(a) Two stable isotopes of lithium $_{3}^{6} Li$ and $_{3}^{7} Li$ have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Boron has two stable isotopes, $_{5}^{10} B$ and $_{5}^{11} B$ . Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of $_{5}^{10} B$ and $_{5}^{11} B$ B.

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Solution

(a)The atomic mass is given by,

$m = \frac{7.5 \times 6.01512 + 92.5 \times 7.01600}{100} = 6.941 u$

(b) Mass of $_{5}^{10} B, m_{1} = 10.01294 u$

Mass of $_{5}^{11} B, m_{2} = 11.00931 u$

Let the abundance percentage of isotopes of boron, $_{5}^{10} B = x %$ and $_{5}^{11} B = (100 - x) %$

The atomic mass of Boron, $m = 10.811 u$

$m = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$

So, $x = 19.89 %; 100 - x = 80.11 %$

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(a)The atomic mass is given by, m=7.5×6.01512+92.5×7.01600100=6.941u(b) Mass of 105B,m1=10.01294u Mass of 115B,m2=11.00931uLet the abundance percentage of isotopes of boron, 105B=x% and 115B=(100−x)%The atomic mass of Boron, m=10.811um=m1x1+m2x2m1+m2So, x=19.89%; 100−x=80.11%