Nuclear & Particle Physics Lecture Notes: PDF

Okay, here is the converted markdown format for the text in the images. # Chapter 11,12,13,14,15 ## Nuclear & Particle Physics (Nuclear = Subatomic physics) ### A Modern Physics Lecture 1 The nucleus holds most of the mass of an atom & can have multiple electric charges. Most properties other than...

Okay, here is the converted markdown format for the text in the images. # Chapter 11,12,13,14,15 ## Nuclear & Particle Physics (Nuclear = Subatomic physics) ### A Modern Physics Lecture 1 The nucleus holds most of the mass of an atom & can have multiple electric charges. Most properties other than mass of atoms & molecules (matter in general) is due to the atomic e's & not nuclei. * However, the existence of elements is due to different charges of the nucleus. Energy involved in natural processes can be traced to nuclear reactions. (nuclear reaction energies is 10^6 - 10^10 larger than chemical energies of reactions. * Nuclear energy in reactors & weapons affects all our lives. * NMR - useful for medical imaging. * Radioactive decay ### Nuclear composition Neutrons, protons - nucleons. Atomic nuclei of the same element have the same no. of protons but can have different numbers of neutrons. Atomic structure was known much before nuclear structure (Why?) (Bohr atomic model - 1913) Because nuclear forces are much stronger than electric forces that hold e's to nucleus, so it is harder to break a nucleus to see what's inside. (charges in nuclear structure ∼ energies in MeV range, charge in e structure of atom ∼ ev range). * p - Proton: charge +e, mass 1836 times e. (difference with Position)) * n - Neutron: Uncharged, mass slightly greater than p. Atomic no.: no. of protons in nucleus = no of e in neutral atom of the element. H → 1, 18 → 2, U → 92. For a given element, there may be different nos. of neutrons (isotopes) → c.g. 999% of H nuclei are single protons, a few also contain a neutron (deuterium), a few 2 neutrons (tritium). deuterium is stable tritium is radioactive & changes into isotope of helium. At any giventive, only 2 kg of tritium of natural origin is present on earth, most in access(??). Heavy water: water in which deuterium atoms instead of hydrogen atoms are combined with oxygen atoms. * * * (Diagram of 3 atoms represented with circles and dots. 🔵 represents Proton, and ⚫ represents neutron. Ordinary hydrogen: one circle with one dot inside it. Deuterium: one circle with one 🔵 and ⚫ inside it. Tritium: one circle with one 🔵 and two ⚫ inside it.) Symbol convention for nuclides (nuclear species): ᴬZ X X = chemical symbol of element. Z = atomic no. of element = no. of protons in nucleus (= no. of e's in neutral atom) N = no of neutrons in nucleus. N = A - Z Hydrogen ¹₁H, deuterium ¹₂H, and isotopes of chlorine are ³⁵Cl, ³⁷Cl, etc. Atomic mass: mass of neutral atoms (nucleus + e's). Expressed in mass units u → defined as 'mass of ¹²C atom (most abundant isotope of carbon) is 12 u'. Atomic mass unit: 1 u = 1.66054 x 10⁻²⁷kg. Energy equivalent: 931.49 MeV/u e.g. Proton mass (kg) = 1.6726 x 10⁻²⁷, mass (u) = 1.007276 Neutron mass = 1.6750 x 10⁻²⁷, mass (u) = 1.008665 Mass of ¹H atom: 1.007825 u Nuclear properties (A) Size: The first estimate of nuclear size was given by the Rutherford scattering experiment & Mass ⇒ Alpha particles, (2p + 2n = helium nucleus), deflected by target nucleus in a thin foil, in a manner consistent with Coulomb's Law provided that they are both small enough to be considered point masses & charges & the distance between them exceeds 10⁻¹⁴m. (for smaller separations the nucleus is no longer a point charge to the alpha particle). **(Diagram)** Current technique for determining nuclear radius: Using particle scattering with fast electrons & neutrons. * e interacts with nucleus only through electric forces ⇒ e scattering tells us about distribution of charge in nucleus, * neutron interacts with nucleus specifically through nuclear forces ⇒ neutron scattering tells us