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.
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.
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.
