At a node the wave function

a. Passes through 0

b. Becomes 0

Which of the following is right? And most importantly why?

See i have seen the graphs and i can see their that it passes through 0 but why cant it become 0?

Also why cant we explain existence of node by assuming particle nature of atom?


closed as off-topic by AccidentalFourierTransform, Cosmas Zachos, CR Drost, sammy gerbil, stafusa Jun 25 '18 at 13:25

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  • $\begingroup$ What about : at a node the wave function is zero. $\endgroup$ – my2cts Jun 24 '18 at 15:13
  • $\begingroup$ Not clear what you are asking. What is the difference between "passes through zero" and "becomes zero"? Where have you read that we cannot explain the existence of a node by assuming the particle nature of the atom? A node is a property of a wave, not a particle. See J Murray's comment below. $\endgroup$ – sammy gerbil Jun 24 '18 at 21:40
  • $\begingroup$ There is a difference which I told in another comment. Yes definitely node is a property of wave but if we were assuming Bohr's model, then how can we discard the idea of node in an atom, that was my question. Anyway it's clear now. $\endgroup$ – RS2000 Jul 22 '18 at 11:12

Passing through zero and becoming zero are the same things, just said in different ways. The crucial thing to note is that the probability associated the w.f is zero at a node. That signifies that the probability to occupy the state described by the node is zero.

Edit: Note the comment made by J. Murray

  • 3
    $\begingroup$ Note that the probability of finding a particle at any individual point is zero, so this is not unique to a node. $\endgroup$ – J. Murray Jun 24 '18 at 15:27
  • $\begingroup$ No I believe they are 2 different things, passing through zero means the probability at 0+ and 0- are finite but becoming 0 means that at both the points it's 0 only which is not the case with node that's y we don't consider infinity a node $\endgroup$ – RS2000 Jul 22 '18 at 11:07

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