# The three stable isotopes of neon:

Question:

The three stable isotopes of neon: ${ }_{10}^{20} \mathrm{Ne},{ }_{10}^{21} \mathrm{Ne}$ and ${ }_{10}^{22} \mathrm{Ne}$ have respective abundances of $90.51 \%, 0.27 \%$ and $9.22 \%$. The atomic masses of the three isotopes are $19.99 \mathrm{u}, 20.99 \mathrm{u}$ and $21.99 \mathrm{u}$, respectively. Obtain the average atomic mass of neon.

Solution:

Atomic mass of ${ }_{10}^{20} \mathrm{Ne}, m_{1}=19.99 \mathrm{u}$

Abundance of ${ }_{10}^{20} \mathrm{Ne}, \eta_{1}=90.51 \%$

Atomic mass of ${ }_{10}^{21} \mathrm{Ne}, m_{2}=20.99 \mathrm{u}$

Abundance of ${ }^{21} \mathrm{Ne} n_{2}=027 \%$

Atomic mass of ${ }_{10}^{22} \mathrm{Ne}, m_{3}=21.99 \mathrm{u}$

Abundance of ${ }_{10}^{22} \mathrm{Ne}, \eta_{3}=9.22 \%$

TheĀ average atomic mass of neon is given as:

$m=\frac{m_{1} \eta_{1}+m_{2} \eta_{2}+m_{3} \eta_{3}}{\eta_{1}+\eta_{2}+\eta_{3}}$

$=\frac{19.99 \times 90.51+20.99 \times 0.27+21.99 \times 9.22}{90.51+0.27+9.22}$
$=20.1771 \mathrm{u}$