# 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.

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 – Mithoron 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. – Martin - マーチン 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. – Nanoputian 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