# Hydrogen probability problem

1. Jan 22, 2008

### 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

2. Jan 22, 2008

### Staff: Mentor

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?

3. Jan 22, 2008

### gazepdapi1

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

4. Jan 22, 2008

### 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

5. Jan 22, 2008

### Staff: Mentor

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.

6. Jan 23, 2008

### gazepdapi1

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

7. Jan 26, 2008

### gazepdapi1

nobody can check for me?