Question

A decay chain of the nucleus $\_{92}^{238}U$ involves eight $\alpha$ -decays and six $\beta$ -decays. The final nucleus at the end of the process will be

• A. $Z=84;A=206$
• B. $Z=84;A=224$
• C. $Z=82;A=206$
• D. $Z=82;A=200$
• E. $Z=84;A=218$

Answer 1

Answer: (C)

(i) When $\alpha$ -particle is emitted from a radioactive nucleus then atomic number decreases by $2$ and atomic mass decreases by $4$ .
(ii) When $\beta$ -particle emitted from a radioactive nucleus then atomic number is increased by $1$ and atomic mass unaffected.
Here, atomic number $\left(Z\right)=92$
After emitting $8\alpha$ -particles and $6\beta$ -particles,
$Z=92-2\times 8+6=82$
and atomic mass, $A=238$
$\therefore A=238-4\times 8=206$