Atomic Structure and Chemical Bonding — Question 1

Given,

49.7% of ⁷⁹₃₅Br and

50.3% of ⁸¹₃₅Br

These isotopes are in the ratio = $\dfrac{49.7}{50.3}$ = $\dfrac{497}{503}$

Average atomic mass = $\dfrac{(79 \times 497)+(81 \times 503)}{1000}$

$= \dfrac{39263+(40743)}{1000} \\[1em] = \dfrac{80006}{1000} \\[1em] = 80.006 \text{ amu} \\[1em] \approx 80 \text{ amu}$

∴ Average atomic mass of bromine = 80 amu