Question

Hydrogen when subjected to photon disocation, yieds one normal atom and atom possessing `1.97 eV` more energy than normal atom .The bond dissociation energy of hydrogen molecule into normal atom is `103 `kcal `mol^(-1)` . Campate the wavelength of effective photon for photo dissociation of hydrogen molecule in the given case

Answer 1

`H_(2) rarr H + H^(o+)`
where H is atom and `H^(o+)` is excited H atom so the energy required to dissociate `H_(2)` in this marmer will be greater then the bond energy of `H_(2)` + extra energy of `E_(("absorbed"))` = dissociate energy of `H_(2)` + extra exrtra energy of excuted atom
Energy required to dissociation in normal mather
`= 103 xx 10^(3) cal`
`= (103 xx 10^(3) xx (4.18))/(6 xx 10^(23))`
`= 7.716 xx 10^(-19) J atom ^(-1)`
The extra energy possessed by the excited atomk in
`1.97 eV= 1.97 xx 1.6 xx 10^(-19) + 3.152 xx 10^(-19)`
`E_(("abesorbed")) = 7.17 xx 10^(-19) + 3.152 xx 10^(-19) = 1.080668 xx 10^(-19) J`
Now calculate the wavelength of photon corresponding this energy
`lambda= (hc)/(E ) = (6.63 xx 10^(-34) xx 3xx 1100^(8))/(1.0868 xx 10^(-18))`
`= 1.830 xx 10^(-7)m = 1830 Å`