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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.
 
On using the given data,
 
Difference in mass number=ii=iDifference in atomic number=ii=i
 
On using the atomic number and mass number of \(α\) and \(β\) particles, respectively.
 
Number of decays=ii=i
 
ΔZ=iα+iβ=i

Number of \(β\) decays \(=\)

Number of \(α\) decays \(=\)