Question

The atomic weight of Ne is 20.2. Ne is mixture of $N{e}^{20}$ and $N{e}^{22}$. The relative abundance of the heavier isotope is:

• A. $90\%$
• B. $20\%$
• C. $40\%$
• D. $10\%$

Answer 1

The correct option is D $10\%$

Let the relative abundance % of $Ne-20=x_1$

So, ​relative abundance % of $Ne-22=x_2$

Now we know that atomic mass = sum of relative abundance in atomic mass of each isotope / 100%

$20.2=\frac{20x_1+22x_2}{x_1+x_2}$

$20.2x_1+20.2x_2=20x_1+22x_2$

$0.2x_1=1.8x_2$

${\frac{x_1}{x}}_2=\frac{1.8}{0.2}=\frac{9}{1}$ Relative abundance

$x_1\%=\frac{9}{10}\times 100=90\%$ lighter isotopes

$x_1\%=\frac{1}{10}\times 100=10\%$ = heavier isotope

Hence, the correct option is $D$