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Question

23892U is known to undergo radioactive decay to form 20682Pb by emitting alpha and beta particles. A rock initially contained 68×106 g of 23892U. If the number of alpha particles that it would emit during its radioactive decay of 23892U20682Pb in three half-lives is Z×1018, then what is the value of Z?

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Solution

92U23882Pb206+82He4+61β0(antineutrino)92U23882Pb206+82He4+61β0(antineutrino)

Initial mole of U238=[68×106238]=x

Mole of U238 decayed in three half-lives =7x/8

In decay from U238 to Pb206, each U238 atom decays and produces 8 α particles and hence, total number of α particles emitted out =(7x8)×8×NA

=78×68×106238×8×6.022×1023=1.2046×1018
Z=1.21


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