Discussion
Given : Masses of \(^{80}_{35}\)Br and \(^{80}_{34}\)Se are respectively 79.918528 amu and 79.916520 amu. Use 1 amu = 931.5 MeV Nucleus \(^{80}_{35}\)Br decays to \(^{80}_{34}\)Se by emitting a positron. The end point energy of the emitted positron is nearly :
Answer: B
The end point energy is nearly 0.84 MeV.
With atomic masses, the Q-value of positron decay must subtract two electron masses: one for the emitted positron and one for the extra atomic electron of the parent:
\[ \begin{aligned} \Delta M c^{2} &= (79.918528 - 79.916520) \times 931.5 \\ &= 0.002008 \times 931.5 = 1.870\ \text{MeV} \\ Q &= 1.870 - 1.022 = 0.848\ \text{MeV} \end{aligned} \]
The maximum (end point) energy of the positron is this Q-value, about 0.84 MeV. Forgetting the 1.022 MeV gives 1.87 MeV, close to option C, while 0.511 and 1.022 MeV are electron rest-mass energies.
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