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Carbon-11 (¹¹₆ C) decays spontaneously into Boron-11 (¹¹₅ B). Given: Atomic mass of ¹¹₆ C = 11.011434 u, Atomic mass of ¹¹₅ B = 11.009305 u, mass of electron = 0.000549 u, 1 u = 930 MeV/c². Which of the following alternatives are correct?

1. A beta⁻ particle is released
2. A beta⁺ particle is released
3. The maximum kinetic energy of the emitted particle is 0.959 MeV
4. An anti-neutrino is also released in the nuclear reaction

Correct answer: A beta⁺ particle is released

Solution

Since Z decreases from 6 to 5, this is beta⁺ (positron) decay. Q = [M(C-11) - M(B-11) - 2*mₑ]*c² = [11.011434 - 11.009305 - 2*0.000549]*930 MeV = [0.002129 - 0.001098]*930 = 0.001031*930 = 0.959 MeV. So both options B (beta⁺ released) and C (max KE = 0.959 MeV) are correct. In beta⁺ decay, a neutrino (not anti-neutrino) is released. So D is wrong.

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