A nuclide \(^{202}_{84} Y\) emits in succession an \(\alpha-particle\) and a \(\beta-particle\). The atomic number of the resulting nuclide is
A
198
B
83
C
82
D
80
Verified Correct Answer: B.
Physical Principles & Numerical Solution
\(^{202}_{84} Y\) → \(^4_2\)He + \(^0_{-1}\)e + \(^{98}_{83}\)X
Therefore, the atomic number of the resulting nuclide = 83.
Source: Myschool Past Questions Archive (Record #33849)