(a) Two stable isotopes of lithium ${ }_{3}^{6} \mathrm{Li}$ and ${ }_{3}^{7} \mathrm{Li}$ have respective abundances of $7.5 \%$ and $92.5 \%$. These isotopes have masses $6.01512 \mathrm{u}$ and $7.01600 \mathrm{u}$, respectively. Find the atomic mass of lithium.
(b) Boron has two stable isotopes, ${ }_{5}^{10} \mathrm{~B}$ and ${ }_{5}^{11} \mathrm{~B}$. Their respective masses are $10.01294 \mathrm{u}$ and $11.00931 \mathrm{u}$, and the atomic mass of boron is $10.811 \mathrm{u}$. Find the abundances of ${ }_{5}^{10} \mathrm{~B}$ and ${ }_{5}^{11} \mathrm{~B}$.
(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 $80.11 \%$.