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Coulomb barrier

  1. Feb 15, 2007 #1
    The Coulomb potential barrier of a system of two nuclei X and Y is approximately given by VC = ZX*ZY*e2/RN where ZX and ZY are the charge numbers of the nuclei, e2 = 1.44 MeV*fm, RN = (AX1/3+AY1/3) × r0 is the sum of the nuclear radii. r0 is a constant usually estimated to 1.2 to 1.3 fm and AX and AY are the mass numbers of the nuclei. Is this right calculation? What does fm stands for?
     
  2. jcsd
  3. Feb 17, 2007 #2

    Meir Achuz

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    It looks rlight. fm stands for femtometer (10^-15 meter).
    The AX1/3 sands for AX to the one third power.
     
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