Q.
Boron has two stable isotopes 5B10 and 5B11. Their respective masses are 10.01294u and 11.00931u and the atomic mass of boron is 10.811u. The abundance of 5B10 and 5B11 are respectively nearing to
Let the percentage of 510B in sample be x.
Then, percentage of 511B is (100−x). So, using formula of average atomic masses of isotopes, 10.811=10010.01294×x+11.00931(100−x) ⇒1081.1=1100.931−0.99637x ⇒0.99637x=19.831 ∴x=0.9963719.831=19.3≈20%