Question

{"given":"Masses: $m(^{64}_{29}\\text{Cu}) = 63.9298 \\text{ u}$, $m(^{64}_{30}\\text{Zn}) = 63.9292 \\text{ u}$. These are isobars (same mass number $A = 64$, different atomic numbers $Z = 29$ and $Z = 30$).","key_observation":"The nuclide with higher atomic mass is energetically less stable and will decay to the nuclide with lower mass. Since $m(^{64}\\text{Cu}) > m(^{64}\\text{Zn})$, copper-64 is radioactive. To decay from $Z=29$ to $Z=30$, the atomic number must increase by 1, which is the hallmark of $\\beta^-$ decay: $n \\rightarrow p + e^- + \\bar{\\nu}_e$.","option_analysis":[{"label":"(A)","text":"Both isobars are stable.","verdict":"incorrect","explanation":"Since Cu-64 has a higher mass than Zn-64, it has excess energy and cannot be stable; it must decay to the lower-mass isobar."},{"label":"(B)","text":"Zn-64 is radioactive, decaying to Cu-64 through beta-minus decay.","verdict":"incorrect","explanation":"Zn-64 has the lower mass (63.9292 u) and is the more stable isobar; it is Cu-64 (higher mass, 63.9298 u) that is radioactive, not Zn-64."},{"label":"(C)","text":"Cu-64 is radioactive, decaying to Zn-64 through gamma decay.","verdict":"incorrect","explanation":"Gamma decay does not change the atomic number $Z$, so it cannot transform Cu ($Z=29$) into Zn ($Z=30$); a change in $Z$ requires $\\beta$ decay."},{"label":"(D)","text":"Cu-64 is radioactive, decaying to Zn-64 through beta-minus decay.","verdict":"correct","explanation":"Cu-64 has higher mass and is unstable; $\\beta^-$ decay converts a neutron to a proton, increasing $Z$ from 29 to 30: $$^{64}_{29}\\text{Cu} \\rightarrow ^{64}_{30}\\text{Zn} + e^- + \\bar{\\nu}_e$$"}],"answer":"(D)","formula_steps":[]}