A nuclide \(^{202}_{84} Y\) emits in succession an \(\alpha-particle\) and a \(\beta-particle\). The atomic number of the resulting nuclide is
Question 1 Report
A nuclide \(^{202}_{84} Y\) emits in succession an \(\alpha-particle\) and a \(\beta-particle\). The atomic number of the resulting nuclide is
Answer Details
The nuclide \(^{202}_{84} Y\) emits an \(\alpha-particle\) which consists of two protons and two neutrons, leading to a decrease in the atomic number by 2 and mass number by 4. Therefore, the resulting nuclide is \(^{198}_{82} Pb\). This nuclide then emits a \(\beta-particle\) which is an electron emitted from the nucleus, leading to an increase in the atomic number by 1 and no change in the mass number. Therefore, the atomic number of the resulting nuclide is 83. So, the answer is (b) 83.
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