radius of a nucleus

Radius of the nucleus

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Radius of the nucleus

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The atomic nucleus is what we can call the core of the atom, with electrons found \textit{orbiting} on the outskirts. It is extremely dense and accounts for more than $99\%$ of the total mass of the atom. It generally comprises of a swarm of two kinds of particles: protons and neutrons (with the exception of the hydrogen nucleus which consists just of a single proton). Protons and neutrons are quite similar in many respects, including their mass, but electrically they are very distinct. Protons carry a positive charge while neutrons are electrically neutral. The number of protons decides the overall behaviour of the atom and is what distinguishes one element from another. The number of neutrons can also vary and will affect the behaviour of the nucleus but it does not interfere much with the properties of the atom.

As an example, the simplest element of all, hydrogen, consists of a single proton as the nucleus and an electron in orbit around it. Now, it is possible to add more neutrons to the nucleus without changing the properties of the atom much. If one neutron is added, we have a nucleus that is given the special name of deuteron. The corresponding atom with this nucleus is called deuterium. If two neutrons are added, we get triton whose corresponding atom is called tritium. Adding any more neutrons will lead to strong instability which we will discuss later. Any of such nuclie that have the same number of protons but different number of neutrons are called isotones.

Moving on along the periodic table, the next element is Helium comprising of two protons and two neutrons in its nucleus. The corresponding helium atom has two electrons orbiting around the nucleus. In any neutral atom, the number of protons in the nucleus equals the number of electrons around it. As before, we can add more neutrons to this system producing isotones of He. The next element Lithium has three protons and three neutrons in its nucleus. And you get the idea: every next element has one more proton than the previous one and the most stable isotone has the same number of neutrons as the protons. But this symmetry between the number of protons and neutrons breaks down for heavier elements (with $A>100$) where the neutron to proton ratio must be greater than one for better stability.

So, to conclude, the atomic nucleus is comprised of protons and neutrons, the numbers of which decides the properties of the nucleus. But you must realise that such a setting is not as natural as it sounds. The protons are positively charged and hence repel one another. While neutrons being electrically neutral do not participate in the electrical interactions and hence cannot apparently prevent protons from flying apart. And yet atomic nuclie exist. We therefore conclude that there must be yet another force different from electrical force that holds the protons and neutrons together. This is the strong force. It acts on protons and neutrons equally well (charge independent) and is over 100 times stronger than the electrical force, hence successfully keeping the nucleus together. Electrons, by virtue of their composition, do not experience strong forces. They are held in their orbits around the nucleus only via electromagnetic forces in conjunction with the quantum rules.

If the strong force is so strong, why is there at all an instability when a large number of protons and neutrons pile together to form larger nuclie? We could even possibly have so large a pile so as to be visible to the naked eye. The answer to this lies in the fact that the strong force is an extremely short range force. It acts on nucleons (a collective label for protons and neutrons) only when they are extremely close to one another. As close as $~10^{-15}$ m. Such a distance is $100,000$ times smaller than the size of the atom. On the other hand if the nucleons are at a separation even slightly larger than this, the strong force between the nucleons is as good as non-existant. This means that inside nuclie that consist of a number of nucleons, every nucleon will tug at only those nucleons that are in their immediate neighbourhood.

Such nature of the strong force leads to two consequences: nucleons in large nuclie can attract through the strong force only those in their vicinity, and not those that are farther, allowing the coulomb repulsion to dominate, making the nucleus ready to split apart and hence unstable. The other consequence is that the density in large nuclie remains almost uniform throughout since nucleons tug at only those in their vicinity. If this were not so, larger the number of nucleons (A), greater would be the strong force on all of them, causing them to cram together more and increasing the density of the nucleus (merely due to its size). But the short range nature of the strong force doesn’t allow for this and hence we observe a consistent density in all nuclie (about 0.15 nucelons/fm$^3$). This constancy of density with increase in $A$ is called the saturation of nuclear forces.

This further leads to a correspondence bewteen the volume of each nucleon of a nucleus and the total volume of the entire nucleus itself. Summing up the volumes of all the nucleons yields the volume of the nucleus. That is, for a nucleus of mass number $A$, the volume of the nucleus is

\[V=AV_0\]

Most stable nuclie are spherical with volume $V={4\over 3}\pi R^3$ and we also assume the nucleons to be spherical with a volume $V_0={4\over 3}\pi r_0^3$. Substituting these, we get

\[{4\over 3}\pi R^3=A{4\over 3}\pi r_0^3\]

Or, taking cube root on both sides,

\[R=r_0A^{1/3}\]

Where it turns out that $r_0=1.2$ fm when the charge distribution of the nucleons is considered (measured by scattering of charged particles like electrons aimed at the nucleus), which tells us the extent to which the nuclear charge is spread and $r_0=1.4$ fm when the mass distribution is considered (scattering of neutrons by the nucleus). In conclusion, larger the atomic mass of the nucleus, larger its radius (or size) is.

As for the universality of the above formula for nuclear radius, it holds good for heavier nuclie with $A>20$. Lighter nuclie do not have a sharp well-defined boundary so that we may speak of the size of the nucleus in terms of its radius. In other words, the nuclear density dies gradually over a region of few fms and hence the boundaries of these lighter nuclie appear fuzzy, making the whole idea of volume, somewhat vague. For heavier nuclei too, the idea of a sharp boundary does not hold well, the boundaries are still fuzzy but there is a finite region over which the nuclear density is constant and begins to die after a certain distance from the centre.

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