Question

In the uranium radioactive series the initial nucleus is \(^{238}U\) and the final nucleus is \(^{206}Pb\). When the uranium nucleus decays to lead, the number of \(\alpha\)-particles and \(\beta\)-particles emitted respectively are:

• A. 4, 6
• B. 8, 6
• C. 5, 8
• D. 7, 4

Hint:

In radioactive decay problems, keep in mind that \(\alpha\)-particles decrease both mass and atomic number, while \(\beta\)-particles increase only the atomic number.

Answer 1

The Correct Option is B

The decay of \(^{238}U\) to \(^{206}Pb\) involves the emission of both \(\alpha\)-particles and \(\beta\)-particles. In the decay process: - Each \(\alpha\)-decay decreases the mass number by 4 and the atomic number by 2. - Each \(\beta\)-decay increases the atomic number by 1, but the mass number remains unchanged. Since the decay process transforms \(^{238}U\) to \(^{206}Pb\), we calculate the number of \(\alpha\)- and \(\beta\)-decays required: - The mass number change: \(238 - 206 = 32\), so 8 \(\alpha\)-particles are emitted. - The atomic number change: \(92 - 82 = 10\), so 6 \(\beta\)-particles are emitted. Thus, the correct number of \(\alpha\)- and \(\beta\)-particles is 8 and 6, respectively.