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

Step 1: Assuming the percentage of the isotopes:

Let $\mathrm{x}\%$ is the percentage of the isotope having the atomic weight = $10.01$

So, $(100-\mathrm{x})\%$ would be the percentage of the isotope having the atomic weight =$11.01$

Step 2: Finding the percentage of each isotope:

Therefore,

$\mathrm{Average}\mathrm{atomic}\mathrm{weight}=\frac{\left[\right(\mathrm{atomic}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{isotope})\times (\mathrm{x})+(\mathrm{atomic}\mathrm{weight}\mathrm{of}\mathrm{second}\mathrm{isotope})\times (100-\mathrm{x}\left)\right]}{100}$

$10.81=\frac{10.01\left(\mathrm{x}\right)+11.01\times (100-\mathrm{x})}{100}$

So, value of x = 20

Hence,

Percentage of isotope having the atomic weight of 10.01 = 20

and, Percentage of isotope having the atomic weight of 11.01 = (100 - 20) = 80