Question

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

[Bond energy of $I_2=240kJmo{l}^{-I}$].

• A. $216\times {10}^{-20}kJ$
• B. $2.16\times {10}^{-19}J$
• C. $0.108\times {10}^{-19}J$
• D. $0.216\times {10}^{-19}J$

Answer 1

Answer: (D)

$0.216\times {10}^{-19}J$

Energy given to $I_2$ molecule $=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{4500\times {10}^{-10}}$$=4.417\times {10}^{-19}J$

Also, energy used for breaking up of $I_2$ molecule is $\frac{B.E}{N_A}=\frac{240\times {10}^3}{6.023\times {10}^{23}}=3.984\times {10}^{-19}J$

$\therefore$ Energy used in imparting kinetic energy to two $I$ atoms $=(4.417-3.984)\times {10}^{-19}J$

$\therefore$ KE/iodine atom $\frac{(4.417-3.984)}{2}\times {10}^{-19}$ $=0.216\times {10}^{-19}J$

Hence, the correct option is $D$