Nuclear Dimension
| From Fig. m1.1 we get an idea that the dimension of a nucleus is of the order of 10-15 m, which is extremely small. Comparing with the size of an atom, nucleus is 1000000 times smaller. You might be surprised how one can determine any object of such small dimension? In fact no microscope exists till date to view such small dimensional object. However, Physicists have other way out to determine such small physical structure. From elementary quantum mechanics it is known that every particle has a wave associated with it, which is known as matter wave or de Broglie wave. The de Broglie wavelength* of 100 mega electron volt (MeV) electron is of similar order of nuclear dimension and hence this can be used as a probe to determine the size of a nucleus. Moreover, the interaction between electron and nucleus is very well known, i.e. electromagnetic and hence electron is chosen as a probe to get the idea about the dimension of a nucleus. Let us see how one gets the idea about nuclear dimension using experimental set up depicted in Fig. m1.2: | ||
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| FIGURE m1.2 Schematic of experimental set up to measure angular deflection of electron from nuclei, where q is the scattering angle. Detector is placed at different q to detect the scattered electrons. Adetector (Faraday Cup for precise measurement of charge) is placed at the back of the target to collect the incident beam of electrons. | ||
| A high-energy electron beam is bombarded on a thin material target under study and the probability of various angular deflections is observed. A typical result of this experiment is shown in Fig.m1.3, which basically shows the angular distribution of electrons scattered from various nuclei. | ||
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| FIGURE m1.3 Angular distribution of electrons scattered from various nuclei. Relative intensity is shown in log scale as a function of scattering angle. | ||
| Now we try to understand how this distribution helps us to obtain the dimension of a nucleus. | ||
| Let us assume that ρ(r) is some density distribution for the nucleons in the nucleus, and it is same for neutrons and protons. Let ρ(0) be the nucleon density near the centre of the nucleus, R the radius at which the density has decreased by a factor 2 below the central value, and a is a measure of the surface thickness such that the distance over which the density falls from 90% of ρ(0) to 10% of ρ(0) is 4.4a. Let ρ(r) is related to all these physical parameters by the relation1 | ||
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| If we plot this function, it will look like the one as shown in Fig. m1.4 below: | ||
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FIGURE m1.4 Plot of equation m1.1 to show the variation of r (r) with r. | ||
| The scattering cross-section# can be calculated using quantum mechanical scattering formalism.$Using ρ(r) as a fitting parameter, the angular deflection of electrons can be calculated and fitted with the plots of Fig. m1.3. The value of ρ(r) is chosen for the best fitted curve and the following information is obtained 1. | ||
| ρ (0) ≅ 1.65 x 1044 nucleons/m3 = 0.165 nucleons/F3. R ≅ 1.07 A1/3 F a ≅ 0.55 F | ———– (m1.2) | |
| where F stands for ‘Fermi’ and is equal to 10-15 m, as the unit of length. The above analysis gives the idea of nuclear dimension, which is almost 106 times less than the atomic dimension. Now just imagine that almost entire mass of an atom is located in such small dimension and therefore the mass density of a nucleus is extremely high. Physical properties of this fascinating object are therefore extremely important from both fundamental understanding and engineering applications point of views. | ||
| *Every particle has an associated wave with it, known as matter wave or de Broglie wave. The wavelength ( λ) of this wave = h/p, where h is the Planck’s constant and p is the momentum of the particle. If you calculate λ value of 100 MeV electron, you will find its wavelength is of the order of nuclear dimension. This is the reason for choosing this as a probe for this experiment. | ||
| #The cross section of any scattering process basically signifies the probability of occurrence of that process. In this case the angular distribution of electrons gives an idea what is the probability of electrons scattered at different angles. For more information about cross section. see Lecture-II of Module-3. | ||
| $For quantum mechanical scattering you have to go through a quantum mechanics course which includes this topic like partial wave analysis, Born’s approximation etc. | ||