If a bromine atom is available in the form of, say two isotopes, $_{35}^{79}Br$ ($49.7\%$) and $_{35}^{81}Br$ ($50.3\%$), calculate the average atomic mass of the bromine atom. [Pg. No. 156]

Isotopes of bromine:

$$_{35}^{79}Br=49.7\%$$

$$_{35}^{81}Br=50.3\%$$

Average atomic mass =

$$\frac{(\text{mass of isotope } 1 \times \% \text{ abundance}) + (\text{mass of isotope } 2 \times \% \text{ abundance})}{100}$$

Average atomic mass:

$$= \frac{(79 \times 49.7) + (81 \times 50.3)}{100}$$

$$79 \times 49.7 = 3926.3$$

$$81 \times 50.3 = 4074.3$$

Average atomic mass:

$$= \frac{3926.3 + 4074.3}{100}$$

$$= \frac{8000.6}{100}$$

$$= 80.006 \text{ u}$$

Therefore, the average atomic mass of bromine is 80.006 u, which is approximately 80 u.