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
The correct answer is: B
Explanation
\(^{202}_{84} Y\) → \(^4_2\)He + \(^0_{-1}\)e + \(^{98}_{83}\)X
Therefore, the atomic number of the resulting nuclide = 83.