Question

A photon of 4000 $\stackrel{o}{A}$ is used to break the iodine molecule, then the % of energy converted to the K.E. of iodine atoms if bond dissociation energy of $I_2$ molecule is 246.5 kJ/mol is:

• A. 8%
• B. 12%
• C. 17%
• D. 25%

Answer 1

The correct option is C 17%

Energy of one photon $=\frac{12400}{4000}=3.1ev=3.11\times 6\times {10}^{-19}Joules$
Energy supplied by one mole photon in KJ/mole $=3.11\times 6\times {10}^{-19}\times 6\times {10}^{23}\times {10}^{-3}=297KJ/mole$
% of energy converted to K.E.$=\frac{297-246.5}{297}=17$