi}{a_{0}}}

I have no way of checking this, so I was wondering if anybody could tell me whether I am right or wrong.

 

[Dick]

Not quite, I don't think. Can you show us how you got that?

 

[White Ink]

\int^{\infty}_{0}\left|\Psi(r)\right|^{2}dr = 1\frac{C^{2}}{4\pi}\int^{\infty}_{0}e^{\frac{-2r}{a_{0}}}dr = 1u = \frac{2r}{a_{0}}

du = \frac{2}{a_{0}}dr

dr = \frac{a_{0}}{2}du\frac{C^{2}}{4\pi}\frac{a_{0}}{2}\int^{\infty}_{0}e^{-u}du = 1

Using:

\int^{\infty}_{0}x^{n}e^{-x}dx = n!

\int^{\infty}_{0}e^{-u}du = 0! = 1

Thus:

\frac{C^{2}}{4\pi}\frac{a_{0}}{2} = 1

C^{2} = \frac{8\pi}{a_{0}}

 

[Proggle]

Think again about the first integral. Even if the wavefunction is only dependent on r, you are still integrating over three-dimensional space. Use the appropriate volume differential for this case (spherical coordinates).

You will notice the equation will reduce to an integral in just r, but it'll have a key difference with the one you're using right now.

 

[White Ink]

Think again about the first integral. Even if the wavefunction is only dependent on r, you are still integrating over three-dimensional space. Use the appropriate volume differential for this case (spherical coordinates).

You will notice the equation will reduce to an integral in just r, but it'll have a key difference with the one you're using right now.

I know exactly what you mean (integrating in the \theta & \phi directions becomes equivalent to multiplying to constant in front of the integration with respect to r by 2\pi^{2}, since the wavefunction is independent of those two variables). I read over the question too quickly. Thank you for your help.

 

[Dick]

I know exactly what you mean (integrating in the \theta & \phi directions becomes equivalent to multiplying to constant in front of the integration with respect to r by 2\pi^{2}, since the wavefunction is independent of those two variables). I read over the question too quickly. Thank you for your help.

In addition there will be an extra factor of r^2 in the integrand, right?

 

[White Ink]

In addition there will be an extra factor of r^2 in the integrand, right?

Now that one beats me. I can't see where an additional r^{2} would come from.

 

[Dick]

dV in spherical coordinates is r^2*sin(theta)*dr*dtheta*dphi. Better look that up to make sure my use of angle names agrees with yours. And you won't get a pi^2 from the integration, do it carefully.

 

[White Ink]

Ooops, I feel stupid now. I forgot about the metric coefficients and took my volume element to be drd\theta . At least I won't be making that mistake again. Thanks.