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 × 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 × 10 18 , then what is the value of Z ?

Chemistry · Thermodynamics & Thermochemistry

JEE Advanced 2020 — Paper 1 — Question 28

${ }_{92}^{238} \mathrm{U}$ 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 ?

Answer: 1.2

Numerical answer — enter this value.

Step-by-step solution

${ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{82}^{206} \mathrm{~Pb}+8\left({ }_{2}^{4} \mathrm{He}\right)+6\left({ }_{-1}^{0} \beta\right)$

Total moles of uranium $=\frac{68 \times 10^{-6}}{238}$ $=0.286 \times 10^{-6} \mathrm{~mol}$

So, after 3 half-lives fraction left $=0.125$

Fraction of uranium reacted $=0.875$

So, number of moles of $\alpha$ - particles $=\frac{8 \times 68 \times 10^{-6}}{238} \times 0.875 \mathrm{~mol}$

So, number of $\alpha$ - particle $=\frac{8 \times 68 \times 10^{-6} \times 0.875 \times 6.023 \times 10^{23}}{238}$

$=12.046 \times 10^{17}$

$=1.20 \times 10^{18}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 1
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Gibbs Free Energy and the Third Law of Thermodynamics