Calculate and compare the energies of two radition one with a wavelength of `300 nm` and the other `600` nm

Explanation

Energy of radiation `:. E_(1) = (hc)/(lambda_(1)) = (6.626 xx 10^(-34) Js xx 3xx 10^(8) m s^(-1))/(300 xx 10^(-9) m ) = 6.626 xx 10^(-19) J`
and `E_(2) = (hc)/(lambda_(2)) = (6.626 xx 10^(-34) Js xx 3xx 10^(8) m s^(-1))/(600 xx 10^(-9) m ) = 3.313 xx 10^(-19) J`
The ratio of `E_(1)` and `E_(2)` is
`(E_(1))/(E_(2)) = (6.262 xx 10^(-19)J)/(3.313 xx 10^(-19)J) = 2`
`:. E_(1) = 2E_(2)`