If bromine is available in form of two isotopes i.e. ${\mathit{Br}}_{35}^{79}$ (49.7%) and ${\mathit{Br}}_{35}^{81}$ (50.3%) then the atomic mass of the bromine is:

If the bromine is available in the form of two isotopes i.e. ${\mathit{Br}}_{35}^{79}$ (49.7%) and ${\mathit{Br}}_{35}^{81}$  (50.3%) then the atomic mass of the bromine 80 amu.

We know that the upper digit denotes the mass of the atoms.

The mass of ${\mathit{Br}}_{35}^{79}$ is 79 amu and its abundance is 49.7% 

The mass of ${\mathit{Br}}_{35}^{81}$ is 81 amu and its abundance is 50.3%
So, the average mass of the bromine =  $79\times \frac{49.7}{100}+ 81\times \frac{50.3}{100}$
⇒ $\frac{3926.3}{100} + \frac{4074.3}{100}$

 $= 39.26+40.74$ $\Rightarrow$ 80 amu