Home/Class 12/Physics/

## QuestionPhysicsClass 12

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

$$M=20.771 u$$
4.6
4.6

## Solution

Atomic mass of $${}^{20}{}_{10}Ne$$ $$m\underset{}{1}=19.99\;u$$
Abundance of the above $$=90.51 \%$$
Atomic mass of $${}^{21}{}_{10}Ne$$ $$m{}_{2}=20.99\;u$$
Abundance of the above isotope, $$n{}_{2}=0.27\%$$
Atomic mass of $${}^{22}{}_{10}\mathrm{N}\mathrm{e},\space m3=21.99\;\mathrm{u}$$
Abundance of the above isotope $$n{}_{3}=9.22\%$$
The average atomic mass of neon is given as
$$M=\frac{{m}_{1}{n}_{1}+m{}_{2}n{}_{2}+m{}_{3}n{}_{3}}{{n}_{1}+{n}_{2}+{n}_{3}}$$
Substituting we get
$$M=\frac{90.51\times 19.99+0.27\times 20.99+9.22\times 21.99}{90.51+0.27+9.22}$$
$$M=20.771\;u$$