Question

Atomic weight of Ne is 20.2. Ne is mixture of $N{e}^{20}$ and $N{e}^{22}$ , relative abundance of heavier isotopes is:

• A. 90
• B. 20
• C. 40
• D. 10

Answer 1

The correct option is D 10

Solution:

Let relative percentage of $N{e}^{20}=X_1$

Let relative percentage of $N{e}^{22}=X_2$

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

$20.2=\frac{20X_1+22X_2}{X_1+X_2}$

On solving we get:

$\frac{X_1}{X_2}=\frac{9}{1}$

Therefore % of $X_1=\frac{9}{10}\times 100$

$=90$% which is lighter isotope

Now % of $X_2=100-90=10$% which is heavier isotope

Hence option D is correct