Question

Mass numbers of two isotopes of an element differ by 2 units (A and A+2). Average atomic mass is 0.5 more than the lower mass number. What could be the ratio of the percentage abundance two isotopes respectively?

• A. 3 : 1
• B. 1 : 2
• C. 1 : 4
• D. 1 : 1

Answer 1

The correct option is A 3 : 1

According to given conditions,
Mass number of isotope of lower mass number be A.
Mass number of $2^{nd}$ isotope $=(A+2)$
Average atomic mass $=(A+0.5)$
If x is the percentage abundance of lower mass number isotope, then $(100-x)$ is the percentage abundance of $2^{nd}$ isotope.
Since $(A+0.5)=\frac{a_1{x}_1+a_2{x}_2}{(x_1+x_2)}=\frac{x\left(A\right)+(100-x)(A+2)}{100}$
$100A+50=xA+(100-x)(A+2)=xA+100A+200-xA-2x$
$2x=150;x=75$
$100-x=25$
so their ratio $=3:1$