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The Strong Force is often, instead, called the Strong Interaction. At atomic scales, it is about 137 times stronger than the Electromagnetic Force, about 1013 times stronger than the Weak Force, and about 1038 times stronger than Gravitation. They are all governed by a similar law, based on Proca's massive electromagnetism, which is a version of Maxwell's theory that includes the term for the Yukawa potential.
At atomic scales, but distances greater than the diametre of one hadron, Strong Interaction approximates to being roughly constant, at about 100000 Newtons. It is, therefore, impossible for a single quark to exist by itself a property connected to QCD confinement and asymptotic freedom (NS, 04-Dec-1993, p25). Any attempt at removing a single quark an infinite distance from its partners would involve applying an infinite amount of energy, and even removing it a short distance would involve such large amounts of energy that new particles would be created instead (NS, 06-Jun-2015, p36). So, from outside the nucleus, we do not feel the effect of the Strong Interaction itself, but instead we just see the lumped effects from the whole nucleus (just as we see gravitation as if the mass were all concentrated at the centre of gravity of the object, and similarly for electrostatic or magnetic forces). This lumped effect is known as the Residual Strong Force, and is to the Strong Interaction as van der Waals forces are to the Electromagnetic (coulomb) Force. |
FissionThe fission event does not always produce the same products, but suppose that it typically produces one barium nucleus and one krypton nucleus (Z=92 breaks into Z=56 plus Z=36). That would mean that A=235 splits into A=137 plus A=84, plus or minus a few isotopes). So, according to this, there would be 235-221=14 surplus neutrons. Not all of these are useful for the chain reaction, being 'non-thermal'. So, a certain proportion, α, of those neutrons will go on to hit other uranium nuclei, while the other proportion (1-α) will escape without doing anything further. This is the essence of the chain reaction. The first neutron will lead to k neutrons (where k=αN) some small time interval later, and hence to k2 electrons one interval later, k3 after that, and so on. The question is, though, whether k<1 or k>1. It is only if you have enough uranium atoms around that you can achieve k>1. So, in a bomb, this is all you would need. In a nuclear power station, though, you must balance k at exactly 1. You do this by starting off with k slightly larger than one (so more than the critical mass is required again), and then moving the neutron-absorbing control rods around (or ultimately moving the fuel rods around) so that k is always hovering around the k=1 figure. |
Inside the atomic nucleus, as a result of their strong attraction, quarks do not go into orbit round each other. That only works for the inverse-square law of gravitation and electromagnetism:
With most of the forces, particles are more attractive (or less repulsive) when they are of opposite polarity. With the strong force, the polarity is three-way, and is referred to by metaphor with colour (with red+green+blue cancelling out in the same way as plus+minus, north+south or up+down).
If three such quarks were to start orbiting at less than 4 nucleon's width, the Residual Strong Force would be so strong that it would attract them even closer together, until some point (at very close range) at which the force becomes repulsive. Thus, the Residual Strong Force is repulsive for distances, r, much less than 1.7fm, but is strongly attractive at r=1.7fm, reducing roughly exponentially, as given by the Yukawa potential, after that:
Since the electrostatic forces fall off with an inverse squared law, there must be a point where the two curves intersect. This seems to be at r=2.5fm, beyond which the Residual Strong Force ends up being weaker than the Electromagnetic Force. It turns out that this is somewhere between r=3.67 to 4 nucleon diameters. Putting this into the usual equation for the volume of a sphere, V=(4/3)π.r3, this means that the crossover is between V=207 and 268 nucleons. Since each nucleon weighs one atomic unit of mass, this means that it happens between A=207 (which is Pb) to 268 (which is Db). So this explains why it is so difficult to find chemical elements above Z=82, and to make them above Z=105: the electrostatic repulsion exceeds the binding forces in the nucleus, and the oversized nucleus ends up disintegrating (or, rather, the probability of it disintegrating increases, and so its half-life reduces).
In the equations above:
According to this approximation (which does not model the repulsive effect of the force) Fr starts smaller than Fe, becomes equal at r=0.76fm, peaking at r=1.47fm, then reduces again until the two forces are equal at f=2.5fm, and then Fr is weaker than Fe for all r greater than that.