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1 Chapter 31 Nuclear Physics and Radioactivity 2 1. Nuclear Structure a) Proton - positive charge - mass 1.673 x 10-27 kg ≈ 1 u b) Neutron - discovered by Chadwick (student of Rutherford) - hypothesized to account for mass of atom - discovered with scattering experiments - zero charge - mass 1.675 x 10-27 kg ≈ 1 u - mass of neutron ≈ mass of proton + mass of electron - neutron can eject electron to form proton, but it’s not a proton and an electron 3 c) Nucleon - constituent of nucleus (neutron or proton) d) Nomenclature A - number of nucleons (atomic mass number) Z - number of protons N - number of neutrons A=Z+N Symbol for nucleus of chemical element X: A Z X 4 Examples: hydrogen nucleus: 11H helium 4: 42 He neutron: 01n electron: -10e aluminum: 27 13 Al 27 Al Since Z determines the element (X), only AX is needed. 5 e) Atomic mass unit, u Define: Mass of 12C = 12 u Then, 1 u = 1.66 x 10-27 kg = 931.5 MeV/c2 mp = 1.00727 u mn = 1.008665 u In chemistry and biology, 1 dalton (Da) = 1 u 6 f) Isotopes; nuclei with the same Z, different N e.g. 35Cl, 37Cl (65%, 35%), 12C, 13C, 14C (99%, 1%, 0.01%) g) Nuclear size and density Close-packed - constant density - Volume proportional to atomic number (A) - Since V = 4/3 πr3, A prop. to r3 - r prop. A1/3 - r ≈ (1.2 x 10-15 m) A1/3 = 1.2 fm A1/3 - density of neutron star = 100 million tonne/cm3 7 2. Nuclear force and stability a) Strong nuclear force - one of the fundamental forces holds protons together in spite of Coulomb repulsion short range: ~ fm (zero for longer range) - - only adjacent nucleons interact acts equally between n-p, n-n, p-p 8 b) Symmetry - Pauli exclusion principle: N=Z gives maximum stability considering only nuclear force c) Coulomb repulsion -long range; all protons interact (only adjacent nucleons feel nuclear force) - repulsion increases with size -- neutron excess needed for stability - above Z = 83 (Bi) stability not possible; larger elements decay emitting radioactivity 9 3. Mass defect and binding energy a) Binding energy energy required to separate constituents of nucleus 10 b) Mass defect From special relativity, adding energy increases mass: B.E. mc2 m1c 2 m2c 2 ... mic 2 B.E. 2 m m c i B.E . mc 2 m mass defect = constituents - composite 11 Example: 4He (alpha particle) m (2m p 2mn m 4 He ) 0.0304u BE mc2 (0.0304u)c2 931.5MeV 2 0.0304 c 2 c 28.3 MeV Compare ionization potential for H atom: 13.6 eV 12 c) Atomic electrons -Masses usually tabulated for neutral atoms (including atomic electrons) - Can use atomic masses if electrons balance: 4 He B.E . 2n 2 H 1 ( 4 He 2e) B.E. 2n 2(1 H e) d) Binding energy per nucleon - determines stability - for 4He, BE = 28 MeV so BE/nucleon = 7 MeV increase in nearest neighbors increase in Coulomb repulsion dominates 13 For a given number of nucleons, - if BE/nucleon increases - mass defect increases - total mass decreases - energy released Energy released Fusion Fission 14 15 Fusion: Potential energy diagram for nucleons: fusion releases energy Energy (high temperature in the sun) required to push nuclei together against the Coulomb force. 16 Fission: Potential energy diagram for two halves of a large nucleus: fission releases energy May occur spontaneously, or be induced by neutron bombardment 17 4. Radioactivity - spontaneous decay of nucleus - releases energy to achieve higher BE/nucleon - mass of parent > mass of products a) - decay 18 - ejection of 4He nucleus A Z 4 P AZ 4 D 2 2 He - transmutation: element changes - Energy released (KE of , daughter, energy of photon) E mc 2 (m p md m )c 2 Use atomic masses for P, D, 4He (electrons balance): E (mp(atom) md(atom) m 4 He(atom))c 2 For 238U, 234Th, 4He, E = 4.3 MeV -decay b) - decay 19 - ejection of electron A Z P A Z 1 D 0 1 - governed by weak nuclear force - transmutation - Energy released, as KE of electron E mc 2 (m p md m )c 2 Use atomic masses for P, D, and add one electron mass: E (mp(atom) md(atom))c 2 For 234Th and 234Pa, - decay E = 0.27 MeV Other modes of beta decay A 0 P D - ejection of positron Z 1 1 A Z - electron capture A Z P 10 AZ 1 D 20 21 c) - decay - emission of a photon Z P Z P - no transmutation - accompanies - decay, fission, neutron decay A A 22 d) Decay series - sequential decays to an eventual stable nucleus - 4 separate series (A can only change by 4) Qu i c k T i m e ™ a n d a T I F F (Un c o m p re s s e d ) d e c o m p re s s o r a re n e e d e d t o s e e th i s p i c t u re . 238U -> 206Pb 235U -> 207Pb 232Th -> 208Pb 237Np -> 209Bi (not obs’d) e) Neutrino, - postulated by Pauli in 1930 to account for missing energy in -decay A Z P AZ 1 D 10 A Z P AZ 1 D 10 - observed in 1956 - mass ~ zero (< ~ eV) (standard model predicts non-zero mass) - could account for missing mass in universe - zero charge - interacts only by weak nuclear force (difficult to detect) 23 24 5. Radioactive decay rate; activity a) Activity Activity is the number of decays per unit time, or where N represents the number of nucleii present. N t For a random process, the activity is proportional to N: N N t This gives (by integration) is the decay constant N N 0e t where N0 is the number of nuclei at t = 0. Units: 1 Bq (becquerel) = 1 decay/s 1 Ci (curie) = 3.7 x 1010Bq (activity of 1 g radium) 25 b) Half-life Exponential decay: For a given time interval, the fractional decrease in N is always the same: N1 e t1 t2 e (t1 t2 ) N2 e Define half-life as the time for activity to reduce by 1/2: 1 e (T1/ 2 ) 2 T1/ 2 ln2 .693 26 Using e t 2 t / ln 2 the exponential can be expressed N N 0 2t / ln 2 so t /T1/ 2 1 N N0 2 number of half-lives 1 N N0 2 27 6. Radioactive dating a) Carbon dating - based on the reaction: 14 C 14 N + T1/2 = 5730 years - 14C/12C ratio constant in atmosphere due to cosmic rays - living organisms ingest atmospheric carbon; dead matter doesn’t - ratio of 14C/12 C in matter gives time since death Equilibrium ratio: 1/8.3 x 1011 ==> 1 g C contains 6 x 1010 atoms of 14C ==> Activity of 1g C (at eq’m) = 0.23 Bq = A0 ==> Activity of 1g C (time t after death) = A= A0e-t 1 A t ln A0 28 b) Dating ancient rocks Age equation: