A certain radioactive material zXA starts emitting α and β particles successively such that the end product is z−3YA−8.The number of α and β particles emitted are
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a
3 and 8 respective
b
2 and 1 respectively
c
3 and 4 respectively
d
4 and 3 respectively
answer is B.
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Detailed Solution
zxA→z−3YA−8 Number of α-particles = Change in mass number 4=84=2Z−2× Number of β-particles =Z−3 Number of β-particles =1