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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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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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