${ }_{92}^{238}U$ is known to undergo radioactive decay to form ${ }_{82}^{206}Pb$  by emitting alpha and beta particles. A rock initially contained  $68\times {10}^{-6}g$ of ${ }_{92}^{238}U$ . If the number of alpha particles that it would emit during its radioactive decay of ${ }_{92}^{238}U$  to  ${ }_{82}^{206}Pb$  in three half-lives is $Z\times {10}^{18},$ then what is the value of Z?

${}_{92}{}^{238}\mathrm{U}\to {}_{82}{}^{206}\mathrm{Pb}+8{}_2{}^4\mathrm{He}^{2+}+6{}_{-1}{}^0\mathrm{e}$

Initial moles of ${ }_{92}^{238}U$ = $\frac{68\times {10}^{-6}}{238}=0.286\times {10}^{-6}$

initial${ }_{92}^{238}U$ nuclides = $0.286\times {10}^{-6}\times 6.023\times {10}^{23}=1.72\times {10}^7$ 

${ }_{92}^{238}U$ present after 3${\mathrm{t}}_{0.5}$=$\frac{1.72\times {10}^7}{8}=$0.215 $\times$10¹⁷

${ }_{92}^{238}U$ decayed = $\left(1.72-0.215\right)\times {10}^{17}=1.5\times {10}^{17}$

Number of $\mathrm{\alpha }-$particles emitted = Number of atoms of ${ }_{92}^{238}U$ decayed $\times$8

Number of $\mathrm{\alpha }-$particles emitted = 1.5$\times$${10}^{17}$ $\times$8 =1.2$\times$10¹⁸