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  4. In nature, ratio of isotopes of Boron, 5 B 10 and 5 B 11, is (given that atomic weight of boron is 10.81 )

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Q. In nature, ratio of isotopes of Boron, ${ }_{5} B ^{10}$ and ${ }_{5} B ^{11}$, is (given that atomic weight of boron is $10.81$ )

Nuclei

Solution:

Atomic weight $=10.81$
Weighted mean is hence $=10.81$
$10.81=\frac{10 \times x+11 \times(100-x)}{100} $
$1081=10 x+1100-11 x $
$x=19$
Hence, $B^{10}: B^{11}=19: 81$