\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + alpha particle
226
220
227
222
Correct answer is D
\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + \(^4_{2}He\)
\(^4_{2}He\) = alpha particle
considering the summation of the mass number
226 = x + 4
x = 226 - 4
x = 222