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U 92238 is known to undergo radioactive decay to form Pb 82206 by emitting alpha and beta particles. A rock initially contained 68 × 10–6g of Pb 92238. If the number of alpha particles that it would emit during its radioactive decay of U 92238 to Pb 82206 in three half-lives is Z × 1018, then what is the value of Z ?

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Detailed Solution

Initial moles of U238=68×10−6238=x Moles of U238 decayed in three half-lives =78xIn decay from U238 to Pb206 , each U238 atom decays and produces 8 α-particles and hence, total number of α-particles emitted out=78x×8×NA=7×68×10−6238×6.022×1023=1.204×1018

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