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
\(^{226}_{80}Z\)
Verified Correct Answer: B.
Physical Principles & Numerical Solution
\(^{226}_{88}X\) \(\to\) \((^{0}_{-1}e)\) + \(^{4}_{2}\alpha\) + \(^{b}_{a}X\)
a + 2 + (-2) = 88 a \(\Rightarrow\) a = 88
b + 4 + 0 = 266 b \(\Rightarrow\) b = 222
Source: Myschool Past Questions Archive (Record #28027)