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²⁹⁰₈₂X →(α) Y →(e⁺) Z →(β⁻) P →(e⁻) Q. In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are:
Only α decay changes mass number (drops by 4 -> 290 - 4 = 286)! Beta particles have mass 0 -> Final A = 286 (eliminates 1 & 3); Z: 82 - 2 - 1 + 1 + 1 = 81 -> (286, 81)!
An alpha particle is a helium-4 nucleus (⁴₂He). Emission of an alpha particle reduces the mass number by 4 and atomic number by 2:
In positron emission, a proton inside the nucleus transforms into a neutron, emitting a positron (⁰₊₁e) and a neutrino. The mass number remains unchanged, while the atomic number decreases by 1:
In beta-minus emission, a neutron transforms into a proton, emitting an electron (⁰₋₁e) and an antineutrino. The mass number is unchanged, while the atomic number increases by 1:
The final emission is an electron (e⁻, identical to β⁻). The mass number remains unchanged, and the atomic number increases by another 1:
The resulting product nucleus Q has mass number A = 286 and atomic number Z = 81.
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Incorrect mass number; subtracted 2 instead of 4 for alpha particle.
Flawless conservation of nucleons and charge: A = 286, Z = 81.
Subtracts 10 from mass number.
Misses the final electron emission (+1 to Z).
Students frequently subtract 1 for both e⁺ and e⁻, forgetting that emitting a negative charge increases the nuclear positive charge (Z increases by 1).
"In any nuclear transformation, total mass number (A) and total electric charge (Z) are conserved strictly."
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