Search any question & find its solution
Question:
Answered & Verified by Expert
In uranium radioactive series, initial nucleus ${ }^{238} \mathrm{U}_{92}$ decays to final nucleus ${ }^{206} \mathrm{U}_{82}$. In this process, the number of $\alpha$-particles and $\beta$-particles emitted are
Options:
Solution:
2832 Upvotes
Verified Answer
The correct answer is:
8 and 6
In decay, ${ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{82}^{206} \mathrm{U}$
Mass number reduces by $\Delta A=32=4 \times 8$ and proton number is reduces by $\Delta Z=10=2 \times 8-6$ So, $8 \alpha$-particles and $6 \beta^{-}$-particles must be emitted.
Mass number reduces by $\Delta A=32=4 \times 8$ and proton number is reduces by $\Delta Z=10=2 \times 8-6$ So, $8 \alpha$-particles and $6 \beta^{-}$-particles must be emitted.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.