The nuclear radius is a very relevant quantity to measure since it determines the scale above which we can offer a chemical description of reality. Given these small scales, measuring a nucleus’s size and shape is difficult, but several methods allow us to do so. This helps us to understand the structure of matter better. Let’s learn more about the nuclear radius and how we can estimate it.
How can we define the nuclear radius?
The nuclear radius is a measure of the size of the nucleus of an atom under the assumption that it is approximately spherical.
Under this assumption, the nuclear radius equals the radius (half the diameter) of the nucleus. We can estimate the nuclear radius with theoretical models and experiments. Below, we analyse examples of both and look at how much they agree on the measured/predicted values.
How can we estimate the nuclear radius?
There are several experimental methods we can use to estimate the nuclear radius. We look at two in this explanation: the closest approach method and the electron diffraction method.
Closest approach method
Rutherford scattering was an experiment carried out by Ernest Rutherford in the early 1900s. Rutherford’s main goal was to investigate the structure of atoms to study the properties of nuclei and provide a reliable atomic model that was based on experiments rather than on theoretical assumptions.
Rutherford performed the experiment by firing alpha particles (two protons and two neutrons) towards a gold foil. The gold foil was surrounded by a screen that detected scattered alpha particles, which allowed Rutherford to measure their amount and deviation pattern.
We have an explanation dedicated just to Rutherford Scattering, so be sure to check that out! Note that Hans Geiger and Ernest Marsden also worked with Rutherford on the gold foil experiment (we’re just telling you so that Rutherford doesn’t get all the credit!).
Since the existing atomic models at the time predicted a very big nuclear radius and almost no space between atoms, Rutherford expected most of the alpha particles to bounce against the foil’s atoms. However, most of them actually landed behind the gold foil, which allowed Rutherford to deduce the following conclusions:
- Matter is almost empty. The distance between atoms is huge compared to their radius.
- The positive charge of an atom is concentrated in a small region that contains most of the mass of atoms, which is the nucleus.
- Electrons orbit the nucleus and contain the negative charge of the atom.
Equation
One way to estimate the nuclear radius is by studying the energy of an alpha particle during the scattering process. Recall that the total energy of a charged particle in the presence of an electric field created by a point-like charge is
\[E = E_p +E_k = k \cdot \frac{Q \cdot q}{r} + \frac{1}{2} \cdot m \cdot v^2\]
where Ep is the potential energy, Ek is the kinetic energy, q is the charge of the particle under study, Q is the charge of the particle creating the electric field, m is the mass of the particle, v is the speed, r is the radial distance between the charges, and k is Coulomb’s constant (with an approximate value of 8.99 · 109 N·m2/C2 ). The units of the energy are measured in joules (J).
It is accurate to note that when the alpha particles are expelled, they are not affected by the electric field of the gold foil’s atoms. This means that they have a total energy determined by their kinetic energy. If we restrict ourselves to particles that are deflected backwards, we know that as they approach the foil, they will slowly lose speed until they are completely stopped by electrical repulsion. At this point, alpha particles only have potential energy with a value we can equate to the initial kinetic energy since energy is conserved at all times.
By equating both contributions and solving for r, we can find the distance between the alpha particle and the nucleus, which allows us to estimate a value for the nuclear radius of gold atoms.
\[r = \frac{2 \cdot k \cdot Q \cdot q}{m \cdot v^2}\]
Problems with the closest approach method
Although this method is simple and intuitive, it is inaccurate (as the name suggests!) for several reasons:
- It is an overestimation of the actual nuclear radius. For it to be accurate in this sense, alpha particles would need to stop their movement at the edge of making contact with gold atoms, which does not happen.
- There are other interactions between alpha particles and nuclei beyond electric interactions. For instance, the strong force affects neutrons and protons.
- It is a considerable simplification to assume that a certain alpha particle is only affected by a gold foil’s atom’s nucleus. Other scattering processes also affect the alpha particle.
- Technically, sending all alpha particles with the same kinetic energy is difficult.
- Neither alpha particles nor the gold foil’s nuclei can be treated as point-like charges.
Rutherford scattering
Electron diffraction method
Diffraction is the bending of waves upon encountering an object or aperture of similar size or bigger than the wavelength.
