Question

4000$\stackrel{\circ }{A}$ photon is used to break iodine molecule, then the % of energy converted to K.E of Iodine atoms, if the bond dissociation energy of $I_2$ molecule is $246.5kJmo{l}^{-1}$

• A. 8%
• B. 12%
• C. 17%
• D. 25%

Answer 1

The correct option is C 17%

Energy of photo = $\frac{hc}{\lambda }=\frac{6.625\times {10}^{-34}\times 3\times {10}^8\times 6.023\times {10}^{28}}{4000\times {10}^{-10}}=299.26KJ/mole$
% of energy converted into K.E = $\frac{299-246.5}{299}\times 100=17\%approx$