Question

A photon with a wavelength 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.

• A. $8\%$
• B. $12\%$
• C. $17\%$
• D. $25\%$

Answer 1

The correct option is C $17\%$

Here the energy of one photon is $\frac{12400}{4000}$

= 3.1 ev

Energy supplied by photon in Kj/mol = $3.1\times 1.6\times {10}^{-19}$ $\times 6\times {10}^{23}\times {10}^{-3}$

= 297 kJmol${}^{-1}$

So the percent of energy converted = 17%