Answer:
Explanation:
In a standing wave function[tex]\psi (x,t) = A sin(kx)[/tex] characterized for x between (0.a). on the off chance that the amplitude of the wave interchange from positive to negative at the interval. there probably been a node at [tex]x_0[/tex], among 0 and a to such an extent that [tex]0<x_0 <a[/tex]. The reasoning is right that the likelihood of discovering the particle at the node [tex]x_0[/tex] is 0 in light of the fact that by definition, the nodes of the wave are the place where the wave function falls and is equivalent to 0. Since the likelihood of discovering a particle at a position [tex]x_0[/tex] at time [tex]t_0[/tex], is provided by [tex]P=|\psi(x_0,t_0)|^2 dx[/tex], this implies that at the nodes of a standing wave,
[tex]P = | \psi (x_0,t_0)|^2 \ dx \\ \\ P = |0|^2 dx \\ \\ P = 0[/tex]
 So the reasoning that the likelihood of the particle being at [tex]x_0[/tex] is 0 is right. Â
However, to examine whether the particle can travel from a position  [tex]x <x_0[/tex] to a position of [tex]x_0>x[/tex]. All together words, can the molecule be found on one or the other side of the node?
The appropriate response is yes. Â
Recall that in quantum mechanics. wave functions at most present with the likelihood of discovering a particle at a specific time inside a time frame. The wave function doesn't present with an old classical actual trajectory that a particle should follow to go in space: all things being equal, it simply yields chances of whether a particle can be found in a specific spot at a specific time. So the reasoning that a particle can't get from a position [tex]x <x_0[/tex] to a position of [tex]x>x_0[/tex], is incorrect.
