1. Nuclear Physics & Society (2009) 1 Radioactivity Three main types, all release energy through the relation Q=  E=  mc 2 –A–Alpha (  ) decay Emission of a 4 He nucleus (i.e. 2 protons and 2 neutrons). –B–Beta (  ) decay Changes proton to neutron (or vice versa). –G–Gamma (  ) decay Protons and neutrons re-arrange to lower energy configuration in same nucleus.

2. Nuclear Physics & Society (2009) 2 For a give fixed A (isobar), we have different combinations of Z and N. e.g., A=Z+N=137 can be from Z=56, N=81 → 137 Ba 81 ; or Z=55, N=82 → 137 Cs 82 (see later)… Mass Parabolas and Radioactive Decays A = constant

3. Nuclear Physics & Society (2009) Another View of the Nuclear Landscape Nuclear masses plotted versus N and Z for the light nuclei Remember  E =  mc 2 The valley represents the nuclei with the lowest total energy. The nuclei up on the sides of the valley are unstable and will decay successively until they reach the bottom and hence stability. Our raw materials for nuclear Physics are the atomic nuclei at the bottom of the valley-there are 283 stable or long-lived isotopes we can find in the Earth’s crust or atmosphere

4. Nuclear Physics & Society (2009) 4 Radioactivity: Part 1- Beta decay  E=  mc 2 determines if nucleus decays by beta-decay. Nuclei try to get to their lowest energy-mass, i.e., lowest mass proton-neutron combination for a given A(=Z+N). Beta decay changes a proton to neutron or vice versa. The effect is to ‘transmute’ from one chemical element to another, changing Z and N but keeping A fixed. An example of beta-decay is 137 Cs 82 → 137 Ba 81 +   + Aside: can leave daughter nucleus in excited state, which decays by Gamma (  ) decay

5. Nuclear Physics & Society (2009) A=N+Z = fixed

6. Nuclear Physics & Society (2009) Example of a mass parabola Mass energy (mc 2 ) A=N+Z=125 p → n +  + + 125 Xe : Z=54; N=71 125 Cs : Z=55; N=70 125 I : Z=53; N=72 125 Ba : Z=56; N=69 125 Te : Z=52; N=73 STABLE ISOBAR FOR A=125 125 In : Z=49 125 Sn : Z=50; 125 Sb : Z=51; n → p +  - +

7. Nuclear Physics & Society (2009) 7 Radioactivity –A–Alpha (  ) decay (cont.) Emission of a 4 He nucleus (i.e. 2 protons and 2 neutrons).  energies are characteristic to the specific decay. Can leave daughter in an excited state which can then decay by.. –G–Gamma (  ) decay Protons and neutrons re-arrange to lower nuclear energy state. No change in N and Z,  ray emitted (high energy/ frequency ‘light’). Typical  energies ~100 – 2000 keV

8. Nuclear Physics & Society (2009) 8 Nuclear Alpha Decay 0+0+ 4+4+ 6+6+ 2+2+ 0+0+ 0.0046 % 0.035 % 25.0 % 74.0 % 242 96 Cm 238 94 PuQ g.s. = 6.216 MeV t 1/2 = 163 d    = ( -1 ) Alpha decay can leave daughter In excited states which can then decay by (characteristic) gamma emission.

9. Nuclear Physics & Society (2009) Alpha-decay of 210 Po: Another example of  E=  mc 2 4 He (Z=2) (N=2) 206 Pb (Z=82) (N=124) 210 Po (Z=84) (N=126) Q = energy release = [ M( 210 Po) – { M( 206 Pb) + M( 4 He)} ] c 2 ~ 5.5 MeV ~1 pJ. i.e., around 0.003% of total 210 Po mass released kinetic energy in decay. (rare occasion, 206 Pb can be left in an excited state….803 keV above the ground state.) Total energy released shared as KINETIC energy between 206 Pb and 4 He. Cons. of linear momentum means most of the energy (5.3 MeV) goes to 4 He.

10. Nuclear Physics & Society (2009) 10 Energy Release in  Decay: A Sum!! Use  E = mc 2 to calculate  particle energies in 210-Polonium decay…… Recall, 1u = 1 atomic mass unit = 931.5 MeV/c 2 (=1.66x10 -27 kg) M( 210 Po) = 209.982848u M( 206 Pb) = 205.974440u M( 4 He) = 4.002603u  E=Q = { M( 210 Po) –[ M( 206 Pb) + M( 4 He)] } c 2  E=Q = [209.982848 – (205.97444 + 4.002603)] x 931.5 MeV  E=Q = 0.005805 x 931.5 MeV = 5.407 MeV BUT energy is split between emitted  particle and 206 Pb such that T  = Q / [1 + M  / M 206Pb ] = 5.407 / [1+4/206] =5.407 / 1.019 = 5.3 MeV

11. Nuclear Physics & Society (2009) 11  spectroscopy is very useful in determining which alpha emitters are present….. Particularly useful in identifying minor actinides from nuclear waste (such as 241 Am)

12. Nuclear Physics & Society (2009) 12 Radiation in our Environment We are all constantly subject to irradiation mainly from natural sources. There are three main sources of such radiation.  a) Primordial -around since the creation of the Earth (  4.5 x 10 9 years)  235,8 U ( and daughters including 210 Po), 232 Th or 40 K (+ 87 Rb, 138 La and others....)  b) Cosmogenic – from interaction of Cosmic rays with Earth and atmosphere.  14 C, 7 Be formed from cosmic ray interactions. Cosmic rays are mostly protons.  c) Produced or enhanced by human activity.  Medical or dental X-rays;  137 Cs (product from nuclear fission, 239 Pu,  241 Am, 239 Pu from weapons fallout

13. Nuclear Physics & Society (2009) 13 Radioactive species in the body Isotope Average amount by weightActivity U-Uranium 90μg 1.1Bq Th-Thorium 30 μg 0.11Bq 40 K 17mg 4.4 kBq Ra 31pg 1.1Bq 14 C 22ng 3.7kBq 3 H-tritium 0.06pg 23Bq Po-Polonium 0.2pg 37Bq Some variation- for example smokers have 4-5 times more Po.

14. Nuclear Physics & Society (2009) Q  210 Pb) = 5.41 MeV E  = 5.30 MeV E( 206 Pb) = 0.11 MeV T1/2 = 138 days. ‘ 218 Po =Radium A’ ‘ 218 At =Radium B’ C D E 210 Po =Radium ‘F’ Radon =‘Emanation’ ‘Radium’ C’ C’’ The Natural Decay Chain for 238 U

15. Nuclear Physics & Society (2009)