WAEC - Physics (2000 - No. 43)
A nuclide \(^{202}_{84} Y\) emits in succession an \(\alpha-particle\) and a \(\beta-particle\). The atomic number of the resulting nuclide is
198
83
82
80
ব্যাখ্যা
\(^{202}_{84} Y\) → \(^4_2\)He + \(^0_{-1}\)e + \(^{98}_{83}\)X
Therefore, the atomic number of the resulting nuclide = 83.