Question

Iodine molecule dissociates into atoms after absorbing light of $4500 Å$. If one quantum of radiation is absorbed by each molecule, calculate the kinetic energy of iodine atoms.

(Bond energy of $I_{2}=240 kJ mol ^{-1}, h=6.626 \times 10^{-34} J sec ^{-c}=3 \times 10^{8} ms ^{-1}$ ).

Answer 1

Energy of the photon, $E=\frac{h c}{\lambda}$

$=\frac{6 \cdot 626 \times 10^{-34} \times 3 \times 10^{8}}{4500 \times 10^{-8} \times 10^{-2}} J =4.42 \times 10^{-19} J$

Bond energy per iodine moleule $=\frac{240 \times 1000}{6 \cdot 02 \times 10^{23}} J =3 \cdot 98 \times 10^{-19} J$

The energy possessed by two iodine atoms $=(4 \cdot 42-3 \cdot 98) \times 10^{-19} J =0 \cdot 44 \times 10^{-19} J$ $\therefore$ The energy possessed by each iodine atom $=\frac{0 \cdot 44}{2} \times 10^{-19} J = 2 \cdot 2 \times 10^{- 2 0 } J$.