Question

Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11 .0 1 . The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron.

Answer 1

Suppose percentage of one isotope = x

.'. Percentage of second isotope = 100 - x

Average atomic weight

percentage abundance of one isotope x Its atomic  weight + percentage abundance of second isotope x Its atomic weight/Percentage abundance of one isotope + percentage abundance of  + second isotope.

Hence, percentage of isotope with atomic weight 10.01 = 20%

.'. Percentage of isotope with atomic weight 11.01 = 100 - x = 100- 20 =  80%