From: email@example.com (Edward C. Wood)
Subject: Re: Temperature dependance of fission in fuel rods??
Date: 18 Dec 1995 18:33:50 -0500
"Thomas L. Billings" <firstname.lastname@example.org> wrote:
>Craig Graham <email@example.com> wrote:
>>Ross Tessien (firstname.lastname@example.org) wrote:
>>> Does anyone know about the reactivity rate for typical fission
>>>reactor fuel rods as a function of temperature during a melt down?
>>>Also, are the products produced typically less radioactive than
>>>the original fuel?
>>I'm not an expert on this, but AFAIK there's no temperature
>>dependence in a fission reaction. Fission and fusion are radically
>>In fission, the reaction is caused by the spontaneous decay of an
>>unstable nucleus- okay, it's accelerated by hitting the thing with
>>a thermal neutron to turn it into a more unstable nucleus, but you
>>get the drift. So the decay is influenced by conditions inside the
>>nucleus- not outside the atom. Obviously, heating the thing up has
>>Where I mentioned thermal neutrons- I just meant relatively low
>There are designs of nuclear reactors, such as the one at Chernobyl,
>that have a "positive thermal neutron gradient". This reactor's
>design allows higher heat to make the moderating material work
>better at producing more neutrons with a higher chance of fissioning
>an atom. It also supposedly allows efficient power production along
>with lots of plutonium production for weaponry. I do not know
>details of this, butothers may be able to give a better description.
>The fission process itself is not affected by the heat though, just
>the moderating material.
In reactors where the fuel contains U-238 (almost all of them) the
fuel gives the reactor a negative fuel temperature coefficient of
reactivity -- the reaction slows down as the fuel heats up. This is
also called the 'Doppler coefficient'. It arises from broadening of
resonances in the capture cross-section of U-238 due to thermal
motion of the nuclei.
What does that mean? Neutrons are 'born' very fast; before they cause
fission they are greatly slowed by bouncing around off of moderator
nuclei. On the way down in speed they must pass a 'picket fence' of
capture peaks in U-238. Each 'picket' or resonance is a small range
of velocity within which the probability of capture goes from small
to enormous and back again. At the peaks the probability of capture
(called cross-section) is so large that any neutron passing by will
be captured -- and will be unavailable to cause fission. The velocity
involved here is the relative velocity between neutron and capturing
nucleus. The nuclei are moving about their mean positions in thermal
agitation. A neutron with, relative to the mean positions, a velocity
that is close to a resonance can be right on it relative to some
moving nucleus. Increasing temperature, which increases thermal
agitation velocities of nuclei, increases the velocity difference
that is described as 'close', thus broadens resonances and reduces
the fraction of neutrons that escape resonance capture.
The Chernobyl reactor had a positive 'void' coefficient when full of
water -- replacing water with steam bubbles caused the fission chain
reaction to speed up. That made the accident possible.
The original question includes the following part: "Also, are the
products produced typically less radioactive than the original fuel?"
No. Collectively they are, initially and for a long time, very much
_more_ radioactive -- meaning number of nuclear disintegrations per
unit time per unit mass -- than the original fuel.
Edward C. Wood