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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.

Answer

\(M=20.771 u\)
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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
\(M=\frac{90.51\times 19.99+0.27\times 20.99+9.22\times 21.99}{90.51+0.27+9.22}\)
\(M=20.771\;u\)
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