ELECTRONS AND PROBABILITY
OBJECTIVES:
1.
2.
3.
To determine the distributions of “hits” about a bullseye.
To obtain and interpret probability information.
To compare the results of this experiment to the quantum theory
prediction of the electron density around an atomic nucleus.
DISCUSSION:
When scientists study the motion of small particles like electrons, they find that
these particles do not seem to behave like the more familiar larger particles such as
billiard balls or bicycle wheels. With the larger particles one can determine the path the
particle will follow, given its initial speed and direction. This does not seem to be
possible with electrons. In order to deal with the rather different properties of electrons, a
science called quantum mechanics was developed. This science is used instead of
classical physics, which works well for larger particles.
According to quantum mechanics, the best we can do in our experiments is to find
the livelihood that an electron will be at a given point in an atom. Say we did a hundred
experiments to locate an electron in an atom, we would find the electron in one hundered
places, some of which were closer to the nucleus than others. Perhaps in ten of tehse
experiments the electron would be found in a certain region in the atom. We would say,
then, that the probability of finding the electron in that region is 10/100, or 0.10. This
kind of information, while useful, is far less specific than knowing the path of the
electron. Quantum mechanics tells us we can find probabilities of the sort just mentioned,
but not the path an electron will follow.
In order to help make clear now one obtains and interprets data on electron
positions, we are going to do an experiment which will furnish very similar data. In our
experiment we will drop darts on a target like the one shown in Figure 1. The target
consists of ten concentric rings one cm wide. If we drop darts on to such a target, we find
that not all of the darts hit the bullseye. There will be a scatter of hits over the target, with
more hits near the bullseye than far from it (if you are good at it, that is).
From the pattern of hits, we can determine the hit density, which is the probability
that a dart will hit a particular unit area located on a given ring on the target. If you think
about this a moment, you can see that the hit density at a given distance from the bullseye
will have to equal the hit probability at that distance divided by the area of the ring at that
distance. If, for example, 16 out of 100 darts hit in ring number 3 in the figure, we can
say that the probability that the next dart will hit that ring is 16/100 or 0.16. That is the hit
probability for that ring. Since the area of that ring is 16 cm2, the hit density would equal
0.16/16, or 0.01. This means that if we placed a square 1 cm on an edge anywhere on ring
3, the odds are that 1 dart out of the next 100 would hit somewhere on that square.
The results of quantum mechanics calculations of electron positions are very
much like those from this experiment. One result from quantum mechanics is shown in
Figure 2 where we have plotted the electron charge density in the H atom as a fucntion of
distance from the nucleus. This graph gives the probability that an eelctron will be found
in a given small volume located at a particular distance from the nucleus. The electron
charge density curve is like the hit density curve in this experiment and essentially tells
us the likelihood of an electron being found at any given point in the atom. From the
graph, we see that the charge density is largest at the nucleus. If we had to select a small
region in the atom where we would be most likely to find the elctron, the best place to put
that region would be at the nucleus. If we put that region anywhere else, we would be less
likely to find the electron there.
Frequently, electron charge densities in atoms are represented in diagrams like
that in Figure 10-10 on page 289 in your text, where the electron charge cloud in a
hydrogen atom is represented. The heavier density at the center of the atom reflects the
high charge density at that point. Electron orbitals used to indicate electron positions are
really charge clouds showing electron density in the atom.
HYDROGEN ATOM TARGET
*Diagram*
*Diagram*
Figure 1. A two-dimensional target for determining
the probability of hits around a bullseye.
*Diagram*
Figure 2. Graph showing electron density
for a 1s electron in a hydrogen atom.
PROCEDURE:
1.
Obtain a target, a piece of plywood, goggles, and a dart. The concentric
cirlces on the target have the radii shown in Figure 1. Tape the target to
the center of the plywood and place it on the floor. Stand on the seat of a
desk and extend your arm so that it is about 2 meters directly above the
ecenter of the target. Drop the dart (DO NOT THROW) onto the target in
such a way as to try to hit the bullseye. The second student should retrieve
the dart and mark the position of the hit with a small x. Repeat this
procedure 99 times for a total of 100 drops. (Students should revrse
positions at the half-way point.) Don’t count any drops which fall outside
the largest circle. Detach the target and return the dart and board.
2.
Count the number of hits in each concentric ring and record this number
on your data table. Divide the number of hits for each ring by its area to
determine the hits per cm2.
POSTLAB ASSIGNMENT
1.
Number of
Concentric
Ring
1
2
3
4
On the table below, place the number of hits in each concentric ring and
the calculated number of hits per unit area.
Avg. distance
of Ring from
Bullseye (cm)
.5
1.5
2.5
3.5
Area of
concentric Ring
(cm2)
3.1
9.4
16
22
Number of
Hits in
Ring
Number of Hits
per Unit Area
(hits/cm2)
5
6
7
8
9
10
2.
4.5
5.5
6.5
7.5
8.5
9.5
28
35
41
47
53
60
Plot the number of hits per unit area (hits/cm2) on the y-axis against the
average distance of the ring from the bullseye on the x-axis. Draw a
smooth curve through the plotted points and extend the curve to x = 0.
How does your graph compare to the graph in Figure 2 in the discussion?
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3.
State in complete sentences how the probability of a hit in any given unit
area on the target varies with the distance of that area from the bullseye.
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4.
Where on the target would you put a one cm square so that it would have
the greatest chance of being hit?
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