Trouble Calculating Hydrogen 1s Electron Probability?

[gazepdapi1]

I am having some trouble starting this one.
Calculate the probability that a hydrogen 1s electron will be found within a distance 2a0 from the nucleus.

Any help is appreciated

 

[jtbell]

To start with, you need the wave function \psi for the hydrogen 1s state. Do you have that? What do you do with a wave function in order to calculate a probability?

 

[gazepdapi1]

So, since the wavefunction for 1s has only the r variable, would I square it first and then integrate from 0 to 2a0?

 

[gazepdapi1]

This is what I have so far. I used the wavefunction (1/sqrt(pi))*(Z/a0)^3/2*exp(-Zr/a0)
I multiplied this by itself and by r^2 and then integrated from 0 to 2a0.
I got a long answer so I'm not sure if this is correct. I got
(exp(-4Z)/8Z^3)*[2(a0)^3*(-8Z^2-5)] + (2a^3/8Z^3)

If someone can check my work, that would be great
Thank you

 

[jtbell]

Remember Z = 1 for hydrogen so you can simplify your equations a little bit. You've got the right \psi . Squaring it gives you the probability density P. This is the probability per unit volume so you have to integrate over all three dimensions (r,\theta,\phi) in spherical coordinates. The volume element in spherical coordinates is r^2 \sin \theta dr d\theta d\phi. So you still need the \theta and \phi integrals which are pretty easy:

P = \int_{r=0}^{r=2a_0} \int_{\phi=0}^{\phi=2\pi} \int_{\theta=0}^{\theta=\pi} P(r,\theta,\phi}) r^2 \sin \theta dr d\theta d\phi

Remember you can check the r integral by taking the derivative of the indefinite integral (before substituting the limits). I haven't had time to work it out myself yet... looks like integration by parts. Actually I normally look it up in a table of integrals.

 

[gazepdapi1]

Okay, I got the probability will be .7616. If someone can check if that's correct, I would appreciate it.

 

[gazepdapi1]

nobody can check for me?