An element X of an atomic number 88 and mass number 226 decays to form...

JAMB 1993

An element X of an atomic number 88 and mass number 226 decays to form an element Z by emitting two beta particles and an alpha particle. Z is represented by

A. \(^{222}_{82}Z\)
B. \(^{222}_{88}Z\)
C. \(^{226}_{86}Z\)
D. \(^{226}_{80}Z\)

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Correct Answer: Option B

Explanation

\(^{226}_{88}X\) \(\to\) \((^{0}_{-1}e)\) + \(^{4}_{2}\alpha\) + \(^{b}_{a}X\)

a + 2 + (-2) = 88 a \(\Rightarrow\) a = 88

b + 4 + 0 = 266 b \(\Rightarrow\) b = 222

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