In a nuclear reaction where polonium-212 (\(^{212}{84}Po\)) emits an alpha particle (\(^{4}{2}\alpha\)), the atomic number and mass number of the resulting element can be determined as follows:
When polonium emits an alpha particle, the reaction can be represented as:
\[ ^{212}{84}Po \rightarrow ^{4}{2}\alpha + ^{A}_{Z}Pb \]
To find the atomic number of lead (\(Pb\)), we can subtract the atomic number of the alpha particle from the atomic number of polonium:
\[ Z_{Pb} = 84 - 2 = 82 \]
The corresponding mass number of lead can be calculated similarly:
\[ A_{Pb} = 212 - 4 = 208 \]
Thus, polonium-212 decays to lead-208, and the atomic number of lead is 82.
Therefore, the correct response is:
82