Q. When a radioactive isotope $\ _{88}R^{228}$decay in series by the emission of $3\alpha$-particles and $\beta$-particle, the isotope finally formed is
A) $\ _{84}X^{228}$
B) $\ _{86}X^{222}$
C) $\ _{83}X^{216}$
D) $\ _{83}X^{215}$
Explanation
(c) Decrease in mass number due to emission of $3\alpha$-particles
and a $\beta - particle = 3 \times 4 = 12$
Decrease in charge number in the process
$= 3 \times 2 - 1 = 5$
$\therefore$ For the resulting element,
$A = 228 - 12 = 216$
$Z = 88 - 5 = 83$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.