1.

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

Answer» Correct Answer - `8.68%`
Energy required to break H-H bond `=(430.53xx10^(3))/(6.023xx10^(23))J//"molecule" =7.15xx10^(-19)J`
Energy of photon used for this purpose `=(hc)/(lambda)`
`=(6.625xx10^(-34)xx3.0xx10^(8))/(253.7xx10^(-9))=7.83xx10^(-19)J`
`therefore` Energy left after dissociation of bond `=(7.83-7.15)xx10^(-19)`
or " "Energy converted into `K.E.=0.68xx10^(-19)J`
`therefore %` of energy used in kinetic energy `=(0.68xx10^(-19))/(7.83xx10^(-19))xx100=8.68%`


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