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Beta Decay

A nucleus wit...

Question

A nucleus with Z = 92 emits the following in a sequence: $α, β^{-}, β^{-}, α, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α$ . Then, the Z of the resultant nucleus is

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Solution

The correct option is C
74

From the question, there are 8 $α$
decays, $4 β^{-}$ decays and $2 β^{+}$ decays

$β^{-}$ decay converts a neutron to proton and $β^{+}$ decay converts a proton to neutron.
$∴ Z_{n e w} = Z_{o l d} - (8 \times 2 + 4 \times 1 - 2 \times 1)$
= 92 - (16 + 4 - 2)
= 74

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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 . Then, the Z of the resultant nucleus is

The correct option is C 74 From the question, there are 8 α decays, 4 β− decays and 2 β+ decays β− decay converts a neutron to proton and β+ decay converts a proton to neutron. ∴Znew=Zold−(8×2+4×1−2×1) = 92 - (16 + 4 - 2) = 74

A nucleus with Z = 92 emits the following in a sequence: $α, β^{-}, β^{-}, α, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α$ . Then, the Z of the resultant nucleus is

The correct option is C
74

From the question, there are 8 $α$
decays, $4 β^{-}$ decays and $2 β^{+}$ decays

$β^{-}$ decay converts a neutron to proton and $β^{+}$ decay converts a proton to neutron.
$∴ Z_{n e w} = Z_{o l d} - (8 \times 2 + 4 \times 1 - 2 \times 1)$
= 92 - (16 + 4 - 2)
= 74