The radionuclide C-11 (atomic number 6) decays by beta+ emission. Given: atomic mass of C-11 = 11.011434 u, atomic mass of B-11 = 11.009305 u, electron mass = 0.000548 u, and 1 u = 931.5 MeV/c². What is the Q-value of this decay?

0.962 MeV
0.962 * 10³ MeV
0.962 eV
0

Correct answer: 0.962 MeV

Solution

In beta+ decay using atomic masses: Q = [m(C-11) - m(B-11) - 2mₑ]*c². Deltaₘ = 11.011434 - 11.009305 - 0.001096 = 0.001033 u. Q = 0.001033 * 931.5 = 0.962 MeV.

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