Diffraction is a phenomenon of wave-like nature. However, one of the important lessons of quantum physics is that the boundary between particles and waves is artificial. We can offer a description of reality in terms of waves. This implies, for instance, that under the appropriate conditions, particles can be affected by diffraction processes.
Before the development of quantum mechanics, a scientist named Louis de Broglie already foresaw this feature of entities in nature and developed a simple mathematical relationship between a particle and its associated wavelength.
The de Broglie wavelength is the wavelength associated with a particle. Its formula is
\[\lambda = \frac{h}{m \cdot v}\]
where m is the particle’s mass, v is the speed of the particle, and h is Planck’s constant (its approximate value is 6.63 · 10-34J·s). The units of the wavelength are in meters.
Imagine a similar scattering experiment like the one described above. However, this time, we use electrons instead of alpha particles. The advantage of using electrons is that they have a relatively small mass (compared to alpha particles), which allows them to accelerate at a speed that yields an associated wavelength of the order of the size of a nucleus.
If the electrons accelerate to this range of speeds, they will exhibit wave-like behaviour when encountering nuclei and form a diffraction pattern (these characteristics have been thoroughly studied). The relevant aspect here is the presence of a peak of diffracted particles that have an amplitude related to the size of the nuclei causing the diffraction.
Equation
Below is the equation showing the relation between the nuclear radius and the angle at which we observe the limit of the diffraction peak:
\[\sin (\theta) = \frac{1.22 \cdot \lambda}{2 \cdot R} = 0.61 \cdot \frac{h}{m \cdot v \cdot R} \rightarrow R = 0.61 \cdot \frac{h}{m \cdot v \cdot \sin(\theta)}\]
R is the size of the diffracting object (the nucleus), and θ is the angle at which we observe the end of the peak of intensity in the diffraction pattern.
This method is much more accurate than the closest approach method and is not influenced by other interactions since the strong force does not affect electrons. The main difficulties of this method include:
- Accelerating the electrons to the needed speed.
- Obtaining a diffraction pattern with enough resolution.
Nuclear radius and nuclear density: what is nuclear density?
As we know, the number of protons in the nucleus determines the element, but the number of neutrons can vary (isotopes). For known elements and their isotopes, we can observe (usually) that as the number of protons increases, there are increasingly more neutrons.
For instance, in light elements, the number of neutrons of different isotopes is close to the number of protons. However, the typical number of neutrons in a nucleus is around 1.5 times the number of protons for heavy elements. Since the number of particles varies in a complex way, it is interesting to study the nuclear mass (which depends on the number of particles) and the nuclear density (which, additionally, depends on the disposition or “packaging” of particles inside the nucleus).
Nuclear mass = number of particles
Nuclear density = disposition of particles
Measuring the nuclear density
Since the nuclear radius can be determined by experimentation and the mass of protons and neutrons is well known, we can estimate the nuclear density under the assumption of uniform spherical spatial distribution.
We can approximate the masses of protons and neutrons by the same value (although they are slightly different): 1.67 · 10-27kg.
Furthermore, we know that the following equation gives the volume of a sphere in terms of radius r:
\[V = \frac{4}{3} \pi \cdot r^2\]
By knowing the number of particles in the nucleus A and their mean mass m, we can estimate the nuclear density ρ as:
\[\rho = \frac{M}{V} = \frac{3 \cdot m \cdot A}{4 \cdot \pi \cdot r^2}\].
Applying this formula to all known atoms allows us to estimate the validity of the uniform spherical distribution we are assuming. It can also explain how the particles are distributed inside the nucleus without carrying out a complex analysis.
The symbol for nuclear density is ρ.
A nuclear density plot for a spherically symmetric nucleus, Wikimedia Commons
Nuclear Radius - Key takeaways
The nuclear radius is a measure of the size of the nucleus of an atom. It is based on a spherical description of nuclei.
The nuclear radius can be estimated using scattering experiments. One of the methods is the closest approach method. It is not very accurate but offers a simple process to measure an overestimation of the nuclear radius.
Another way to measure the nuclear radius is with the electron diffraction method. It is accurate and powerful but relies on a more complex experimental setting and mathematical reasoning.
The nuclear density is another relevant quantity to characterise nuclei, which allows us to understand how particles distribute themselves in the nucleus.
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