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Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsSecond & nth Order ReactionsMedium2 minPYQ_2008

ChemistryMediumsingle choice

In successive emission of $α$ -and $β$ -particles, the number of $α$ -and $β$ -particles should be emitted for the conversion of $_{92} U^{238}$ to $_{82} Pb^{206}$ are

Options:

Answer:

Solution:

$_{92} U^{238} ⟶_{82} Pb^{206} + m_{2} He^{4} + n_{- 1} e^{0}$ On comparing both sides $238 \Rightarrow m 92 2 m - n \Rightarrow n = 206 + 4 m = 8 = 82 + 2 m - n = 10 = 6$ $∴ α$ -particles emitted, $m = 8$ $β$ -particles emitted, $n = 6$

Stream:MHT_CETSubject:ChemistryTopic:Chemical KineticsSubtopic:Second & nth Order Reactions

⏱ 2mℹ️ Source: PYQ_2008

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