The dissociation energy of `H_(2)` is `430.53kJ//mol`.If `H_(2)` is exposed to radiant energy of wavelength `253.7nm`.What `%` of radiant energy will be converted into K.E?

Answer

`9(8.68%)`

Explanation

`E_(H-H)` bond dissocation `=(430.53xx10^(3))/(6.023xx10^(23)) J` per molecule `=7.15xx10^(-19) J` per molecule
`E_("Photon")=(hc)/lambda=(6.626xx10^(-34)xx3.0xx10^(8))/(253.7xx10^(-9))=7.83xx10^(-19) J`
Energy converted into kinetic energy = Energy left after dissociation of bond.
`:.` Energy converted into `KE=(7.83-7.15)xx10^(19) J=0.68xx10^(-19) J`
`:. %` of energy converted into `KE=(0.68xx10^(-19))/(7.83xx10^(-19))xx100=8.68%`