Answer:
[tex]_{84}^{208}Po[/tex]
B is correct.
Explanation:
Given the equation
[tex]_{86}^{212}Rn\rightarrow_{2}^{4}He[/tex]
According to alpha emission
Alpha emission is a radioactive decay in which the atomic nucleus emit the alpha particle and formed a new nucleus whose mass number reduced by four and atomic number reduced by two.
Therefore,
Po is the new nucleus.
[tex]_{86}^{212}Rn\rightarrow_{2}^{4}He+_{84}^{208}Po[/tex]
Where, Rn = Radon
Po= polonium
