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\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + alpha particle

\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + alpha particle

A.

226

B.

220

C.

227

D.

222

View answer & explanation

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

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