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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Advanced 2024 · Paper 2 · Chemistry · Chemical Kinetics

A sample initially contains only U-238 isotope of uranium. With...

A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into Pb−206\mathrm{Pb}-206 while the rest of it remains undisintegrated. When the age of the sample is P×108\mathbf{P} \times 10^8 years, the ratio of mass of Pb−206\mathrm{Pb}-206 to that of U−238\mathrm{U}-238 in the sample is found to be 7 . The value of P\mathbf{P} is ______ [Given: Half-life of U-238 is 4.5×1094.5 \times 10^9 years; log⁡c2=0.693\log _c 2=0.693 ]

Correct answer

0

Step-by-step solution

{ }_{92} \mathrm{U}^{238} \quad \longrightarrow \quad{ }_{82} \mathrm{~Pb}^{206}+8{ }_2 \mathrm{He}^4+6_{-1} \beta^0 \mathrm{t}=0 \quad \mathrm{~N}_0=(1238+7206)\left(\frac{1}{238}+\frac{7}{206}\right) moles t=tNt=1238\mathrm{t}=\mathrm{tN}_{\mathrm{t}}=\frac{1}{238} moles x=7206\mathrm{x}=\frac{7}{206} moles Life of sample (\rightarrow t) years ([A]_0 \propto) Initial mole of (\mathrm{U}-238) ([A]]_t \propto) Final mole of U-238 (∣A∣0∣A∣t=1228+72061238=0.0042+0.03400.0042=9.1=2.303log⁡2×t4.5×109=2.303log⁡9.1t=14.27×109 years =142.7×108 years P=142.7P=143\begin{aligned} & \frac{|A|_0}{|A|_t}=\frac{\frac{1}{228}+\frac{7}{206}}{\frac{1}{238}} \\ & =\frac{0.0042+0.0340}{0.0042} \\ & =9.1 \\ & =\frac{2.303 \log 2 \times t}{4.5 \times 10^9}=2.303 \log 9.1 \\ & t=14.27 \times 10^9 \text{ years } \\ & =142.7 \times 10^8 \text{ years } \\ & P=142.7 \\ & P=143 \end{aligned})
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