# What is the correctly diagrammed electron probability distribution for p orbital?

This is what my book shows (the darker the more probable):

The obvious places where the electron probability is zero is the midsection of the dumbbell.

My question is how the probability distributed inside the dumbbell.

## 1 Answer

The probability distribution of an electron in the p sub-shell is determined from its wavefunction. The wavefunction of a system contains all the information about the quantum states of the system. Once of the most important properties of the wavefunction is that the squared modulus of the wave function, $|ψ|^2$, is a real number, interpreted as the probability density of the particle. Here is the wavefunction for the 2p orbital:

So if you square this function you will get the probabily distribution which looks like this:

As you can see, this corresponds with the p orbital drawn in your book, as there is a node in the middle, on either side is high probability density and at minus and positive infinity probability density is actually very, very small despite being shown as 0 on the wavefunction. The reason for this is that you need to remember that the orbital shows where an electron will be 90% of the time, not where it will be at all times. This means that the electron can be found outside the orbital.

• I'm afraid your last paragraph is wrong. Radial density without angular won't give you such shape Aug 31 '15 at 23:34
• Indeed, the angular distribution is missing.
– Greg
Sep 1 '15 at 4:05
• At infinity the probability density is not zero. It is infinitely small, but significantly different from zero. Sep 1 '15 at 7:01
• Thanks for pointing that out. I wasn't sure to say if the probability density is non-zero at infinity as the graphs that I have uploaded go to zero at infinity rather form an asymptote. Sep 1 '15 at 7:10
• @Martin-マーチン That expression means it converges to zero (asymptotic behavior)...
– Greg
Mar 5 '17 at 14:24