Q. When $\ _{88}Ra^{236}$ decays in a series by emission of $3\alpha$-particles and one $\beta$-particle, isotope $X$ formed is
A) $\ _{83}X^{224}$
B) $\ _{84}X^{218}$
C) $\ _{84}X^{220}$
D) $\ _{82}X^{223}$
Explanation
(a) $(3_{2}^{4}He + 1_{- 1}e^{0})$ result in decrease in mass number
$= 3 \times 4$ and
Decrease in charge number $= 3 \times 2 + 1( - 1) = 5$
$\therefore$ Isotope $(X)$ has mass number $= 236 - 12 = 224$
and charge number $= 88 - 5 = 83$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.