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I think we should take another look at the original question:
On 01/12/2011 03:38 PM, Jeff Loats wrote:
In discussing Rutherford scattering I ask students to use the simple caseof
an alpha particle colliding head on with an electron at rest. The idea isto
use conservation of energy and momentum to show that in a classicalpath
"billiard ball" model, the alpha particle can ignore electrons in its
to a good approximation.
This term a curious student asked some great questions about what would
happen when such a collision took place.
This exceeded my knowledge a bit, so I thought I would ask here.
What would happen if an alpha particle was fired head-on at an
electron.
(I know the question is posed in an incorrect "billiard ball" fashion.)
Actually, let's go back one step further. As always, we should ask
where the student is coming from, i.e. ask what prompted the question.
I suspect this is one of those cases where the student's reach exceeds
his grasp, i.e. he has identified a truly profound issue, but cannot
adequately articulate the question. And the rest of us are in the
same boat! In particular, the question quoted above is not really the
key question. We have not yet asked the key question.
IMHO here is the question that the student should have asked ... and
the rest of us should have asked:
By way of background: Rutherford and colleagues were working in the
1909-1911 timeframe. That means they started out with no clue about
atomic structure. The plum-pudding model was consistent with prior
experimental observations. They famously discovered that matter
contained nuclei, which were /not/ well described by the plum-pudding
model. Using 20/20 hindsight we understand this observation in terms
of modern atomic theory.
But as Paul Harvey would ask, what about the rest of the story? Why
did not Rutherford et al. discover that the plum-pudding model was
no good with respect to electrons (not just with respect to nuclei)?
More specifically, why was the plum-pudding model ever even remotely
consistent with *any* experimental observations? (Even the observations
of Rutherford et al. are "mostly" consistent with the plum-pudding
model; the famous large-angle scattering events are rare.) In terms
of modern physics, how do we understand the /difference/ between the
alpha/nucleus interaction and the alpha/electron interaction?
(I know the question is posed in an incorrect "billiard ball" fashion.)
Indeed. I recommend that in the future you not attempt any "billiard
ball" explanations for why we more-or-less ignore the alpha/electron
interactions in the context of these experiments. For one thing, even
if we could ignore the effect on the alpha particle, we could not ignore
the effect on the electrons. A billiard-ball interaction would produce
loads of multi-MeV scattered electrons ... which are not observed.
The correct answer lies not in billiard-ball mechanics but rather in
wave mechanics. The electron is not a hard little ball but rather a
large fluffy cloud, i.e. a large fluffy wave packet. If you shoot an
alpha particle at it -- even dead-center head-on -- the alpha will go
straight through, to a good approximation. This will make a small hole
in the cloud, but most of the cloud will still be there.
In particular, the Coulomb force will transfer remarkably little
momentum. As the alpha approaches from the left, there will be a
leftward force on the electron ... but as the alpha departs to the
right, there will be a rightward force on the electron, and these
cancel to first order.
Normally we think of the Rutherford experiment as having the electron
at rest and the alpha in motion, but with a Galilean transformation
we can equally well think of the alpha at rest and the electron in
motion. First of all, this is closer to being the center-of-mass
frame. More importantly, in this frame the student can understand
the situation in analogy to an ordinary sound wave interacting with
a flagpole. Even if the wave-packet of sound hits the pole head-on,
the wave packet will not bounce back toward the source. Most of the
wave packet will just diffract around the flagpole and keep on going.
There will be /some/ sound reflected back toward the source, but to
a first approximation (namely the first Born approximation) the amount
of energy scattered in the back-to-source direction is the same as the
amount of energy scattered in any(*) other direction in the XY plane.
(*) There is an exception for the forward direction, i.e. the
direction diametrically away from the source, associated with
the interference between the scattered wave and the other 99%
of the wave, i.e. the part of the wave that didn't scatter and
just kept going. We have a name for this: It is the /shadow/.
Serious suggestion: Have the student set up a ripple tank and play
with it. That's what I did when I was in high school. Hours and
hours and hours of gazing at ripples in tanks. There are also
"ripple tank" applets these days, some of which are good and should
be used /in addition/ to doing the experiment (not instead of doing
the experiment).
In this particular case, CW mode is easy to set up, and allows you
to see the shadow, and is fine as a starting point ... but if you
really want to see the scattered wave, and appreciate how small it
is, you need a wave packet, i.e. a finite-length wave train.
====
Note that the Coulomb scattering cross-section goes like Z _squared_
so this is another reason why the plum-pudding model works for the
electron cloud but not (always) for the nuclei. Z^2 for a gold nucleus
is more than six thousand times bigger than Z^2 for an electron. Even
given that the electrons are 79 times more numerous than the nuclei,
the nuclei still do most of the scattering.
To summarize: Hitting an electron is different from hitting a gold
nucleus:
++ It is different by a factor of Z^2 which sets the cross section.
++ It is different by factors involving √m (where m is the mass)
which are (inversely) related to the size and fluffiness of the
wave packet.
-- AFAICT no part of this can be properly explained by analogy to
billiard balls. Yeah, billiards conserve energy and momentum,
but so does everything else. The thermal de Broglie length for a
billiard ball is very small compared to the classical size of the
ball, but this is not true for electrons. I do not recognize any
distinction between waves and particles, but I do recognize the
difference between the short-wavelength limit (billiards) and the
long-wavelength limit (sound meets flagpole, and electron meets
alpha). The /way/ in which the wave conserves energy and momentum
(including the aforementioned shadow) is very different in the two
limits.
-- The suggestion that the alpha might "cleanly miss" the electrons
is also AFAICT not any part of the explanation. Not only is the
Coulomb interaction an infinite-range interaction, but the electron
cloud is a huge target. There is electron density everywhere in
the gold foil, filling the entire volume. You can't miss. Instead,
you punch a small hole in a big fluffy cloud.
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