On 27/5/2003, Kevin
Phyland posted:
One of my more
astute Physics students asked me a question today that I couldn't
readily answer (no great feat...it was the asking that
probably floored me)...
As I
*understand* it an alpha particle is a Helium-4 nucleus...my student
asked whether the emission of the alpha particle left the parent atom
negatively-charged, to which I knee-jerkingly responded, "Obviously!"
But it got me
wondering...it seems like it should be so...but no (high-school) text I
had could either confirm or deny this, or whether such a
charging of the parent atom had any ramifications chemically...
While you ponder
that (as I did) my student then proceeded to ask me why a Helium
nucleus was the "optimum" way for the parent nucleus to
emit a particle!
I vaguely recall
from Physics 303 a million years ago that it is an extremely stable
nuclide but still couldn't answer with any certainty as to
why it was preferable to emitting say just any number of nuclides with
same proton-neutron numbers (i.e. an oxygen-16 nucleus
etc...)
Any
clarification possible?
Donald
Lang replied:
First
question first. Charge
is conserved. So the atom is left with two more electrons than belong
to it. It also has enough momentum so that it probably collides with
several other atoms almost immediately and there is considerable
rearrangment.
A
metal container full of Uranium is predicted to stay uncharged. If you
can get rid of other radiations, it should be a fairly easy
experiment....
The
answer to the second question is that everything that is not forbidden
is compulsory, but some compulsory things take longer... And so
something else that is more compulsory happens first. If you know
the masses of the atoms (a study in itself) then you can compare the
mass of say Uranium 238 with the sum of the masses of Thorium 234 and a helium
atom. U238 is heavier, so in due time, which averages out in the
billions of years, it splits that way. Hacking off a neutron or a proton
by itself you have to add energy to get any known isotope with mass
237. It don't go. Not at all. Likewise a deuteron does not have enough
binding energy to allow you to hack it out and leave anything of mass
236..
There
is enough energy left over to make it possible to split U238 into
two atoms much lower in mass. The barrier against this happening
fast involves nuclear and electrostatic forces. It is compulsory in the
long run but alpha emission is a lot faster. Spontaneous fission is one
decay mode of Uranium 238. Thorium 234 is heavier than Uranium
234. So the nucleus of Thorium 234 emits two electrons one after the
other, and a few other things happen as well and you land up with
Uranium 234, which needed those two electrons to balance out its
charge..
That
concludes our lesson on radioactivity. There will be a compulsory exam
later today.
Ray
queried:
Along with alpha
particle emissions, aren't beta particles (electrons) also emitted?
If so, then the
charge would be balanced back to neutral or base charge for the
particular element involved.
Donald
answered:
Nice
try but not in the ball park.
In
the various Uranium decay sequences there are alphas, betas and gammas,
sometimes in competition. Charge balances are immediate. You can't wait
for the next burp.
When
a Uranium 238 has been gone long enough, all the daughters will also be
gone. They all have shorter half lives. So eventually there will
be lead 206 with 82 electrons, eight helium atoms, each with two
electrons, and quite a lot of energy, mostly distributed in very
small packets.
You
can discover by simple arithmetic that there are now 98 electrons in
place of the 92 on U238. So there must be six beta decays on any path
from Go to Whoa. There will also be six neutrinos long gone from this
earth.
The
beta decays occasionally compete with possible alpha decays but most
events on the chain involve nuclei that just do alpha decay or just do
beta decay.
Emphatically
--- Charge is conserved in each event. To avoid one simple question: In
any nuclear process, in any observed physical process to date, the
algebraic sum of the initial charges is the same as that algebraic sum after
the event.
The
statement includes beta decay. Emission of a negative electron involves
leaving a nucleus with one extra positive charge. One negative emitted,
one positive left behind: net effect no change in charge.
If
you insist, I will admit that people are unlikely to conduct serious
checks on charge conservation in routine nuclear work these days. Until
someone gives a good reason otherwise, it is not an experiment that will
be easy to get funded.
After an amusing digression involving President
Bush and the Colliding
Pants, originally published in The
Onion, Kevin Phyland posted:
Yeah...I know
that within a given alpha particle event charge is conserved. What I
meant was, since the remaining nuclide is *now* charged and the
positive charges (i.e. alpha particles) have headed off to other
regions, what happens to the extant negative charge on
the parent nuclide?
Consider: We
have a (gedanken) blob of U-238...it undergoes alpha decay, produces
lots (I know, half-life = long) of alpha particles...so now we
have lots of negatively-charged U-234?
*If* this is the
case, why don't the negatively-charged nuclides repel each other or
(failing that theory) attract electrons from other atoms?
Sorry, I didn't really
follow the original explanation(s)...He-4 is obviously the ejaculation
of choice but what happens to the (I presume) IONS
produced?
Donald
Lang replied:
In
our blob, as described below, the alphas set off as naked He 4 nuclei.
Rutherford and Co studied their history out in the open and found
they could pick up and shed electrons until they came to rest, at which
time they became He 4 atoms. Meanwhile the Th 90 /234 items left behind
with 90 nuclear charges and atomic mass 234 have two electrons over and
above their usual supply on hand. The atoms as a whole have each been
given a jolt and push off in the opposite direction from their
corresponding alphas.
Eventually
they too come to rest and need 90 electrons to settle down. Each alpha
picks up two electrons eventually. Each Th 234 sheds two from its
nascent state. In a blob of Uranium metal the surplus electrons at one
point will migrate to where the deficit occurs. Even in a large blob of
some non conducting compound of Uranium surplus and deficit atoms are
produced in equal quantities and should cancel out in quite small
regions.
Hope
the above helps
