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Alpha Decay

A nucleus wit...

Question

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

A

76

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B

78

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C

82

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D

74

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Solution

The correct option is C $78$
$\begin{matrix} Given - & * A nucleons with z = 92 * Sequence \to 2 α, 2 β^{- 1}, 4 α, 2 β^{- 1}, α, 2 β^{+}, α * Total = 8 α particles, 4 β^{- 1}, 2 β^{+} particles emitted. \end{matrix}$

$\begin{matrix} we know - * 1 α particle decreases z by (2) * 1 β^{+} decay decreases z by (1) * 1 β^{-} decay increases z by (1) \Rightarrow z_{final} = 92 - 8 α_{decay} - 2 β_{decay}^{+} + 4 b e t a^{-} decay = 92 - 16 - 2 + 4 z_{f} = 78 \end{matrix}$

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