Why is boron so good at neutron absorption?

Question

Why is boron so good at absorbing neutrons? Why does it have such a large target area compared to the size of its nucleus?

Answer 1

It's boron-10 that is the good neutron absorber. Boron-11 has a low cross section for neutron absorption.

The size of the nucleus isn't terribly relevant because neutrons are quantum objects and don't have a precise position. The incident neutron will be delocalised and some part of it will almost always overlap the nucleus. What matters is the energy of the reaction:

$$ ^{10}\text{B} + n \rightarrow ^{11}\text{B} $$

and the activation energy for the reaction.

I'm not sure we understand nuclear structure well enough to give a quantitative answer to this. However neutrons, like all fermions, like to be paired and $^{10}$B has 5 neutrons while $^{11}$B has 6 neutrons. So by adding a neutron we are pairing up the neutrons and completing a neutron shell. We would expect this to be energetically favourable.

This argument would apply to any nucleus with an odd number of neutrons, but $^{10}$B is a light nucleus so we expect the effect to be particularly big. The lightest such nucleus is $^{3}$He, with one neutron, and that has has an even bigger neutron absorption cross section. However practical considerations rule out the use of $^{3}$He as a neutron absorber. $^{6}$Li, with three neutrons, also has a reasonably high cross section, though it is less than boron and helium.

Answer 2

One thing which John Rennie hasn't mentioned is that boron gets a certain amount of "help" from the fact that neutrons are deliberately retained within a nuclear reactor.

This is the function of the moderator. Neutrons ejected from a decaying uranium travel fast (as is typical of products of nuclear reactions) and if the reactor was filled with vacuum, they would just escape through the walls. But it's filled with some low mass material like water or graphite, which slows down and scatters the neutrons (without reacting with them) until they become "thermal neutrons" (i.e with the same kinetic energy as the surrounding material at the same temperature), so that they just wander about aimlessly within the reactor. Besides the reduced velocity, they also have a much increased path length.

Once this has been done, we can to a reasonable extent ignore the loss of neutrons through the reactor wall, so there are only three other things that can happen:

1. It can undergo beta decay, becoming an electron, proton and neutrino. The half life for this is about 10 minutes.

2. It can collide with a fissile atom such as uranium and propagate a nuclear reaction

3. It can be absorbed by a material like boron.

So you see, once the neutron has been slowed down, the small cross section of nuclei becomes something of a non-issue: it has an average of 10/ln(2)=15 minutes to find and interact with a nucleus, which is absolutely ages on an atomic timescale.

As to why boron in particular is good at absorbing neutrons, John already covered that.

https://en.wikipedia.org/wiki/Neutron_moderator

As an intuitive way of understanding the role of scattering, consider the effect cloud cover has on visible light. During the day, it prevents the sunlight reaching the Earth's surface and reflects it back into space, even though it does not absorb much light itself. At night, it prevents the light from buildings and streets from being released into space, and reflects it back to the Earth's surface where it is absorbed.

According to https://en.wikipedia.org/wiki/Neutron_cross_section boron-10 has an absorbtion cross section of 200 barn for thermal neutrons and 0.4 barn for fast neutrons. A barn is 1E-28m. For comparison, the van de waals radius of boron is 192E-12m https://en.wikipedia.org/wiki/Atomic_radii_of_the_elements_(data_page)

The following is a back of the envelope calculation showing the value of slowing the neutrons down.

The probability of interaction B-10 with a slow neutron is about 200E-28/pi*4/(192E-12)^2=7E-7 for a one atom thick sheet of boron atoms. For a 1mm (5208000 atom) thick sheet, the probability of the neutron surving is (1-7E-7)^5208000 = 0.026 which is pretty low. For fast neutrons a one atom thick sheet interacts with 0.4E-28/pi*4/(192E-12)^2= 1.4E-9, and the probability of surviving a 1mm thick sheet is (1-1.4E-9)^5208000 =0.9928 which is pretty high. Disclaimers: this assumes pure B-10; the Van Der Waals radius is not necessarily the best one to use, depending on the speciation of boron.