1.

If the energy of a photon corresponding to a wavelength of 6000Å is 3.32xx10^(-19)J, the photon energy in J) for a wavelength of 4000Å will be :

Answer»


SOLUTION :`E_(1)=(HC)/(lambda_(1)),E_(2)=(hc)/(lambda_(2))`
`:.(E_(2))/(E_(1))=(lambda_(1))/(lambda_(2)) rArr E_(2)=(lambda_(1))/(lambda_(2))E_(1)`


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