An element of nucleon number P and atomic number Q emits an alpha particle from its nucleus. The resultant numbers of the new element formed are respectively

Analyze the question: An alpha particle is emitted from the nucleus of an atom. We need the new nucleon number (mass number) and atomic number (proton number) of the daughter element.

Concept: An alpha particle is a helium nucleus, ⁴₂He. It consists of 2 protons and 2 neutrons, so:

• Nucleon number = 4
• Atomic number = 2

When the nucleus emits an alpha particle, it loses both of these quantities. The nucleon number and atomic number are each conserved in the reaction.

Solution:

Parent nucleus: ᴾ_Q X

Alpha emission:

ᴾQ X → ᴾ⁻⁴(Q−2) Y + ⁴₂He

Check the conservation:

• Nucleon number: P = (P − 4) + 4 ✔
• Atomic number: Q = (Q − 2) + 2 ✔

So the new element has nucleon number P − 4 and atomic number Q − 2.

Why the others are wrong:

• A: Atomic number would increase, which happens in beta-minus decay, not alpha decay.
• C and D: The nucleon number cannot increase when a particle is emitted.

Exam tip: Remember "alpha takes away 4 and 2": the mass number drops by 4 and the atomic number drops by 2. For comparison, beta-minus decay leaves the mass number unchanged and raises the atomic number by 1. Always check that the top numbers and bottom numbers balance on both sides.