Binding Energy of a Nucleus
| Binding Energy of a Nucleus | ||
| Why some nuclei are highly stable and some are not? … Nuclear binding energy | ||
| It has been seen from several observations that some nuclei are extremely stable and they don’t transform to other nuclei by any radioactive decay process.* On the other hand, some nuclei are not stable and they undergo radioactive decay process and transform naturally to other nuclei. The solution to this problem lies in the understanding of nuclear binding energy. It has been observed that when a nucleus is formed from its constituent particles, its mass is less than the sum of the masses of the individual particles. This difference in mass (∆m) also known as mass defect gets converted into an amount of energy ∆E = (∆m)c 2. This energy is known as the binding energy (B.E.) of the nucleus. Therefore, one can write the following relation: | ||
B.E. = (Zmp + Nmn)C2 – MNC2 —————– (m1.22) | ||
where, | Z = no. of protons N = no. of neutrons m p = rest mass of a proton m n = rest mass of a neutron M N = nuclear rest mass | |
| But the nuclear rest mass is MN = Ma – Zme, where Ma represents the atomic mass corresponding to the specified nuclear mass, Zme is the total mass of the orbiting electrons. The binding energy of the electrons is neglected, as it is quite small compared to the nuclear binding energy. | ||
∴ B.E. = (Zmp + Nmn)C2 – (Ma – Zme)C2 —————– (m1.23) | ||
Again proton mass m p = m H –m e, where m H is the atomic mass of hydrogen atom. | ||
∴ B.E. | = Z(mH – me)C2 + NMnC2– (MaC2– Zme)C2 = Z(mH+N me)C2 + MaC2 —————– (m1.24) | |
| In atomic mass units the binding energy can be expressed as: | ||
| B.E. = (ZmH + Nmn) – Ma —————– (m1.25) | ||
| The nuclei having positive binding energies are stable and in order to disrupt the nucleus into its constituents, energy is to be supplied from outside. However, the nuclei whose binding energy is negative are unstable and get disintegrated by themselves. The maximum binding energy per nucleon corresponds to A = 56 (Fe) and 62 (Ni). The binding energy/nucleon for 56Fe and 62Ni are 8.790 MeV/nucleon and 8.795 MeV/nucleon respectively. In fact, 62Ni is the most stable nucleus in the periodic table. However, stars are generally enriched with 56Fe. Why? You will find its answer in Shurtleff and Derringh’s article.3 The light and very heavy nuclei contain less bound nucleons. Thus the source of energy production in fusion of light nuclei or fission of very heavy nuclei can be a source of energy. They are basically the basis of nuclear energy production in reactors and weapons. We will learn about nuclear reactors in module IV. | ||
| ————————————————————————————————————————– * Radioactivity is a physical process by which a nucleus disintegrates naturally by emitting different particles, like alpha, beta and gamma. For details see module V of this course. ————————————————————————————————————————– | ||
| Let us compute binding energy of a very small nucleus in the following example. | ||
| —————————————————————————————– me1.1: Compute binding energy for 2He 4. The He-nucleus is made up of 2-protons and 2-neutrons. Solution: Here, Zm H = 2 x 1.007825 amu = 2.015650 amu and Nm n = 2 x 1.008665 amu = 2.017330 amu ∴ Zm H + Nm n = 4.032980 amu Again, atomic mass of 2He 4 = 4.002603 amu Difference = + 0.030377 amu Since 1 amu = 931 Mev, B.E. = 0.030377 ´ 931 = 28.28 MeV ————————————————————————————————————————– | ||
| What holds nucleons together? — strong interaction | ||
| Now a natural question arises: what holds the nucleons together in a nucleus? It is known that systems are bound together by forces, as for example, masses are attracted by gravitational force, and electrical charges are attracted or repelled by electromagnetic force. Can these forces do the job of binding nucleons? The electromagnetic force most certainly cannot. Neutrons have no electrical charge, so that they don’t experience the electromagnetic force at all, and the principal electromagnetic force between protons is a strong Coulomb repulsion, which tends to tear the nucleus apart. On the other hand, gravitational force which is an attractive force between every pair of nucleons, is smaller by a factor of about 1039 than the electrical force between two protons. Its effect is completely negligible in such low dimensional objects. Therefore, the only explanation is to consider another force which is strong, attractive and short range. This is known as nuclear force. Yukawa was the first person to give an explanation for generation of this force in a nucleus considering exchange of mesons between nucleons. The exchange of meson is possible between two protons, two neutrons and also between one proton and one neutron. All these interactions are attractive and confined within the nucleus. If you want to learn more about this force, study some course on particle physics/ phenomenology. | ||