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Distance between electron and nucleus

Question:
What force (or theory) keeps the electrons in the 1s orbital from 
hitting the nucleus or eventually running out of energy?
Also how does that relate or does it ever relate to the
decay of an atom?
 dcheng

Answer 1:  This is a fantastic question -- try the Physics section.
Seriously, I think the folks in physics would be more capable
of addressing this particular subject.  I will point out that the
electron has very low probability of ever being in the nucleus.  As
you know there is no node at the nucleus of a 1s orbital and its
amplitude is greatest there but the probability of it being there
is very low because the nucleus occupies such a small volume.
The probability of finding the electron at the center of the nucleus
is zero (for an infinitesimally small space) and the distance from
the origin at which you are most likely to find a 1s orbital is
_FAR_ outside of the nucleus.
Be sure and post your question in the Physics section.  I'd like to
see more details on this too.
 gregory r bradburn

Answer 2:
*Ahem*....I'd like to mention to gregory that there are
two full-time theoretical physical chemists answering
questions on NEWTON (frank brown and myself)...plus,
I have a degree in physics. So atomic questions can be posted
here to good effect.
 
Greg's points are all correct, but I'd like to add something.
Fundamentally, the student's question is not answerable
in the sense that atoms do not behave like macroscopic objects.
If they did, the electron would collapse onto the nucleus due to
their mutual electrostatic attraction....
 
One of the predictions of quantum mechanics is that any particle
(or system of particles) which experience some sort of attractive
interaction will, instead of settling down to a motionless "collapsed"
state, settle into a state which has a distribution of momentum
and position values. If this were not the case it would be inconsistent
with the fact that particles have wavelike properties (electrons
are diffracted by crystals, for example)...and any system that behaves
like a wave must satisfy an uncertainty principle for the measurement
of its position and momentum (or group velocity if you prefer).
So, in an idealized sense, if an electron were to spiral down
onto the nucleus and stick to it, there would be virtually no
uncertainty in its position or in its momentum...which violates the
uncertainty principle.
 rtopper

Answer 3:
   Electrons with net velocities along paths circling the nucleus (those
not in "s orbitals") don't hit the nucleus for the same reason that a rock
you threw from a moving car wouldn't hit a billboard if you aimed directly
at it: The sideways part of the velocity would cause it to miss.  (A closer
analogy to the electron is trying to bean the calliope of a carousel while
riding one of the horses.)
   What about electrons in s orbitals?  Well, they can and do hit the
nucleus.  Not often.  Although the probability per unit volume of an s
electron being at the nucleus is very high, the volume of the nucleus is
only about a trillionth the volume occupied by the typical electron.  But
occasionally one strays in, and then (if certain factors relating to the
physics of the nucleus are right) it gets absorbed by a proton, which turns
into a neutron and emits a neutrino.  This, as you've guessed, is a form of
radioactive decay of elements, called "K-shell capture" because it normally
only happens for electrons in the lowest (K) shell.  The atom loses 1 in
atomic number (e.g. a chlorine atom turn into a sulfur atom) but stays
neutral.  The neutrino is hard to detect, but the hole the captured
electron leaves is filled by an electron "falling down" from a higher
energy level in the atom, which results in an X-ray being emitted that you
can detect.
     Incidentally the s electrons that come very close to the nucleus and
don't get captured are the source of "spin-spin coupling" in nuclear
magnetic resonance (NMR) spectra.  This effect basically allows you to
count the number of hydrogens bonded to each carbon atom in a molecule.
Pretty helpful if you're trying to figure out what the molecule looks like.
 christopher grayce


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