Till now we have discussed the discovery of the nucleus and the different hypotheses of the constituents of the nucleus. We saw that the nucleus is made of Protons and Neutrons. But we didn’t talk much about the size of the nucleus. So, in this post on nuclear physics, we are going to discuss the size of the nucleus.
The nucleus is very small compared to its atom. It is around 10^5 times smaller than the size of the atom. The atomic sizes are in the order of Angstrom. That is in the order of 10^{-10}\rm \ m. Whereas the nuclear sizes are in the order of Fermi. That is in the order of 10^{-15}\rm \ m. It is quite small and hence we cannot measure its size even by using the world’s best microscope. So how did scientists measure the size of the nucleus? Let’s understand it.
Suppose you have a ball in your hand and if I ask you about the diameter of the ball then you can easily estimate it. Right? But do you know how did you calculate that? Most of you might don’t know about it but yes you have calculated the diameter of the ball because of light. Actually, the light which is falling on the ball gets diffracted and scattered by it. The diffracted and scattered light waves fall on your eyes and then you are able to analyze it. Actually, your eyes play the role of the detector and the light plays the role of probe here and by means of that, you are able to measure the diameter of this ball.
Now, if I choose the sound wave as a probe and your ears as a detector then can you able to estimate the size of the ball that is in your hand? I mean if I ask you to close your eyes and I put a sound source before the ball then can you able to estimate the size of this ball?
In this case, the sound source sends some sound waves towards you which first falls on the ball and then moves toward you, and you hear it. Now I ask you the size of the ball just by listening to this sound wave. Will you be able to estimate that? The answer should be no. But why? Why the size of this ball could not be analyzed by a sound wave whereas it can easily be analyzed by a light wave? It’s due to the difference in their wavelengths. The wavelength of visible light ranges from 380 to 750 nanometers which is much smaller than the diameter of this ball. Whereas, the wavelength of audible sound ranges from 5 cm to 15 m which is quite more than the size that you want to investigate and that is why you are not able to get any useful information about the size of this ball.
Now let’s come to the nucleus. Since the nuclear size is expected to be something like 10^{-14} or 10^{-15}\rm \ m and the light has a wavelength in the order of 10^{-12}\rm \ m which is around 1000 times the expected size of the nucleus. Therefore, the light cannot be used as a probe to find the size of a nucleus.
So, to find the size of a nucleus, you need a probe with a wavelength of the order of ~10 ^{-15}\rm \ m or smaller than that. If you take high energetic electrons as a probe then it can be possible to find the size of the nucleus. Let’s understand how?
The wavelength of a moving particle depends on its momentum by using the de Broglie relation;
\begin{aligned}\lambda&=\dfrac{h}{p}\\\end{aligned}
So, from this relation, if you have high energetic electrons then its momentum will be large and hence the wavelength will be small. And due to its high momentum and low mass, it will be considered a relativistic particle.
Now, for a relativistic particle, we have a famous formula that can nicely relate the total energy, momentum, and rest mass of the particle. That formula is;
\begin{aligned}E^2&=(pc)^2+ (m_0 c^2)^2\\\end{aligned}
Here, E is the total energy, p is momentum, c is speed of light, m_0 is the rest mass, and m_0c^2 is the rest mass energy of the particle.
For an electron, the total energy is given by;
\begin{aligned}E&=\sqrt{(pc)^2+ (m_0 c^2)^2}\\\end{aligned}
And the Kinetic energy is given by;
\begin{aligned}K&=E-m_0c^2\\\end{aligned}
Since, the rest mass energy of the electron is around 0.511\rm \ MeV so, for high energetic electrons, having energy in hundreds of MeV, we have;
\begin{aligned}pc&\gg m_0 c^2\\\end{aligned}
And we can write;
\begin{aligned}K &\approx E\\&\approx pc\\\end{aligned}
So, now let’s come back to the wavelength calculation. The wavelength of the energetic electron will be;
\begin{aligned}\lambda&= \dfrac{h}{p}\\&= \dfrac{hc}{K}\\\end{aligned}
Here, h is the Planck’s constant, c is the speed of light and if you multiply these two things they come out to be 1240\rm \ MeV-fm.
So, the wavelength \lambda of the energetic electron is given by;
\fcolorbox{black}{aqua}{$\lambda=\dfrac{1240\rm \ MeV-fm}{K}$}
\textcolor{red}{(1)}
Now, if you want to have the wavelength in the order of the femtometer, say 5 femtometers then what should be the kinetic energy of the electron?
In equation 1, if you put \lambda = 5\rm \ fm then the kinetic energy will be K = 248\rm \ MeV.
So, the conclusion is you need to have hundreds of \rm \ MeV's of kinetic energy to be given to the electron so that its wavelength becomes comparable to or smaller than the size of the nucleus.
In 1958, an experiment was performed in a lab at Stamford University, California. Electron beams of energy 420\rm \ MeV were produced with the linear accelerator and the beam was allowed to fall on carbon and oxygen targets. The results were published in the journal “Physics Review” in 1959.
A graph between the differential cross-section per unit solid angle and the scattering angle is shown above that is taken from the paper. The differential cross-section per unit solid angle is nearly equal to the number of electrons that are coming at a particular angle. You can see here that when you go from 30^0 of scattering angle towards 80^0 then first the number of electrons decreases and at around 53^0 or 54^0 of scattering angle you get the first minima. After that it increases and then it goes through a maximum and then decreases.
Let’s compare this graph with the single slit diffraction pattern for circular aperture and try to find out the size of the carbon nucleus. From the single slit diffraction pattern for a circular aperture, the first minimum at \theta occurs, when
\fcolorbox{black}{aqua}{$\sin\theta=\dfrac{1.22\lambda}{D}$}
\textcolor{red}{(2)}
Where \lambda is the wavelength of incident light and D is the diameter of the circular aperture.
In this electron scattering experiment, an electron in the beam may collide either with electrons of the carbon atom or with the carbon nucleus. When it collides with the electrons of the carbon atom then most of its energy is imparted to the electrons of the carbon atom and it becomes less energetic. But on the other hand when it collides with the carbon nucleus then it scattered with almost the same energy because of the heavy nucleus. So, you need to analyze only those scattered electrons which have energy close to 420\rm \ MeV.
For 420\rm \ MeV electron beam, the wavelength will be;
\begin{aligned}\lambda&=\dfrac{1240\rm \ MeV-fm}{420\rm \ MeV}\\&\approx3\rm \ fm\\\end{aligned}
Now if you plug the value of \lambda=3\rm \ fm and \theta=53^0 in equation 2, we have;
\begin{aligned}D&=\dfrac{1.22 \times3\rm \ fm}{\sin53^0}\\&=4.6\rm \ fm\\\end{aligned}
It would be the diameter of the carbon nucleus and hence the radius of the carbon nucleus would be 2.3\rm \ fm.
This is just an analogy, not a perfect mathematical correspondence, but still, it works in this case because after all electron beams can be considered as electron waves that are getting diffracted in different directions.
So, this is just an estimate. But the electron scattering can be taken further ahead. And with a little bit more calculations, you can get more information about the structure of the nucleus with the help of quantum mechanics. We will see it in the next post on nuclear physics.