JEE Advance - Chemistry (2020 - Paper 1 Offline - No. 17)
$$_{92}^{238}U$$ 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 $$ \times $$ 1018, then what is the value of Z?
Answer
1.2
Explanation
$$_{92}^{238}U\buildrel {} \over
\longrightarrow _{82}^{206}Pb + 8_2^4He + 6_{ - 1}^0\beta $$
Number of moles of $$_{92}^{238}U$$ present initially
= $${{68 \times {{10}^{ - 6}}} \over {238}}$$
After three half-lifes, moles of $$_{92}^{238}U$$ decayed
$$ = {{68 \times {{10}^{ - 6}}} \over {238}} \times \left( {1 - {1 \over {{2^3}}}} \right)$$
$$ = {{68 \times {{10}^{ - 6}}} \over {238}} \times {7 \over 8}$$
Therefore, number of $$\alpha $$-particles emitted
$$ = {{68 \times {{10}^{ - 6}}} \over {238}} \times {7 \over 8} \times 8 \times 6.023 \times {10^{23}}$$
$$ = 1.204 \times {10^{18}}$$
$$ \approx 1.2 \times {10^{18}}$$
Thus, the correct answer is 1.2.
Number of moles of $$_{92}^{238}U$$ present initially
= $${{68 \times {{10}^{ - 6}}} \over {238}}$$
After three half-lifes, moles of $$_{92}^{238}U$$ decayed
$$ = {{68 \times {{10}^{ - 6}}} \over {238}} \times \left( {1 - {1 \over {{2^3}}}} \right)$$
$$ = {{68 \times {{10}^{ - 6}}} \over {238}} \times {7 \over 8}$$
Therefore, number of $$\alpha $$-particles emitted
$$ = {{68 \times {{10}^{ - 6}}} \over {238}} \times {7 \over 8} \times 8 \times 6.023 \times {10^{23}}$$
$$ = 1.204 \times {10^{18}}$$
$$ \approx 1.2 \times {10^{18}}$$
Thus, the correct answer is 1.2.
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