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The longest wavelength doublet absorption is observed at `589` and `589.6 nm`. Caiculate the frequency of each transition and energy differebce between two excited states. |
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Answer» `lambda_(1)=589 nm=589xx10^(-9)m` `:. V_(1)=c/lambda_(1)=(3.0xx10^(8)ms^(-1))/(589xx10^(-9)m)=5.088xx10^(14) s^(-1)` `lambda_(2)=589.6 nm =589.6xx10^(-9) m` `:.v_(2)=c/lambda_(2)=(3.0xx10^(8)m s^(-1))/(589.6xx10^(-19)m)` `=5.088xx10^(14) s^(-1)` `DeltaE=E_(2)-E_(1)` `=h(v_(2)-v_(1))` `=(6.626xx10^(-34)J s)(5.093-5.088)xx10^(14) s^(-1)` `3.31xx10^(-22)J` |
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