Nuclear Chemistry Part-2 -CHEMISTRY.pptx

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1. NUCLEAR CHEMISTRY PART-2 Prof. Nandhakumar C Assistant Professor Department o f Chemistry Sri Ramakrishna College of Arts and Science Coimbatore - 641 006 Tamil Nadu, India 1

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4. Radioactive equilibrium Let a radioactive substance A decay to give another radioactive substance B which decays to form substance C. If λ A and λ B are the decay constants for the two changes, we can write λ A λ B A B C The rate of disintegration of A is also the rate of formation of B. When the rate of disintegration of A(or formation B) is equal to the rate of disintegration of B, the amount of B does not change with lapse of time. Then the radioactive equilibrium is said to be established between the substance A and the substance B. 4 Sri Ramakrishna College of Arts and Science

5. Radioactive equilibrium At this stage, dN A /dt = dN B /dt Where N A and N B are atoms of A and B present at the equilibrium. Since λ A N A = λ B N B N A / N B = λ B / λ A But λ α 1 / t 1/2 Therefore N A / N B = t 1/2 of A / t 1/2 of B Thus the atoms of A and B are present in the ratio of their half-lives. The radioactive equilibrium differs from a chemical equilibrium in that it is irreversible . 5 Sri Ramakrishna College of Arts and Science

6. Transmutation reactions Transmutation reactions is the conversion of one chemical element (or) an isotope into another chemical element. Using Protons: 15 1 12 4 (i) N + H → C + He 7 1 6 2 23 1 23 1 (ii) Na + H → Mg + n 11 1 12 0 6 Sri Ramakrishna College of Arts and Science

7. Transmutation reactions using protons 40 1 40 1 (iii) Ar + H → K + n 18 1 19 0 7 1 4 4 (iv) Li + H → He + He 3 1 2 2 Proton - H¹ (or) ‘p’ (atomic number = 1, atomic mass = 1) 7 Sri Ramakrishna College of Arts and Science

8. Transmutation reactions using deuterons Deuteron - H² (or) ‘d’ (atomic number = 1, atomic mass = 2) 1 2 3 (i) H + H → He 1 1 2 3 2 4 1 (ii) H + H → He + n 1 1 2 0 37 2 38 1 (iii) Cl + H → Ar + n 17 1 18 0 8 Sri Ramakrishna College of Arts and Science

9. Transmutation reactions using α -particles 27 4 30 1 (i) Al + He → P + n 13 2 15 0 209 4 211 1 (ii) Bi + He → At + 2 n 83 2 85 0 14 4 17 1 (iii) N + He → O + H 7 2 8 1 9 Sri Ramakrishna College of Arts and Science

10. Transmutation reactions using neutrons 23 1 24 0 (i) Na + n → Mg + e 11 0 12 -1 27 1 24 4 (ii) Al + n → Na + He 13 0 11 2 14 1 14 1 (iii) N + n → C + H 7 0 6 1 10 Sri Ramakrishna College of Arts and Science

11. Nuclear chemistry reactions 23 1 24 0 (i) Na + n → Mg + e 11 0 12 -1 This nuclear reaction is written by 23 24 Na ( n, β ) Mg 11 12 14 4 17 1 (ii) N + He → O + H 7 2 8 1 This nuclear reaction is written by 14 17 N ( α , p ) O 7 8 11 Sri Ramakrishna College of Arts and Science

12. Artificial Radioactivity Many stable nuclei when bombarded with high speed particles produce unstable nuclei that are radioactive. The radioactivity produced in the manner by artificial means is known as artificial radioactivity (or) induced radioactivity . The artificial isotopes disintegrate in a definite fashion and have specific half-life. For example, aluminium-27 when bombarded with a neutron emits an alpha particle and forms sodium-24 which is radioactive. It disintegrates spontaneously by emission of a beta particle and the product is magnesium-24. Sodium-24 has half-life of 24 hours. 12 Sri Ramakrishna College of Arts and Science

13. Nuclear Stability and Neutron-Proton Ratio Nuclei are composed of protons and neutrons. The protons would tend to fly apart due to repulsive forces between them. But the neutrons in some way hold the protons together with in the nucleus. The stability of a nucleus seems to depend on the neutron to proton ratio ( n/p) in the nucleus. The diagram shows the neutron to proton ratios for all known stable elements. Each point on the graph indicates the number of protons and neutrons in a particular stable nucleus. It is clear from the graph that: 13 Sri Ramakrishna College of Arts and Science

