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Beta Decay

The isotope ...

Question

The isotope $_{92}^{238} U$ decays successively to form $_{90}^{234} T h,$ $_{91}^{234} P a,$ $_{92}^{234} U,$ $_{90}^{230}$ $T h$ and $_{88}^{226} R a$ . The radiations emitted in these five steps are

A

α, α, α, β, β

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B

α, α, β, β, α

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C

α, β, β, α, α

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D

β, β, α, α, α

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Solution

The correct option is C $α, β, β, α, α$
$_{92}^{238} U \to_{90}^{234} T h \to_{91}^{234} P a \to_{92}^{234} U \to_{90}^{230} T h \to_{88}^{226} R a$
A -4 0 0 -4 -4
Z -2 +1 +1 -2 -2

$α β β α α$

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