The radioactive nuclide 88228 Ra decays by emission of three α-particles. The nuclide "X" finally formed is:
Correct answer: C. 21682X
- A. 22084X
- B. 22286X
- C. 21682X
- D. 21588X
Explanation
According to the equation: 22888Rn → 21682X + 3 42He
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