A nuclide 220R86 decays to a new nuclide by two α- emissions, the nuclide S is:
Correct answer: C. 220S80
- A. 212S84
- B. 212S82
- C. 220S80
- D. 210S82
Explanation
After two α-emissions, the values for nuclide S would be: Atomic number (Z): 86 - (2 x 2) = 86 - 4 = 82 Mass number (A) 220 - (2 x 4) = 220 - 8 = 212.
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