Uranium-238 (²³⁸₉₂U) decays through a series to stable lead-206 (²⁰⁶₈₂Pb). How many alpha and beta-minus particles are emitted in total?
Correct answer C. 8 alpha and 6 beta
Explanation
Number of alpha = (238 − 206)/4 = 8, lowering Z by 16 to 76. To reach Z = 82, six β⁻ decays are needed, each raising Z by 1.
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