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Standard X
Physics
Question

A nucleus with Z=92 emits the following in a sequence; α, α, β−, β−, α, α, α, α, β−, β−, α, β+, β+, α. The Z of the resulting nucleus is :

  1. 76
  2. 78
  3. 82
  4. 74

A
76
B
82
C
78
D
74
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Solution
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Each α particle emission removes 2 protons from the decaying compound.

Each β+ particle emission decreases the atomic number by 1.

Each β− particle emission adds 1 atomic number.

Thus the Z of resulting nucleus is 92−(8)(2)−(2)(1)+(4)(1)=78

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