Question

Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron.

Answer 1

Let the % of isotope with at. wt. 10.01 = x

\(\therefore \) % of isotope with at. wt. 11.01 = (100 - x)

Now, since,

At. wt. = \(\frac{x \times 10.01 + (100 - x)\times 11.01}{100}\)

\(10.81 = \frac{x \times 10.01 + (100 - x)\times 11.01}{100}\)

x = 20

Hence 

% of isotope with at. wt. 10.01 = 20

% of isotope with at. wt. 11.01 = 100 - 20 = 80