Question

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. $81:19$

Answer 1

The correct option is A $19:81$

Let the percentage of $B^{10}$ atoms be $x$, then average atomic weight
$=\frac{10x+11(100-x)}{100}=10.81$
$\Rightarrow x=19$
$\therefore \frac{N_{B^{10}}}{N_{B^{11}}}=\frac{19}{81}$.