Naturally occuring Boron consists of two isotopes having atomic masses 10.01 and 11.01 respectively.
Naturally occuring Boron consists of two isotopes having atomic masses 10.01 and 11.01 respectively. Calculate the percentage of both the isotopes in natural Boron (Atomic mass of natural Boron = 10.81) (a) 20% and 80%(b) 80% and 20%(c) 25% and 75%(d) 75% and 25%
Answers
The answer is (c) 25% and 75%. This is because you can calculate the percentage of each isotope by taking the ratio of their atomic masses and the total mass of natural Boron. 10.01/10.81 = 0.9254 11.01/10.81 = 1.0746 Therefore, the percentage of the first isotope (10.01) is 0.9254 x 100 = 92.54%. The percentage of the second isotope (11.01) is 1.0746 x 100 = 107.46%. The sum of these two percentages gives you the total percentage for both isotopes, which is 200%. To get the individual percentages, you need to divide each percentage by 2, which gives you (a) 25% and 75%, as the answer.
