Here is a sketch of a 2py orbital. The sketch is about 800 pm wide, the coordinate (x,y,z) axis are all shown. Suppose an atom with its nucleus at the origin has an electron in a 2py orbital. Complete each row of the table below by deciding how PA , the probability of finding the electron at point A, compares to PB, the probability of finding the electron at point B.

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Here is a sketch of a 2py orbital. The sketch is about 800 pm wide, the coordinate (x,y,z) axis are all shown. Suppose an atom with its nucleus at the origin has an electron in a 2py orbital. Complete each row of the table below by deciding how P, the probability of finding the electron at point A, compares to PB, the probability of finding the electron at point B.

-
Transcribed Image Text:-
point A
point B
compare PA to PB
P
< P
A
В
P
= P
100 pm above the nucleus, along the +z axis.
100 pm below the nucleus along the -z axis.
A
P
> P
A
В
< P
A
В
100 pm in front of the nucleus, along the -y axis.
100 pm to the left of the nucleus, along the -x axis.
P
= P
A
В
Р >Р
A
В
Transcribed Image Text:point A point B compare PA to PB P < P A В P = P 100 pm above the nucleus, along the +z axis. 100 pm below the nucleus along the -z axis. A P > P A В < P A В 100 pm in front of the nucleus, along the -y axis. 100 pm to the left of the nucleus, along the -x axis. P = P A В Р >Р A В
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