about distribution of nuclear matter. For this study, the de Broglie wavelength of scattered particle should be smaller than radius of nucleus. ($\lambda= h/mv$). Results: * Volume of nucleus is directly proportional to no. of nucleons it contains (uts mass no. A) ⇒ density of nucleons is nearly same in interior of different nuclei. * Nuclear radius: R = R₀ A¹/³, where R₀ = 1.2 x 10⁻¹⁵m ≈1.2 fm (femtometer) or fermi! (since vol = ⁴⁄₃πR³ R α A ⇒ R α R₀A¹/³) * R₀ is approx. since nuclei don't have sharp boundaries * R₀ is a little less when deduced from e scattering, & more from neutron scattering ⇒ nuclear mass and nuclear charge are not identically distributed through a nucleus e.g. R for ¹²C nucleus is (1.2) (¹²C)/³ fm ≈ 2.7 fm. Fig:Density & nucleons in ⁵⁹₂₇Co & ¹⁹⁷₇₉Au vs. radial distance from centre. **(Graph)** R₀ ≈ 4.5 fm Rᴀᴜ ≈ 7 fm Nuclear radii R = 1.2 A¹/³ fm are as indicated. * * * **Question:** Find density of ¹²C nucleus. **Solution:** Atomic mass of ¹²C is 12 u ⇒ nuclear density p = m/v = (12 u) /(⁴⁄₃πR³) => (12 u) (1.66 x 10⁻²⁷ kg/u) / ⁴⁄₃ π(2.7 x 10⁻¹⁵)³ = p = 2.4 x 10¹⁷ kg/m³ Since The figure is essentially same for all nuclei, since R³ scales as mass no. A which is directly proportioned to mass m of a nucleon. (B) Spin and magnetic moment of nucleus Like e's, p & n are fermions with spin quantum nos. s = 1/2 ⇒ Spin angular momentum **S** has magnitude S = √(s(s+1)) h = √3/2 ħ with spin magnetic quantum nos. mₛ = +1/2 S²|s,m> = s(s+1)ħ²|s,m> * As for e's, magnetic moments are associated with spins of p & n. In nuclear physics, magnetic moments are expressed in nuclear magnetons µɴ, Where, Nuclear magneton µɴ = eh/2mp = 5.051 x 10⁻²⁷ J/T = 3.152 x 10⁻⁸ eV/T (mₚ = proton mass) This is smaller than Bohr magneton µʙ = eh/2me ≈ 9.274 x 10⁻²⁴ J/T (mₑ electron mass) * The spin magnetic moments of p & n have components in any z direction & Proton: µₚ₂ = +2.793 µɴ Neutron: µɴ₂ = -1.913 µɴ The 2 possibilities for signs are fermions = -1/2, +1/2 ± for µₚ₂ because µₚ₂ is in same direction as spin S, ⇒ for µɴ₂ have + since & S opposite to S. Spin magnetic moments of p & n, and spin angular momentum S in both cases eg hydrogen nucleus ¹₁H has single proton ⇒ total angular momentum **J** is given by S = √3/2 ħ. Nucleons in more complex nuclei may have orbital angular momentum due to their motion inside the nucleus, as well as spin angular momentum. For such a nucleus, total angular momentum = vector sum of spin & orbital angular momenta of its nucleons. Nuclear Binding Energy - the missing energy that helps a nucleus together. eg Deuterium ²₁H has a neutron & proton in nucleus; & its mass should be equal to mass of ²₁H atom + mass of neutron. Expected mass of ²₁H = mass of ²₁H = 1.007825 u + mass of neutron =1.008665 u = 2.016490u But the measured mass is 2.014102 u = 0.002388 u less than its above combination. The 'missing mass' corresponds to energy given off when a ²₁H nucleus is formed from free proton & neutron. The energy equivalent is ΔE = (0.002388 u) (931.49 MeV/u) = 2.224 MeV * * * This is tested by experimentally breaking a deuterium nucleus into separate n & p the required energy is actually 2.224 MeV! When less energy than 2.224 MeV is given to a ²₁H nucleus the nucleus stays together. When more energy is given, the extra energy becomes K.E. of n & p as they fly apart. Binding energy = energy equivalent of missing mass of a nucleus or amount of energy lost when they are brought together & nucleus is formed or amount of energy required to break a nucleus. Greater the B.E., more is energy to be supplied to break the nucleus. Eʙ = \[Zm(¹₁H) + Nm(n) - m(ᴬ₂X) (931.49 MeV/u) mass defect Where Eʙ = B.E. in MeV of a nucleus ᴬ₂X, with N = A -Z neutrons (Z = at no., A = no. of nucleons) m(¹₁H) atomic mass of ¹₁H, m(n) = neutron mass, m(ᴬ₂X) atomic mass of ᴬ₂X, all in mass units (u). (Electron causes -) For ²₁H (deuterium), Eʙ = 2. 