${}_{92}^{238} \mathrm{U}$ is known to undergo radioactive decay to form ${}_{82}^{206} \mathrm{~Pb}$ by emitting alpha 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?

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Chemical Kinetics

_{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 1 0^{- 6} g

_{92}^{238} U

. If the number of alpha particles that it would emit during its radioactive decay of

_{92}^{238} U

_{82}^{206} Pb

in three half-lives is

Z \times 1 0^{18}

, then what is the value of Z?

Answer: 1.2

Initial moles of

X_{238} X_{2}^{2238} U

\frac{68 \times 1 0 ^{- 6}}{238}

mol

After three half-lives, fraction decayed =

1 - \frac{1}{2 ^{3}} = \frac{7}{8}

Moles of

X_{238} X_{2}^{2238} U

decayed =

\frac{7}{8} \times \frac{68 \times 1 0 ^{- 6}}{238}

mol

In the decay

X_{238} X_{2}^{2238} U X_{206} X_{2}^{2206} Pb

Total α-particles emitted:

N = \frac{7}{8} \times \frac{68 \times 1 0 ^{- 6}}{238} \times 8 \times 6.022 \times 1 0^{23}

N = 7 \times \frac{68 \times 1 0 ^{- 6}}{238} \times 6.022 \times 1 0^{23} = 1.204 \times 1 0^{18}

Therefore, Z = 1.20

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Numerical answer