13.4 Obtain approximately the ratio of the nuclear radii of the gold isotope 197 79 Au and the silver isotope 10747 Ag .

13.2 Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units of MeV from the following data: m ( 5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u

13.3 SIZE OF THE NUCLEUS As we have seen in Chapter 12, Rutherford was the pioneer who postulated and established the existence of the atomic nucleus. At Rutherford’s suggestion, Geiger and Marsden performed their classic experiment: on the scattering of a-particles from thin gold foils. Their experiments revealed that the distance of closest approach to a gold nucleus of an a-particle of kinetic energy 5.5 MeV is about 4.0 × 10–14 m. The scattering of a-particle by the gold sheet could be understood by Rutherford by assuming that the coulomb repulsive force was solely responsible for scattering. Since the positive charge is confined to the nucleus, the actual size of the nucleus has to be less than 4.0 × 10–14 m. If we use a-particles of higher energies than 5.5 MeV, the distance of closest approach to the gold nucleus will be smaller and at some point the scattering will begin to be affected by the short range nuclear forces, and differ from Rutherford’s calculations. Rutherford’s calculations are based on pure coulomb repulsion between the positive charges of the a- particle and the gold nucleus. From the distance at which deviations set in, nuclear sizes can be inferred. By performing scattering experiments in which fast electrons, instead of a-particles, are projectiles that bombard targets made up of various elements, the sizes of nuclei of various elements have been accurately measured. It has been found that a nucleus of mass number A has a radius R = R 0 A1/3 (13.5) where R0 = 1.2 × 10–15 m (=1.2 fm; 1 fm = 10–15 m). This means the volume of the nucleus, which is proportional to R 3 is proportional to A. Thus the density of nucleus is a constant, independent of A, for all nuclei. Different nuclei are like a drop of liquid of constant density. The density of nuclear matter is approximately 2.3 × 1017 kg m–3. This density is very large compared to ordinary matter, say water, which is 103 kg m–3. This is understandable, as we have already seen that most of the atom is empty. Ordinary matter consisting of atoms has a large amount of empty space. 309 Reprint 2025-26 Physics Example 13.1 Given the mass of iron nucleus as 55.85u and A=56, find the nuclear density? Solution mFe = 55.85, u = 9.27 × 10–26 kg mass 9.27 × 10 − 26 1 13.1 Nuclear density = = −15 3 × volume (4 π /3)(1.2 × 10 ) 56 = 2.29 × 1017 kg m–3 The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects EXAMPLE has been compressed to such an extent that they resemble a big nucleus. 13.4 MASS-ENERGY AND NUCLEAR BINDING ENERGY 13.4.1 Mass – Energy Einstein showed from his theory of special relativity that it is necessary to treat mass as another form of energy. Before the advent of this theory of special relativity it was presumed that mass and energy were conserved separately in a reaction. However, Einstein showed that mass is another form of energy and one can convert mass-energy into other forms of energy, say kinetic energy and vice-versa. Einstein gave the famous mass-energy equivalence relation E = mc 2 (13.6) Here the energy equivalent of mass m is related by the above equation and c is the velocity of light in vacuum and is approximately equal to 3×108 m s–1. Example 13.2 Calculate the energy equivalent of 1 g of substance. 13.2 SolutionEnergy, E = 10–3 × ( 3 × 108)2 J E = 10–3 × 9 × 1016 = 9 × 1013 J Thus, if one gram of matter is converted to energy, there is a release EXAMPLE of enormous amount of energy. Experimental verification of the Einstein’s mass-energy relation has been achieved in the study of nuclear reactions amongst nucleons, nuclei, electrons and other more recently discovered particles. In a reaction the conservation law of energy states that the initial energy and the final energy are equal provided the energy associated with mass is also included. This concept is important in understanding nuclear masses and the interaction of nuclei with one another. They form the subject matter of the next few sections. 