Q. Atomic weight of boron is 10.81 and it has two isotopes$\ _{5}B^{10}\ \ and\ _{5}B^{11}$ . Then ratio of $\ _{5}B^{10}\ :\ \ _{5}B^{11}$ in nature would be
A) 19:81
B) 10:11
C) 15:16
D) 81:19
Explanation
(a) Let the percentage of $B^{10}$ atoms be $x$, then average atomic weight
$\ \ \ \ \ \ \ \ \ \ = \frac{10x\ + 11(100 - x)}{100} = 10.81$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ = 19$
$\therefore\ \ \ \ \ \ \ \frac{N_{B^{10}}}{N_{B^{11}}} = \frac{19}{81}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.