Structure of the Atom — Question 11

Let the percentage of ₈¹⁶X be a and that of ₈¹⁸X be 100 - a.

16.2u = $\dfrac{\text{16a}}{100}$ + $18\dfrac{\text{(100-a)}}{100}$

⇒ 1620 = 16a + 1800 – 18a

⇒ 1620 = 1800 – 2a

⇒ 2a = 1800 - 1620

⇒ a = $\dfrac{1800 - 1620}{2}$

⇒ a = $\dfrac{180}{2}$

⇒ a = 90

∴ Percentage of ₈¹⁶X = 90%

and

Percentage of ₈¹⁸X = 100 - a = 100 - 90 = 10%

Hence, the percentage of the isotope ₈¹⁶X in the sample is 90% and that of ₈¹⁸X is 10%