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Question

If by successive disintegration of 92U238, the final product obtained is 82Pb206, then how many number of α and β particles are emitted?

A
8 and 6
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B
6 and 8
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C
12 and 6
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D
8 and 12
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Solution

The correct option is A 8 and 6
We know that atomic number of the parent nucleus decreases by 2 and mass number decreases by 4 in each alpha decay whereas atomic number of the parent nucleus increases by 1 while mass number remains the same in each beta minus decay.
Change in mass number ΔA=238206=32
Number of alpha particles Nα=324=8
Decrease in mass number due to alpha decays only =2×8=16
Net decrease in mass number ΔZ=9282=10
So increase in mass number due to beta particles =1610=6
Number of beta particles Nβ=61=6

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