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If { }_{92} U^{238} undergoes successively 8 -decays and 6 -de

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Solving time: 1 mins

If $_{92} U^{238}$ undergoes successively $8 α$ -decays and $6 β$ -decays, then resulting nucleus is

_{82} U^{206}

_{82} P b^{206}

_{82} U^{210}

_{82} U^{214}

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Camila Discussed

_{92} U^{238}

undergoes successively

8 α

-decays and

6 β

-decays, then resulting nucleus is

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

(Z = 92) U^{(A = 238)} (8 α, 6 β) \to Z^{\cdot} X^{A^{'}}

A^{'} = A - 4 n_{α} = 238 - 4 \times 8 = 206

and

Z^{'} = n_{β} - 2 n_{σ} + z = 6 - 2 \times 8 + 92 = 82

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Question Text	If $_{92} U^{238}$ undergoes successively $8 α$ -decays and $6 β$ -decays, then resulting nucleus is
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	93