Decay identification: in the nuclear transformation 93Np239 → 94Pu239 + ?, identify the missing particle.

Difficulty: Easy

Correct Answer: electron

Explanation:


Introduction / Context:
Recognising decay modes from changes in atomic number Z and mass number A is a foundational nuclear skill. The transformation 93Np239 → 94Pu239 keeps A constant and increases Z by 1, pointing to a specific beta process.


Given Data / Assumptions:

  • Parent: neptunium-239 (Z = 93, A = 239).
  • Daughter: plutonium-239 (Z = 94, A = 239).
  • Unknown emitted particle to be identified.


Concept / Approach:
When Z increases by 1 and A remains constant, a neutron in the nucleus converts to a proton. That is β− decay (electron emission): n → p + e− + anti-ν. β+ decay or electron capture would decrease Z by 1; α decay would reduce A by 4 and Z by 2; neutron emission would reduce A by 1.


Step-by-Step Solution:
Compare Z and A before and after: ΔZ = +1, ΔA = 0.Match to β− decay signature.Therefore the emitted particle is an electron (beta minus), with an antineutrino also emitted.


Verification / Alternative check:
Decay chains from activation of U-238 (via n,γ to U-239 → β− to Np-239 → β− to Pu-239) illustrate this exact step and particle identity.


Why Other Options Are Wrong:

  • Proton or neutron emission would alter A.
  • Positron emission (β+) would reduce Z by 1.
  • Alpha emission changes both Z and A significantly (−2, −4).


Common Pitfalls:
Forgetting that mass number counts nucleons only; electrons do not change A, so β− keeps A constant while increasing Z.


Final Answer:
electron

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