A nucleus with Z -92 emits the following in a sequence- - - - - - - - - - - - - - - The Z of the resultant nucleus is -AIEEE-2003-A- 82B- 74C- 76D- 78

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Beta Decay

A nucleus wit...

Question

A nucleus with Z = 92 emits the following in a sequence: $α, β^{-}, β^{-}, α, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α,$ The Z of the resultant nucleus is [AIEEE-2003]

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Solution

The correct option is C 78
From the question, there are 8 $α$ decays, $4 β^{-}$ decays and $2 β^{+}$ decays

$∴ Z_{n e w} = Z_{o l d} - 8 \times 2 + 4 \times 1 - 2 \times 1$
= 92 - 16 + 4 - 2
= 78

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A nucleus with Z = 92 emits the following in a sequence: α, β−, β−, α, α, α, α, α, β−, β−, α, β+, β+, α, The Z of the resultant nucleus is [AIEEE-2003]

The correct option is C 78From the question, there are 8 α decays, 4 β− decays and 2 β+ decays ∴Znew=Zold−8×2+4×1−2×1 = 92 - 16 + 4 - 2 = 78

A nucleus with Z = 92 emits the following in a sequence: $α, β^{-}, β^{-}, α, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α,$ The Z of the resultant nucleus is [AIEEE-2003]

The correct option is C 78
From the question, there are 8 $α$ decays, $4 β^{-}$ decays and $2 β^{+}$ decays

$∴ Z_{n e w} = Z_{o l d} - 8 \times 2 + 4 \times 1 - 2 \times 1$
= 92 - 16 + 4 - 2
= 78