The atomic weight of Ne is 20.2. It is a mixt | vedclass

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(D) Let the abundance of $Ne^{20}$ be $x_1$ and $Ne^{22}$ be $x_2$. Given the average atomic mass is $20.2$. The formula is $20.2 = \frac{20x_1 + 22x_

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$10$

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(D) Let the abundance of $Ne^{20}$ be $x_1$ and $Ne^{22}$ be $x_2$. Given the average atomic mass is $20.2$. The formula is $20.2 = \frac{20x_1 + 22x_2}{x_1 + x_2}$. Assuming $x_1 + x_2 = 100$,we have $20.2 = 0.20x_1 + 0.22x_2$. Since $x_1 = 100 - x_2$,we substitute: $20.2 = 0.20(100 - x_2) + 0.22x_2$. $20.2 = 20 - 0.20x_2 + 0.22x_2$. $0.2 = 0.02x_2$. $x_2 = \frac{0.2}{0.02} = 10$. Thus,the abundance of the heavier isotope $Ne^{22}$ is $10\%$.

The atomic weight of $Ne$ is $20.2$. It is a mixture of $Ne^{20}$ and $Ne^{22}$. What is the relative abundance of the heavier isotope (in $\%$)?

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