A radioactive nucleus ²⁵²₉₈X, initially in its ground state, undergoes the following sequence: it emits three α-particles, then two β⁻-particles, then one β⁺-particle, and finally one γ-photon. No other particles are emitted. Which pair gives the mass number A and atomic number Z of the final daughter nucleus?
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Question

A radioactive nucleus ²⁵²₉₈X, initially in its ground state, undergoes the following sequence: it emits three α-particles, then two β⁻-particles, then one β⁺-particle, and finally one γ-photon. No other particles are emitted. Which pair gives the mass number A and atomic number Z of the final daughter nucleus?

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G1 Physics · Accepted Answer

Track mass number and atomic number separately. Each α-decay changes A → A−4 and Z → Z−2. After three α-decays: A = 252−12 = 240, Z = 98−6 = 92. Each β⁻-decay converts a neutron into a proton, so A remains unchanged and Z increases by 1. After two β⁻-decays: Z = 92+2 = 94. A β⁺-decay converts a proton into a neutron, so A remains unchanged and Z decreases by 1: Z = 94−1 = 93. A γ-decay only removes excitation energy; it does not change A or Z. Therefore the final nucleus has A = 240, Z = 93. Co…