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

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$_{82} U^{210}$

$_{82} U^{206}$

$_{82} U^{214}$

$_{82} {Pb}^{206}$

answer is B.

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Detailed Solution

$_{(z = 92)} U^{(A = 238)} {\overset{_{- 1} β^{0}}{\to}}_{z'} X^{A'}$

$So A' = A - 4 n_{α} = 238 - 4 x 8 = 206 and Z' = n_{β} - 2 n_{α} + z = 6 - 2 x 8 + 92 = 82 .$

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