Resonances — the Effect of Temperature

The cross-sections of many nuclei exhibit sharp resonances over part of the energy range. Heavy nuclei such as uranium and plutonium have resonances in the range from about 1 eV up to 500 eV or more. Lighter elements have resonances mostly at higher energies of 100 keV or more, although there is an isolated resonance of 23Na at 3 keV. It would be impossible to do a fundamental mode calculation for all these resonances even if they had been resolved experimentally because there would have to be far too many fine groups. Fortunately it is possible to calculate the reaction rate in a resonance directly, but in doing this it is important to allow for the effect of temperature. In outline the method is as follows.

In the vicinity of an isolated resonance the microscopic cross­section for a certain reaction, neutron capture say, is given by

image031(1.32)

image032

Figure 1.1 The effect of temperature on one component of the velocity of a 238U nucleus.

Here ac0 is the capture cross-section at E0, the energy of the resonance peak, and Г is the “width” of the resonance (the full width in energy at half the maximum cross-section). Ec is the kinetic energy of the neutron and nucleus relative to the centre of mass of the neutron — nucleus system. If the nucleus is stationary Ec = E/(l + 11 A) where A is the ratio of the mass of the nucleus to that of the neutron and E is the neutron energy.

Unless the reactor is at absolute zero temperature, however, the nucleus is unlikely to be stationary. It moves at random due to thermal agitation and this affects Ec, which is increased if the nucleus hap­pens to be moving towards the approaching neutron or decreased if it is moving away. The effect on the apparent cross-section can be calculated if the probability distribution of the velocity of the nucleus is known. It is usually satisfactory to assume it to have a Maxwell — Boltzmann distribution. The resulting distribution of the component of velocity parallel to a certain direction (which we can take as the direction of the neutron) is shown in Figure 1.1 for an atom of 238 U at various temperatures.

From this the probability distribution of Ec can be deduced, and hence the mean cross-section for a neutron of energy E. It is given by

1

ac (E, T) « Oc0 EE f (z, x), (1.33)

Подпись: f (Z,x) Подпись: 1 exp( (x - y)2/4z2)d 2Zfn -TO (1 + y2) Подпись: (1.34)

where

Z 2 = 4E0bT /AY2, (1.35)

and

x = 2(E — E0)/Y. (1.36)

T is the absolute temperature and b is Boltzmann’s constant. The approximations used in reaching equation 1.33 introduce negligible errors in most fast reactor applications.

Figure 1.2 indicates how the effective cross-section depends on temperature. The parameters f, Z and x are proportional to the cross­section, the absolute temperature and energy respectively.