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This is what my book shows (the darker the more probable): enter image description here

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.

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most venerable sir
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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:

enter image description here

So if you square this function you will get the probabily distribution which looks like this: enter image description here

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.

Nanoputian
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    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 01 '15 at 04:05
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    At infinity the probability density is not zero. It is infinitely small, but significantly different from zero. – Martin - マーチン Sep 01 '15 at 07:01
  • 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 01 '15 at 07:10
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    @Martin-マーチン That expression means it converges to zero (asymptotic behavior)... – Greg Mar 05 '17 at 14:24
  • FWIW, I would also say that it simply refers to the limit which is exactly zero. Strictly speaking, saying that a function has a value "at infinity" is not the best practice (usually). The infinities are sometimes appended to the real number line making something like $f(\infty)$ bearable but even then in this case the value would again be exactly zero. Happy to be proven wrong though. – Linear Christmas Mar 05 '17 at 20:42