MHT CET · Physics · Nuclear Physics
In the uranium radioactive series, the initial nucleus is \({ }_{92}^{238} \mathrm{U}\) and that the final nucleus is \({ }_{82}^{206} \mathrm{~Pb}\). When uranium nucleus decays into lead, the number of \(\alpha\)-particles and \(\beta\)-particles emitted are
- A \(4 \alpha, 5 \beta\)
- B \(5 \alpha, 3 \beta\)
- C \(6 \alpha, 7 \beta\)
- D \(8 \alpha, 6 \beta\)
Answer & Solution
Correct Answer
(D) \(8 \alpha, 6 \beta\)
Step-by-step Solution
Detailed explanation
For \(\alpha\) emission, the atomic number decreases by 2 and mass number decreases by 4.
\(\therefore \quad\) No. of \(\alpha\) particles emitted \(=\frac{238-206}{4}=8\)
\(\therefore \quad\) New atomic number \(=92-(8 \times 2)=76\)
For \(\beta\)-emission, only the atomic number increases by 1.
\(\therefore \quad\) No. of \(\beta\)-particles emitted \(=82-76=6\)
\(\therefore \quad\) No. of \(\alpha\) particles emitted \(=\frac{238-206}{4}=8\)
\(\therefore \quad\) New atomic number \(=92-(8 \times 2)=76\)
For \(\beta\)-emission, only the atomic number increases by 1.
\(\therefore \quad\) No. of \(\beta\)-particles emitted \(=82-76=6\)
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