The relative atomic mass of boron (which exists in two isotopic forms

The relative atomic mass of boron (which exists in two isotopic forms ¹⁰B and ¹¹B) is 10·81. What will be the abundance of ¹⁰B and ¹¹B, respectively (consider a sample of 100 atoms) ?

19% and 81%

81% and 19%

38% and 62%

62% and 38%

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Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2024

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Let the abundance of ¹⁰B be x and the abundance of ¹¹B be y. The total abundance is 100%, so x + y = 1 (or 100 if using percentages). The average atomic mass is calculated as the weighted average of the isotopic masses: (mass of ¹⁰B * x) + (mass of ¹¹B * y) = Average Atomic Mass. Using approximate integer masses (10 for ¹⁰B and 11 for ¹¹B), we have: 10x + 11y = 10.81. Since x + y = 1, we can write y = 1 – x. Substituting this into the equation: 10x + 11(1 – x) = 10.81 => 10x + 11 – 11x = 10.81 => 11 – x = 10.81 => x = 11 – 10.81 = 0.19. So, the abundance of ¹⁰B is 0.19 or 19%. Then, y = 1 – x = 1 – 0.19 = 0.81 or 81%. The abundance of ¹⁰B and ¹¹B, respectively, is 19% and 81%.

The average atomic mass of an element is the weighted average of the masses of its isotopes, where the weights are their natural abundances. This principle is used to determine the relative abundance of isotopes if the average atomic mass and isotopic masses are known.

Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons, resulting in different atomic masses. Boron’s two stable isotopes are ¹⁰B (with 5 protons and 5 neutrons) and ¹¹B (with 5 protons and 6 neutrons). The natural abundance of isotopes is relatively constant for a given element across the Earth.

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