The dissociation energy of  $H_2$  is $430.53KJ/mol.$  If $H_2$  is exposed to radiation energy of wavelength $253.7nm.$   What % radiant energy will be converted into kinetic energy

Dissociation in energy for H-H bond  $=\frac{430.53\times {10}^3}{6.023\times {10}^{23}}J/molecule$
 $=7.15\times {10}^{-19}J$
E/photon  $=\frac{hc}{\lambda }=7.83\times {10}^{-19}J$
 $E_{used}=(7.83-7.15)\times {10}^{-19}$
Energy convention to K.E $=0.68\times {10}^{-19}J$  
$\%$  energy converted into K.E = $\frac{0.68\times {10}^{-19}}{7.83\times {10}^{-19}}\times 100=8.68\%$