13.4.2 Nuclear binding energy In Section 13.2 we have seen that the nucleus is made up of neutrons and protons. Therefore it may be expected that the mass of the nucleus 310 is equal to the total mass of its individual protons and neutrons. However, Reprint 2025-26 Nuclei the nuclear mass M is found to be always less than this. For example, let us consider 168 O ; a nucleus which has 8 neutrons and 8 protons. We have Mass of 8 neutrons = 8 × 1.00866 u Mass of 8 protons = 8 × 1.00727 u Mass of 8 electrons = 8 × 0.00055 u Therefore the expected mass of 168 O nucleus = 8 × 2.01593 u = 16.12744 u. The atomic mass of 168 O found from mass spectroscopy experiments is seen to be 15.99493 u. Substracting the mass of 8 electrons (8 × 0.00055 u) from this, we get the experimental mass of 168 O nucleus to be 15.99053 u. Thus, we find that the mass of the 168 O nucleus is less than the total mass of its constituents by 0.13691u. The difference in mass of a nucleus and its constituents, ∆M, is called the mass defect, and is given by ∆ M = [ Zm p + ( A − Z ) m n ] − M (13.7) What is the meaning of the mass defect? It is here that Einstein’s equivalence of mass and energy plays a role. Since the mass of the oxygen nucleus is less that the sum of the masses of its constituents (8 protons and 8 neutrons, in the unbound state), the equivalent energy of the oxygen nucleus is less than that of the sum of the equivalent energies of its constituents. If one wants to break the oxygen nucleus into 8 protons and 8 neutrons, this extra energy ∆M c2, has to be supplied. This energy required Eb is related to the mass defect by ∆ M c2 (13.8) Eb = Example 13.3 Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of 168 O in MeV/c2. Solution 1u = 1.6605 × 10–27 kg To convert it into energy units, we multiply it by c 2 and find that energy equivalent = 1.6605 × 10–27 × (2.9979 × 108)2 kg m2/s2 = 1.4924 × 10–10 J 1.4924 × 10 −10 = −19 eV 1.602 × 10 = 0.9315 × 109 eV = 931.5 MeV or, 1u = 931.5 MeV/c2 For 168 O , ∆M = 0.13691 u = 0.13691×931.5 MeV/c2 = 127.5 MeV/c 2 EXAMPLE The energy needed to separate 168 O into its constituents is thus 13.3 127.5 MeV/c2. If a certain number of neutrons and protons are brought together to form a nucleus of a certain charge and mass, an energy Eb will be released 311 Reprint 2025-26 Physics in the process. The energy Eb is called the binding energy of the nucleus. If we separate a nucleus into its nucleons, we would have to supply a total energy equal to Eb, to those particles. Although we cannot tear apart a nucleus in this way, the nuclear binding energy is still a convenient measure of how well a nucleus is held together. A more useful measure of the binding between the constituents of the nucleus is the binding energy per nucleon, Ebn, which is the ratio of the binding energy Eb of a nucleus to the number of the nucleons, A, in that nucleus: Ebn = Eb / A (13.9) We can think of binding energy per nucleon as the average energy per nucleon needed to separate a nucleus into its individual nucleons. Figure 13.1 is a plot of the binding energy per nucleon Ebn versus the mass number A for a large number of nuclei. We notice the following main features of the plot: (i) the binding energy per nucleon, Ebn, is practically constant, i.e. practically independent of the atomic number for nuclei of middle mass number ( 30 < A < 170). The curve has a maximum of about 8.75 MeV for A = 56 and has a value of 7.6 MeV FIGURE 13.1 The binding energy per nucleon for A = 238. as a function of mass number. (ii) Ebn is lower for both light nuclei (A<30) and heavy nuclei (A>170). We can draw some conclusions from these two observations: (i) The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon. (ii) The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short-ranged. Consider a particular nucleon inside a sufficiently large nucleus. It will be under the influence of only some of its neighbours, which come within the range of the nuclear force. If any other nucleon is at a distance more than the range of the nuclear force from the particular nucleon it will have no influence on the binding energy of the nucleon under consideration. If a nucleon can have a maximum of p neighbours within the range of nuclear force, its binding energy would be proportional to p. Let the binding energy of the nucleus be pk, where k is a constant having the dimensions of energy. If we increase A by adding nucleons they will not change the binding energy of a nucleon inside. Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. The binding energy per nucleon is a constant and is 312 approximately equal to pk. The property that a given nucleon Reprint 2025-26 Nuclei influences only nucleons close to it is also referred to as saturation property of the nuclear force. (iii) A very heavy nucleus, say A = 240, has lower binding energy per nucleon compared to that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightly bound. This implies energy would be released in the process. It has very important implications for energy production through fission, to be discussed later in Section 13.7.1. (iv) Consider two very light nuclei (A ≤10) joining to form a heavier nucleus. The binding energy per nucleon of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion. This is the energy source of sun, to be discussed later in Section 13.7.2.