Question

(a) Two stable isotopes of lithium 63 Li and 73 Li have respectiveabundances of 7.5% and 92.5%. These isotopes have masses6.01512 u and 7.01600 u, respectively. Find the atomic massof lithium. (b) Boron has two stable isotopes, 105B and 115B. Their respectivemasses are 10.01294 u and 11.00931 u, and the atomic mass ofboron is 10.811 u. Find the abundances of 105B and 115 B.

Answer 1

a)

Given: The masses of lithium isotopes L 3 6 i and L 3 7 i are 6.01512 u and 7.01600 u and the abundances of L 3 6 i and L 3 7 i are 7.5% and 2.5% respectively.

The atomic mass of lithium atom is given as,

m= m 1 η 1 + m 2 η 2 η 1 + η 2

Where, The masses of lithium isotopes L 3 6 i and L 3 7 i are m 1 and m 2 and the abundances of L 3 6 i and L 3 7 i are η 1 and η 2 respectively.

By substituting the given values in the above formula, we get

m= 6.01512×7.5+7.016×92.5 92.5+7.5 = 694.0934 100 =6.940934 u

Thus, the atomic mass of Lithium is 6.940934 u or 6.941 u.

b)

Given: The mass of boron isotope 5 10 B is 10.01294 u, the mass of boron isotope 5 11 B is 11.00931 u and the atomic mass of boron is 10.811 u.

Let the abundance of boron isotope 5 10 B be x% and the abundance of boron isotope 5 11 Bis ( 100−x )%.

The atomic mass of boron atom is given as,

m= m 1 η 1 + m 2 η 2 η 1 + η 2

Where, The masses of boron isotopes 5 10 B and 5 11 B are m 1 and m 2 and the abundances of 5 10 B and 5 11 B are η 1 and η 2 respectively.

By substituting the given values in the above formula, we get

10.811= 10.01294×x+11.00931×( 100−x ) x+100−x 1081.11=10.01294x+1100.931−11.00931x 11.00931x−10.01294x=1100.931−1081.11 x=19.89%

The abundance of boron isotope 5 11 B is,

100−19.89=80.11%

Thus, the abundance of boron isotope 5 10 B is 19.89% and the abundance of boron isotope 5 11 B is 80.11%.