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The dissociation energy of H(2) is 430.5...

The dissociation energy of `H_(2)` is `430.53 kJ mol^(-1), `If `H_(2)` is of dissociated by illumination with radiation of wavelength `253.7 nm` , the fraction of the radiant energy which will be converted into ikinetic energy is given by

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The correct Answer is:
`9(8.68%)`

`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%`
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