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प्रश्न
Mass number of a radioactive element is 232 and its atomic number is 90. When this element undergoes certain nuclear reactions, it transforms into an isotope of lead with a mass number 208 and an atomic number 82. Determine the number of alpha and beta decay that can occur.
उत्तर
Mass number A = 232
Atomic number Z = 90
Daughter element:
Mass number A = 208
Atomic number Z = 82
Difference in mass number = 232 – 208 = 24
Difference in atomic number
= 90 – 82 = 8
Atomic number of α = 2
Atomic number of β = -1
Mass number of α = 4
Mass number of β = 0
Difference in mass number in transformations
= 24
Number of a decays = `24/4` = 6
Difference in atomic number = 8
ΔZ = 6α + 4β
= 6(2) + 4(-1)
= 12 – 4
= 8
∴ Number of β decays = 4
∴ Number of α decays = 6
∴ Number of β decays = 4
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