14. Nuclear Stability and Neutron-Proton Ratio The lower elements (up to Z=20) , the stable nuclei have about equal number of protons and neutrons i.e., n/p = 1 . For higher elements to be stable , there must be more neutrons than protons i.e., n/p > 1 . The shaded portion in the diagram represents the region or belt of stability . The element whose n/p ratios lie inside the belt are stable. A nucleus whose n/p lies above or below the stability belt is radioactive or unstable on account of unfavourable n/p ratio. It emits α (or) β -particles so as to move into the stability range. 15 Sri Ramakrishna College of Arts and Science

15. Nuclear Stability and Neutron-Proton Ratio a. A nucleus that is above the stability belt emits a β -particle where by a neutron is converted to proton. Thus n/p decreases and becomes more stable or enters the stability belt. For example, 14 14 0 C → N + β 6 7 -1 n/p 1.33 1.0 16 Sri Ramakrishna College of Arts and Science

16. Nuclear Stability and Neutron-Proton Ratio b. A nucleus whose n/p lies below the stability belt emits an α -particle and loses 2 protons and 2 neutrons. This results in a net increase of neutrons. This results in a net increase of n/p and the new nucleus may enter the stability belt. For example, 238 234 4 U → Th + α 92 90 2 n/p 1.565 1.60 The radioactive nuclei continue to emit α (or) β -particles, one after the other, till a stable nucleus is end-product. 17 Sri Ramakrishna College of Arts and Science

17. Odd-Even rule for Nuclear Stability Odd-Even rule : Proton Neutron Mass number Stability Odd Odd Even Unstable Odd Even Odd Stable/Unstable Even Odd Odd Stable/Unstable Even Even Even Highly Stable 18 Sri Ramakrishna College of Arts and Science

18. Packing Fraction The ratio of the difference of isotopic mass and mass number of the nuclide to its mass number is called packing fraction. Packing fraction = (Isotopic mass – Mass number) × 10⁴ Mass number Some important facts about packing fraction are: Quite often the packing fraction is multiplied by 10⁴ for convenience. Packing fraction is negative or zero for stable nuclei Packing fraction is positive for unstable nuclide. 19 Sri Ramakrishna College of Arts and Science

19. Packing Fraction Packing fraction for 18 Ar-40 Packing fraction = (39.9648 – 40) × 10⁴ / 40 = -9.405 20 Sri Ramakrishna College of Arts and Science

20. Mass Defect The difference between the experimental and calculated masses of the nucleus is called the mass defect. We know that atomic nucleus consists of protons and neutrons, collectively known as nucleons. It is found that the measured mass of nucleus is always less than the sum of the masses of the individual protons and neutrons which make it up. Let us take the example of helium ₂He⁴, it consists of two protons and two neutrons. Its mass may be calculated as mass of the protons = 2 × 1.00815 = 2.01630 amu mass of the neutrons = 2 × 1.00899 = 2.01798 amu Total mass = 4.03428 amu 21 Sri Ramakrishna College of Arts and Science

21. Mass Defect However, the experimental mass of the helium nucleus is only 4.00388. Mass Defect = Experimental mass – Calculated mass of nucleus of nucleus Mass Defect ( Δ m) = 4.00388 – 4.03428 = 0.03040 amu 22 Sri Ramakrishna College of Arts and Science

22. Nuclear Binding Energy Atomic nucleus is made of protons and neutrons closely packed in a small volume. Although there exist repulsive forces between the component protons , the nucleus is not split apart. This is so because the nucleons are bound to one another by very powerful forces. The energy that binds the nucleons together in the nucleus is called the Nuclear Binding Energy . When a nucleus is formed from individual protons and neutrons , there occurs a loss of mass (mass defect). According to Einstein’s theory, it is this mass defect which is converted into binding energy. Hence binding energy is the energy equivalent of the mass defect. 23 Sri Ramakrishna College of Arts and Science

23. Nuclear Binding Energy Δ E = Δ mC² Δ E – Energy difference Δ m – Mass defect C – Velocity of light (3 × 10⁸ ms⁻¹ ) 24 Sri Ramakrishna College of Arts and Science

