1.

During photosynthesis, chlorophyll absorbs light of wavelength 440 nm and emits light of wavelength 670 nm. What is the energy available for photosynthesis from the absorption-emission of a mole of photons?

Answer» `DeltaE=[(Nhc)/(lamda)]_("absorbed")-[(Nhc)/(lamda)]_("evolved")`
`=Nhc[(1)/(lamda_("absorbed"))-(1)/(lamda_("evolved"))]`
`=6.023xx10^(23)xx6.626xx10^(-34)xx3xx10^(8)[(1)/(440xx10^(-9))-(1)/(670xx10^(-9))]`
`=0.1197[2.272xx10^(6)-1.492xx10^(6)]`
`=0.0933xx10^(6)J//mol=93.3kJ//mol`


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