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

Chemistry JAMB 2017

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

  • 226
  • 220
  • 227
  • 222 checkmark

The correct answer is: D

Explanation

\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + \(^4_{2}He\)

\(^4_{2}He\) = alpha particle

considering the summation of the mass number

226 = x + 4

x = 226 - 4

x = 222

There is an explanation video available .

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