224 MeV, For ²⁰⁹₈₃Bi (isotope Bismuth) Eʙ = 1640 MeV. ∫ range/eichode For comparison = Typical BE is 8 x 10¹⁰ kJ/kg, while heat of vaporization of water is 2.26 x 10³ kJ/kg, heat given off by burning gasoline is 47 x 10⁶ kJ/kg (17 million times smaller!!) * * * Binding energy per nucleon - Indicates stability of nucleus. It is the average found by dividing total binding energy by no. of nucleons in a given nucleus. es. for ²₁H: BE/nucleon = 2.224 MeV / 2.017 = 1.112 MeV / nucleon for ²⁰⁹₈₃Bi, it is 1640 MeV / 209 = 7.8 MeV / nucleon, at mass of ²⁰⁹₈₃Bi : 208.980 u Plot of BE / nucleon VS, Mass no. A **(Graph)** The greater the BE/nucleon, the more stable is the nucleus, since more energy is needed to break it. The maximum is at 8.8 MeV / nucleon when the total no. of nucleons is ⁵⁶₂₆Fe (²⁶Fe, an iron isotope). ⇒ this is the most stable nucleus. Two important conclusions from this plot: (1) If we somehow split a heavy nucleus into 2 medium-sized ones, each of the 2 new nuclei will have more binding energy per nucleon than the original nucleus the extra energy will be given off. eg. ²³⁵₉₂U breaking into 2 smaller nuclei: the BE difference | nucleon is about 0. 8 MeV Total energy given off is ((0.8 Mev / nucleon) ( 235 nucleons)) =188 MeV. * * * This is known as nuclear fission. This is an enormous amount of energy for a single atomic event / Ordinary chemical reactions (rearrangements & e's) Liberate eV /atom energies. The diagram depicts: A = 10 This fusion is in nucleus fusion; 2 = 10. Seven smaller times gives low million times energy per atom than burning of wood or oil (2) Joining 2 light nuclei to give single medium nuclei will also increase BEI nucleon in the new nucleus.es two ²₁H (deuterium)) Combine ⁴₂He (helium) nucleus, over 23 Mev energy is released This is known as nuclear fusion.. It is the main source of energy in sun & stars. The above graph is the most significant one in all of science! The fact that BE exists at all is that nuclei more complex than single proton of hydrogen can be stable, thus this stability accounts for existence of all elements and all matter. The peak in the middle explains the energy that powers evolution of universe, since it explains fusion of light nuclei to heavier nuclei.. e.gal Be of ²⁰₁₀Ne is 160.647 MeV. Find its atomic mass, Solution Z = 10, N=10 → m(ᴬ₂X) = [Zm(¹₁H) + Nm(n)) - Eʙ m( ²⁰₁₀Ne) = [10(1.007825 u)+ 10(1.008665)] - 160. 647 Mev/(931. 49 Mew/a) equals 19.992u 5) Find energy needed to remove a neutron from nucleon of 21. 1. Find energy to remove a proton Question. Why are they different?! Solution. C. 21) (4) neutron → Ca - (5) 21 50 m +m From table of mass(App) Therefore M(CO) 40.9627895640.962278 neutron Neutron is equal to 21(49 mev/u) or 11MEV 28ca. Ca to Remove 20 = 2.01 * 931 * * * * # Nuclear Physics history: * Becquerel (1896) → radioactivity was observed * J.J.Thomson (1897) → discovery of e, internal structure of atom (Early 20th century) Curie, Rutherford, Chadwick - radioactivity (α,β,γ radiation) studied. Rutherford(1906) - α -particle scattering (helium nuclei or α particles thru Al foil or gold leaf). Geiger-Marsden. ⇒ Rutherford model of atom with a small, dense nucleus (foremost) with most of atomic mass. (metal leaf) with embedded electrons, → heavy tive charged particles (neutron was unknown!) eg. N-14 had nucleus with 14p, 7e, & 7 orbital e. (Total 21 nuclear particles.) (Total 21 nuclear particles.) But Spin studies in 1929 showed e has spin -1/2, also ⇒ in above model, out of 21 nuclear particles, Prespin-20 should pair up leaving spin 1/2 of nucleus. But expt.ally N-14 has spin 1. Chadwick (1932) → neutron discovered (neutral particle with some mass as proton)(h spin 1/2). It was suggested that nucleus has neutrons not electrons. ⇒ N-14 ther has 7 protons, 7 neutrons ⇒ one unpaired p, one unpaired n adding to spin 1. (½ + ½). This led to current model of nucleus. * * * 10. Discovery to election. TT thomsen al experiment. Charge - oil drop experment. : (09) scattering experimnet nucleus showd dense center and cles than required change led = nude heliun = nuclei - (2) (2)