

How does an electron on a 2p orbital cross the node of nucleus?
To Fermilab,
Thank you very much.
Hi Danny, It does have something to do with the wave properties of the electron. If you measure the position of the 2p electron you will find that it is located on either side of the node. If you measure the position enough times you will get the "dumbell structure", ie two "lobes" of probability and a region in the middle with zero probability. This structure is known as the probability density of a 2p electron. If the electron behaved strictly as a particle getting from one lobe to the other without crossing the node of zero probablity violates common sense. At the scale of an atom the electron has the properties of a wave, so you can think of the probability density of a 2p electron in a hydrogen atom as a standing wave with a node in the middle. For example, take a jump rope or a long spring have one person hold one end and another move the other end up and down rapidly. You will see that the wave travels down the rope and reflects off the stationary point and creates a return wave. If I remember correctly, you should have a node in the middle, a point that does not move. It looks like a sine wave. You will notice that the rope moves the most half way between the node, at the crest or trough, and does not move at all at the node. This is analogous to the probablity density of a 2p electron. So, there is no probablity of measuring the electron at the node just as there is no perpendicular motion of the rope at the jump rope node. This is 2 dimensional example for a 3d problem. The main thing to remember is the electron behaves like a wave until its position is measured and the wave function collapses. If you measure the location enough times you will get the probability density for the electron. In this case there are two "lobes" with a "node" or region of zero probablity in the middle. I would like to add, that you can never measure the position exactly, there is always some uncertainty, ie, if you could measure the position exactly you would not be able to measure the momentum with any certainty. This is known as Heisenberg's uncertainty principle. The uncertainty in x (position) times the uncertainty in p (momentum) is greater than or equal to Planck's constant (h) divided by 2 pi. I hope this answered you question. If not please send me an email.
Sincerely, 
last modified 9/13/1999 physicsquestions@fnal.gov 
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