The radionuclide 11 C decays according to 11 C 11 B + e + + v,
Question:
The radionuclide 11C decays according to 11C → 11B + e+ + v, T1/2 = 20.3. The maximum energy of the positions is 0.960MeV.
(a) Show that the designation energy Q for this process is given by Q = (mC - mB – 2me) c2 where mC and mB are the atomic masses of 11C and 11B, respectively, and me is the mass of a positron.
(b) Given the mass values mC = 11.011 424 u, mB = 11.009 305 u, and me = 0.000 548 6 u, calculate Q and compare it with the maximum energy of the emitted positron given above.
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a Since the positron has the same mass as an electron and ...View the full answer
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Related Book For 
Fundamentals of Physics
ISBN: 978-0471758013
8th Extended edition
Authors: Jearl Walker, Halliday Resnick
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