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Manipal MET2014PhysicsNuclear Physics

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

Options

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

Correct answer

D. 4 , 1

Step-by-step solution

Z=90 Th ^ A=228 Z^ =83 Bi ^ A^ =212 Number of -particles emitted n_ = A-A^ 4 = 228-212 4 =4 Number of -particles emitted n_ =2 n_ -Z+Z^ =2 4-90+83=1 .

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