1. 1.Kinetic Stability : probability that an unstable nucleus will decompose into more stable species through radioactive decay. 2.All nuclides with 84 or more protons are unstable and will decay. Light nuclides where Z = A-Z (neutron/proton ratio is 1). Nuclides with even numbers of neutrons and protons are unusually stable. Especially stable nuclides exhibit magic numbers, 2,8,20,28,50,82,126 of neutrons or protons. A nuclide is a unique atom of the type. Alpha decay, emits a NUCLEAR CHEMISTRY: INTRO X A Z He 4 2

2. NUCLEAR CHEMISTRY: STABILITY GRAPH 120 110 50 60 70 80 90 100 10 20 30 40 #n 20 p40 p60 p80 p100 p

3. NUCLEAR CHEMISTRY: BETA DECAY C 14 6 e  + N 0 7 BETA DECAY 1.THE ATOMIC NUMBER OF THE PRODUCT INCREASES. 2.NUCLIDES ABOVE THE PENNINSULA (ZONE) OF STABILITY DECAY WITH BETA DECAY(SEE GRAPH ON OTHER SLIDE). 3.PENETRATING RADIATION. 4.THE BETA PARTICLE CONES FROM THE DECOMPOSITION OF A NEUTRON TO A PROTON AND BETA PARTICLE. THE BETA PARTICLE IS AN ELECTRON “BORN” IN THE NUCLEUS. 5.BETA DECAY IS SPONTANEOUS, NOTICE ONLY ONE REACTANT. p 1 0 e  + n 0 1 1 BETA PARTICLE

4. NUCLEAR CHEMISTRY: ALPHA DECAY U 238 92 Th  + He 234 2 4 90 ALPHA DECAY 1.THE ATOMIC NUMBER OF THE PRODUCT DECREASES BY 2. 2.THE MASS NUMBER OF THE PRODUCT DECREASES BY 4. 3.ALPHA RADIATION IS NON PENETRATING TO HUMAN SKIN, HOWEVER IT CAN BE INGESTED. 4.COMMON MODE OF DECAY FOR HEAVY RADIOACTIVE NUCLIDES. 5.NEUTRON/PROTON RATIO INCREASES. 6.THE ALPHA PARTICLES ARE POSITIVE, THESE HIGH ENERGY He ATOMS HAVE LOST THE ELECTRONS, ARE REPELLED BY POSITIVE ELECTRODES AND ARE AFFECTED BY MAGNETIC FIELDS. 7.ALPHA DECAY IS SPONTANEOUS. ALPHA PARTICLE

5. NUCLEAR CHEMISTRY: POSITRON EMISSION Na 22 11 Ne  + e 22 +1 0 10 POSITRON EMISSION 1.DECAY MODE FOR NUCLIDES BELOW ZONE OF STABILITY. 2.CHANGES A PROTON TO A NEUTRON. 3.PRODUCT HAS A HIGHER NEUTRON TO PROTON RATIO. 4.THE POSITRON IS THE ANTIPARTICLE TO AN ELECTRON THE REACTION OF A POSITRON WITH A BETA PARTICLE PRODUCES GAMMA RADIATION. POSITRON e +1 0 e 0 +  GAMMA 0 0

6. NUCLEAR CHEMISTRY: ELECTRON CAPTURE Hg 201 80 Au + e 201 0 79 ELECTRON CAPTURE 1.AN INNER SHELL ELECTRON IS CAPTURED BY THE NUCLEUS. INNER ORBITAL SHELL ELECTRON e +1 0 e 0 +  GAMMA 0 0  + 0 0

7. ln (N 0 /N) = kt t 1/2 =0.693/k NUCLEAR CHEMISTRY: ELECTRON CAPTURE E = c 2 m Log N 0 = kt N 2.303 ( )

8. Example problem, binding energy m = (7(1.0072765) + 7(1.0086649)) – 13.999234 = 0.112536 amu E = 0.112536 amu 1 g x 1kg x 8.987551*10 16 m 2 6.0221 * 10 23 amu 1000g s 2 = 1.67682 * 10 -11 kg m 2 /s 2 = 1.67682 * 10 -11 J Binding energy/nucleon = 1.67682 * 10 -11 J/ 14 = 1.19773 * 10 -12 J/nucleon E = c 2 m Mass defect equation OBJECTIVE calculate the binding energy per nucleon of N –14, nuclear mass is 13.999234 mass of protonmass of neutronmass nucleus For 14 N, A=14 Avagodro’s #

9. Example problems:Half Life Calculate the mass of Co-60 that remains from a 0.0100 g sample after 1.00 year has elapsed. Log N 0 = kt N 2.30 ( ) k= 0.693 = 0.693 = 0.132y t 1/2 5.27/y Co-60 Half life, from tables 1-FIRST FIND k FROM HALF LIFE. Log N 0 = (0.132/y)(1.00y) = 0.0570, ANTILOG OF 0.0570 IS 1.14 N 2.303 ( ) 2-FIND N 0 /N RATIO N 0 = 1.14 = 0.0100g ; N = 0.00877g OF AFTER 1 YEAR N C 60 27