QJEEAdv20Chf8d7: JEE Advanced 2020 Chemistry - Class 12 JEEAdv20Chf8d7 Solution: is known to undergo radioactive decay to form by emitting alpha …

Question

Verified Solution • 2025–26 Session moderate

_{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?

Accepted Answer

_{92}^{238} U {\overset{}{⟶}}_{82}^{206} P b + 8_{2}^{4} H e + 6_{- 1}^{0} β

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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