Two stable isotopes of lithium

Solution:

(a) Mass of lithium isotope ${ }_{3}^{6} \mathrm{Li}, m_{1}=6.01512 \mathrm{u}$

Mass of lithium isotope ${ }_{3}^{7} \mathrm{Li}, m_{2}=7.01600 \mathrm{u}$

Abundance of ${ }_{3}^{6} \mathrm{Li}, \eta_{1}=7.5 \%$

Abundance of ${ }_{3}^{7} \mathrm{Li}, \eta_{2}=92.5 \%$

The atomic mass of lithium atom is given as:

$m=\frac{m_{1} \eta_{1}+m_{2} \eta_{2}}{\eta_{1}+\eta_{2}}$

$=\frac{6.01512 \times 7.5+7.01600 \times 92.5}{92.5+7.5}$

$=6.940934 \mathrm{u}$

(b) Mass of boron isotope ${ }_{5}^{10} \mathrm{~B}, m_{1}=10.01294 \mathrm{u}$

Mass of boron isotope ${ }_{5}^{11} \mathrm{~B}, m_{2}=11.00931 \mathrm{u}$

Abundance of ${ }_{5}^{10} \mathrm{~B}, \eta_{1}=x \%$

Abundance of ${ }_{5}^{11} \mathrm{~B}, \eta_{2}=(100-x) \%$

Atomic mass of boron, m = 10.811 u

The atomic mass of boron atom is given as:

$m=\frac{m_{1} \eta_{1}+m_{2} \eta_{2}}{\eta_{1}+\eta_{2}}$

$10.811=\frac{10.01294 \times x+11.00931 \times(100-x)}{x+100-x}$

$1081.11=10.01294 x+1100.931-11.00931 x$

$\therefore x=\frac{19.821}{0.99637}=19.89 \%$

And 100 − x = 80.11%

Hence, the abundance of ${ }_{5}^{10} \mathrm{~B}$ is $19.89 \%$ and that of ${ }_{5}^{11} \mathrm{~B}$ is $80.11 \%$.