Lecture_39

Lecture_39 - PHYS 342 Fall 2011 Lecture 39 Radioactive...

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PHYS 342 Fall 2011 Lecture 39: Radioactive Decay: , , Ron Reifenberger irck Nanotechnology Center Birck Nanotechnology Center Purdue University Because there are no external forces, momentum and energy are conserved in radioactive decay process Lecture 38/39 a radioactive decay process 1
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Alpha decay – an example http://phet.colorado.edu/simulations/sims.php?sim=Alpha_Decay When nucleus has too many protons compared to the number of eutrons the repulsive neutrons, the repulsive electrostatic force begins to dominate; nucleus becomes unstable nd emits an article 226 88 Ra Radium and emits an α particle. 22 4 2 He Radon Stopped by paper 222 86 Rn 226 222 4 88 86 2 Ra Rn He  2
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Why alpha decay? What is binding energy of ? 4 2 He ) y t m Chem Symbol nuclear mass 2 ) 2 2 / HB ( . (. ) x y x let m Chem Symbol M Chem Symbol atomic mass 42 2 2 () 2 2 / 22 pn ee m H emm B c mH e m m Bc neglect electron binding energies mm       2 24 2 2 2 2 / /( ) 2 2 2 4.002603 2 1.007276 1.008665 0.000549 pne MH e m m m B c Bc M H e m m m uu u u   4.002603 2 2.016490 0.0303 2 931.5 / 77 Mev c u 3 1 28.3 u B MeV
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Why Alpha Decay? alpha particle large nucleus radial separation, r 8 3 radial separation, r 28.3 MeV p n n p N Z If the last two protons and lasr two neutrons in a ucleus re ound y ss an 8 eV en p n 4 nucleus are bound by less than 28.3 MeV, then it is energetically favorable to form an alpha particle that is emitted from the nucleus.
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The Energetics of Alpha Decay 230 4 226 92 2 90 230 2 4 2 226 2 ) () ( ) UH e T h m U c m H e c m T h c
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This note was uploaded on 12/08/2011 for the course PHYS 342 taught by Professor Staff during the Fall '08 term at Purdue.

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Lecture_39 - PHYS 342 Fall 2011 Lecture 39 Radioactive...

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