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  4. The natural boron of atomic weight 10.81 is found to have two isotopes 10 B and 11 B. The ratio of abundance of isotopes of natural boron should be:

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Q. The natural boron of atomic weight 10.81 is found to have two isotopes ${ }^{10} B$ and ${ }^{11} B$. The ratio of abundance of isotopes of natural boron should be:

Nuclei

Solution:

Let abundance of $B ^{10}$ be $x \%$
$\therefore $ Abundance of $B ^{11}(100- x ) \%$
$10.81=\frac{(10 \times x)+11(100-x)}{100}$
or $1081=10 x+1100-11 x $
$\Rightarrow x=19$
$\therefore $ Ratio of abundance
$=\frac{19}{100-19}=\frac{19}{81}$