Question

Boron has two isotopes \[B - 10\] and \[B - 10\] . If the average atomic mass of boron is $10.80u$ , what is the percentage of a heavier isotope?
A. 80
B. 20
C. 30
D. 70

Answer 1

Hint: Isotopes are varieties of a particular chemical element which differ in the number of neutrons, and consequently in the number of nucleons. All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom. Thus, the mass percentage of an isotope is calculated by determining the ratio of one isotope in a mixture of the two isotopes.

Complete step by step answer:
As per the question, the average atomic mass of boron is = $10.80u$.
Let the mass percentage of \[B - 10\]isotope of boron = $x\% $
Accordingly, the mass percentage of \[B - 10\] isotope of boron = $(100 - x)\% $
The average atomic mass is the mean of the sum of individual masses.
The mathematical formula of average atomic mass = $\dfrac{{{n_1}{M_1} + {n_2}{M_2}}}{{100}} = {M_{avg}}$
Where, ${n_1} = $ number of nucleons of \[B - 10\] isotope$ = 10$
${n_2} = $ number of nucleons of \[B - 11\] isotope$ = 11$
${M_1} = $ Atomic mass percentage of \[B - 10\] isotope
${M_2} = $ Atomic mass of percentage \[B - 11\] isotope
Average atomic mass = $\dfrac{{10x + 11(100 - x)}}{{100}} = 10.80$
Solving for $x$ , we have:
$ \Rightarrow 10x + 1100 - 11x = 1080$
$ \Rightarrow x = 20$
Thus, the mass percentage of \[B - 10\] isotope is found to be $20\% $ .
Now, in order to determine the mass percentage of the \[B - 11\] isotope, we substitute the value of $x$in $(100 - x)\% $. Thus, we have the mass percentage of \[B - 11\] isotope as $(100 - 20)\% = 80\% $

So, the correct answer is Option A .

Note:
The boron exists in nature in the form of two isotopes. One is ${}^{10}B$ and the other is ${}^{11}B$. The later one is a stable isotope and exists in a major amount in nature. The ${}^{10}B$ isotope consists of five protons, five electrons and five neutrons whereas the ${}^{11}B$ isotope consists of five protons, five electrons and six neutrons.