A photon with a wavelength of `4000A^@` 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

Options

A. `8%`
B. `12%`
C. `17%`
D. `25%`

Answer

C

Explanation

Energy of one photon `=(12400)/(4000) = 3.1 eV`
Energy supplied by one mole photon in KJ/mole `=3.1 xx 1.6 xx 10^(-19) xx 10^(23) xx 10^(-3) = 297 `KJ `"mol"^(-1)`
`therefore` of energy converted to
K.E. `=(297 - 246.5)/(297)= 17%`