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

A

Explanation

`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)`