An element X of an atomic number 88 and mass number 226 decays to form
PHYSICS
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. \(^{226}_{80}Z\)
Correct Answer: B. \(^{222}_{88}Z\)
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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