Q. The end product of the decay of $\ _{90}Th^{232}\ \ is\ _{82}Pb^{208}$. The number of $\alpha\ and\ \beta$-particles emitted are respectively

A) 6,4
B) 3,3
C) 4,6
D) 6,0
Explanation

(a) Let $\alpha$-particles emitted are $x$ and $\beta - particles\ $ emitted are y
$\ _{90}Th^{232}\ \rightarrow \ _{82}Pb^{208}\ \ + \ x\ _{2}He^{4}\ + \ y\ \ _{- 1}e^{0}$
On comparing atomic number
$\ \ \ \ \ \ 90 = 82 + 2x - y$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x - y = 8\ \ \ \ \ \ \ \ \ \ \ $ …(i)
On comparing mass number
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 232 = 208\ + \ 4x$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ x = 6\ \ \ \ \ \ \ \ $
Putting the value of $x$ in Eq.(i), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = 4$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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