Calculate and compare the energies of two radiations one with wavelength 800 pm and the other with wavelength 400 pm.

Explanation

Energy of photon,
`E = hv = (hc)/(lambda)`
Here `" " c = 3.0 xx 10^8 ms^(-1)`
In first case , `" " lambda = 800 nm = 800 xx 10^(-9) m`
`:. " " E_1 = ((6.626 xx 10^(-34) Js)xx (3 xx 10^8 ms^(-1)))/(800 xx 10^(-9) m)`
`= 2.48 xx 10^(18) J`
In second case,
`lambda = 400 nm = 400 xx 10^(-9) m`
`E_2 = ((6.626 xx 10^(-34) Js) xx (3 xx 10^8 ms^(-1)))/(400 xx 10^(-9) m)`
`= 4.91 xx 10^(-19) J`
Ratio of energy of first and secnd radiations,
`(E_1)/(E_2) = (2.48 xx 10^(-19) J)/(4.97 xx 10^(-19) J) = 1/2`
`:. E_1 : E_2 = 1:2 "or" E_2 = 2E_1`
Thus, energy of the radiation with wavelength 400 nm is twice that of the radiation of wavelength 800 nm.