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When 90 Th 228 transforms to 83 Bi 212 , then the number to the emitted α and β -particles are, respectively

Options

  1. A8 α , 7 β
  2. B4 α , 7 β
  3. C4 α , 4 β
  4. D4 α , 1 β

Correct answer

D. 4 α , 1 β

Step-by-step solution

Z = 90 T h A = 228 → Z ′ = 83 B i A ′ = 212 Number of α -particles emitted n α = A - A ′ 4 = 228 - 212 4 = 4 Number of β -particles emitted n β = 2 n α - Z + Z ′ = 2 × 4 - 90 + 83 = 1 .

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