Question

a nucleus with Z = 92 emits the following in a sequence alpha, beta- , alpha , alpha, alpha , alpha , alpha, beta - , beta- , beta+, alpha, beta+, alpha. the Z of the resulting nucleus is​

Answer 1

Answer:

For each α emission z decreases by 2; 

For each β+ emission, z decreases by 1  and that for each β− it increases by 1.

No of α emitted =8

No of β+ emitted =2

No of β− emitted =4

∴  The resulting nucleus has atomic number  z=z(original)−2(8)−2(1)+1(4)

 ⟹z=92−16−2+4=78