In the Uranium radioactive series, the initial nucleus is \({ }_{92}^{238} \mathrm{U}\) and final nucleus is…
In the Uranium radioactive series, the initial nucleus is \({ }_{92}^{238} \mathrm{U}\) and final nucleus is \({ }_{82}^{206} \mathrm{~Pb}\). When the Uranium nucleus decays to lead, the number of \(\alpha\)-particles emitted is .......... and the number of \(\beta\)-particles emitted is ____.
6,8
8,6
16,6
32,2
Solution
According to question,
Initial nuclei \(={ }_{92}^{238} \mathrm{U}\)
Final nuclei \(={ }_{82}^{206} \mathrm{~Pb}\)
Change in atomic number \(=92-82=10\)
When an \(\alpha\)-particle emits from a nuclei, then its atomic number is decreased by 2 units and atomic mass is decreased by 4 units. When a \(\beta\)-particle emits from a nuclei, then atomic number of parent nuclei is increased by 1 unit whereas its atomic mass remains same.
Change in atomic mass \(=238-206=32\)
Since, atomic mass changes only due to emission \(\alpha\)-particle, hence number of emitted \(\alpha\)-particle
\(=\frac{32}{4}=8\)
Number of emitted \(\beta\)-particles = total change in atomic number due to emission of \(\alpha\)-particles total change in atomic number
\(=8 \times 2-10=6\)