\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + alpha particle
\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + \(^4_{2}He\) \(^4_{2}He\) = alpha particle considering the summation of the mass number 226 = x + 4 x = 226 - 4 x = 222
There is an explanation video available below.
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