# Particle Equations

1. May 8, 2010

### Karmic Leprec

Is there an equation for the numerical probability of an electron being present as a function of it's radial distance from the nucleus of an atom?
Or am I missing something?

2. May 8, 2010

### phyzguy

Yes, that's what the wave function tells you. The absolute square of the wave function tells you the probability that the electron is at a given point in space. If you just want the radial distance, you can just look at the radial part of the wave function. For a hydrogen atom in the lowest energy state (ground state), the wave function falls off as exp(-r/a0), where a0 is the Bohr radius, which is about 0.5 Angstrom units of .05 nm. The probability drops off as the square of this, or exp(-2r/a0).

For higher energy levels the wave function is more complicated, but you can look it up here:

http://en.wikipedia.org/wiki/Hydrogen_atom

3. May 8, 2010

### Dickfore

When calculating the probability of being at a particular distance r from the nucleus, there is an extra factor of $r^{2}$ due to the surface area of a sphere.

4. May 8, 2010

### Karmic Leprec

I don't understand how the wave function works. But I'll figure that out in my own time.
However, I could use help on this follow up question:
I'm assuming, since protons are generally located closer to the center of an atom, that protons' probability density become much higher, in comparison to an electron, the closer you look to the center of an atom. Is this correct?

5. May 8, 2010

### phyzguy

That's basically correct. The radial extent over which the proton wavefunction is significantly different from zero is approximately 2000 times smaller than the radial extent of the electron wavefunction.

6. May 8, 2010

### Karmic Leprec

And the radial extent over which the proton wavefunction is significantly different from zero; that distance is about the size of the nucleus of the atom, right?

7. May 8, 2010

### phyzguy

Correct. That distance is about 1 fm = 10^-15 m.

8. May 8, 2010

thank you