Question

Naturally occurring boron has two isotopes whose atomic weights are 10.01 (I) and 11.01 (II). Atomic weight of natural boron is 10.81 The percentage of isotopes (I) and (II) respectively are :

• A. 20 and 80
• B. 10 and 20
• C. 15 and 75
• D. 30 and 70

Answer 1

The correct option is A 20 and 80

$10.81=\frac{10.01x_1+11.01x_2}{x_1+x_2}$
$10.81x_1+10.81x_2=10.01x_1+11.01x_2$
$10.81x_1-10.01x_1=11.01x_2-10.81x_2$
$0.80x_1=0.20x_2$
$\frac{x_1}{x_2}=\frac{20}{80}$

$x_1=20$

$x_2=80$