Question

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

Answer 1

Step 1: Given data

Mass of ${}_{92}{}^{238}\mathrm{U}$$=$$68\times {10}^{-6}\mathrm{g}$

Number of half-lives$=3$

Step 2: Calculating the value of Z

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

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

Number of molecules of ${}_{92}{}^{238}\mathrm{U}$ produced$=0.286\times {10}^{-6}\times {\mathrm{N}}_{\mathrm{A}}$

$=0.286\times {10}^{-6}\times 6.022\times {10}^{23}\Rightarrow 1.72\times {10}^{17}$

After three half-lives number of molecules$=\frac{\mathrm{Number}\mathrm{of}\mathrm{mlecules}}{{\left(2\right)}^3}$

$=\frac{1.72\times {10}^{17}}{8}\Rightarrow 0.215\times {10}^{17}$

So, the number of molecules of uranium decayed$=1.72\times {10}^{17}-0.215\times {10}^{17}\Rightarrow 1.5\times {10}^{17}$

From the reaction, it produces eight alpha particles, so the number of alpha particles produced$=8\times 1.5\times {10}^{17}\Rightarrow 12\times {10}^{17}\mathrm{or}1.2\times {10}^{18}$

Therefore, the value of Z$=1.2$