JAMB - Physics (1993 - No. 37)
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
\(^{222}_{82}Z\)
\(^{222}_{88}Z\)
\(^{226}_{86}Z\)
\(^{226}_{80}Z\)
Förklaring
\(^{226}_{88}X\) \(\to\) \((^{0}_{-1}e)\) + \(^{4}_{2}\alpha\) + \(^{b}_{a}X\)
a + 2 + (-2) = 88 a \(\Rightarrow\) a = 88
b + 4 + 0 = 266 b \(\Rightarrow\) b = 222
a + 2 + (-2) = 88 a \(\Rightarrow\) a = 88
b + 4 + 0 = 266 b \(\Rightarrow\) b = 222