The atomic mass of boron is 10.81 u and it has two isotopes, \( {}^{10}_{5}\mathrm{B} \) and \( {}^{11}_{5}\mathrm{B} \). The ratio \( {}^{10}_{5}\mathrm{B} : {}^{11}_{5}\mathrm{B} \) in nature would be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 19 : 81
The atomic mass of an element is the average mass of its isotopes. Each isotope counts according to how common it is, that is, its relative abundance. Take the isotope masses as 10 u and 11 u. Suppose a fraction \( x \) of the boron atoms is \( {}^{10}_{5}\mathrm{B} \). Then a fraction \( 1-x \) is \( {}^{11}_{5}\mathrm{B} \).
\( \begin{aligned} 10.81 &= 10x+11(1-x) \\ &= 11-x \end{aligned} \)
\( \Rightarrow\; x=0.19 \)
So 19% of the atoms are \( {}^{10}_{5}\mathrm{B} \) and 81% are \( {}^{11}_{5}\mathrm{B} \). Therefore, the ratio \( {}^{10}_{5}\mathrm{B} : {}^{11}_{5}\mathrm{B} \) is 19 : 81.
NCERT (2026-27):
Summary – “Most elements are mixtures of two or more isotopes. The atomic mass of an element is a weighted average of the masses of its isotopes and calculated in accordance to the relative abundances of the isotopes.”