24. Nuclear Binding Energy 25 Sri Ramakrishna College of Arts and Science

25. Nuclear Binding Energy 26 Sri Ramakrishna College of Arts and Science

26. Sums 1. Calculate the mass defect, binding energy and binding energy per nucleon for ₂He⁴ isotope mass of 4.00260 amu. Solution: Experimental (Given) mass M = 4.00260 amu protons = 2, electrons = 2, neutrons = 2 p = 1.007825 × 2 = 2.01565 amu n = 1.008665 × 2 = 2.01733 amu Calculated mass M’ = 2.01565 + 2.01733 M’ = 4.03298 amu 27 Sri Ramakrishna College of Arts and Science

27. Sums Mass defect Δ m = M’ – M = ( 4.03298 – 4.00260) amu = 0.03038 amu Binding energy (BE) = Δ m × 931 MeV [1 amu = 931 MeV] = 0.03038 × 931 MeV = 28.28378 MeV Binding energy per Nucleon (BE/Nu) = 28.28378 / 4 = 7.070945 MeV = 7.1 MeV 28 Sri Ramakrishna College of Arts and Science

28. Sums 2. Calculate the mass defect, binding energy and binding energy per nucleon for ₁₈Ar⁴⁰ isotope mass of 39.96238 amu. Solution: Experimental (Given) mass M = 39.96238 amu protons = 18, electrons = 18, neutrons = 22 p = 1.007825 × 18 = 18.14085 amu n = 1.008665 × 22 = 22.19063 amu Calculated mass M’ = 18.14085 + 22.19063 M’ = 40.33148 amu 29 Sri Ramakrishna College of Arts and Science

29. Sums Mass defect Δ m = M’ – M = ( 40.33148 – 39.96238 ) amu = 0.3691 amu Binding energy (BE) = Δ m × 931 MeV [1 amu = 931 MeV] = 0.3691 × 931 MeV = 343.6321 MeV Binding energy per Nucleon (BE/Nu) = 343.6321 / 40 = 8.5908025 MeV = 8.6 MeV 30 Sri Ramakrishna College of Arts and Science

30. Nuclear forces The nucleus of an atom consists of positively charged protons and uncharged neutrons. According to Coulomb’s law, protons must repel each other with a very large force, because they are close to each other and hence the nucleus must be broken into pieces. But this does not happen. It means that, there is some other force in the nucleus which overcomes the electrostatic repulsion between positively charged protons and binds the protons and neutrons inside the nucleus. This force is called nuclear force. 31 Sri Ramakrishna College of Arts and Science

31. ∏ - meson theory (Exchange theory) Yukawa(1935) suggested that the nuclear force existing between any two nucleons may be due to the continuous exchange of particles called mesons. just as photons, the exchange particle in electromagnetic interactions. ∏ - meson theory (Exchange theory) Neutrons and protons can be converted into each other by the exchange of a new particle meson. p → n + ∏⁺ p + ∏⁻ → n Charge +1 0 +1 Charge +1 -1 0 Mass 1 1 0 Mass 1 0 1 32 Sri Ramakrishna College of Arts and Science

32. ∏ - meson theory (Exchange theory) Proton from change to be a neutron at very fast. Because emit ∏⁺, like the speed of light 3 × 10⁸ ms⁻¹. A meson may be ∏⁺, ∏⁻ (or) ∏⁰. A neutron by accepting a ∏⁺ meson converted into a proton. A proton by ejecting a ∏⁺ meson converted into a neutron. A neutron by ejecting a ∏⁻ meson converted into a proton. A proton by accepting a ∏⁻ meson converted into a neutron. 33 Sri Ramakrishna College of Arts and Science

33. Nuclear fluid theory One property of nuclei is that the average binding energy per nucleon is approximately the same for all stable nuclei, which is similar to a liquid drop . The liquid drop model treated the nucleus as a drop of incompressible nuclear fluid, with nucleons behaving like molecules in a liquid . Niels Bohr and John A. Wheeler explained the nuclear fission process the help of liquid drop model. A Liquid drop has a spherical shape due to surface tension. On applying external force the sphere changes into ellipsoid. Which may change into a dumb bell shape when the force is larger. This may break at narrow end into two portions. 34 Sri Ramakrishna College of Arts and Science

34. Nuclear fluid theory In the same way, when the heavier nucleus absorbs a neutron, a compound nucleus is formed and is left in an excited state. The excitation energy sets up a series of rapid oscillations. The compound nucleus undergoes distortion from spherical to dumb bell shape. Each portion of the dumb bell has a positive charge and one repels the other. This results in fission and the formation of fission fragments. 35 Sri Ramakrishna College of Arts and Science