If the energy of a photon corresponding to a wavelength of `6000 Å is 3.32 xx 10^(-19) J` , the photon energy for a wavelength of `4000 Å` will be

Options

A. `1.4 eV`
B. `4.9 eV`
C. `3.1 eV`
D. `1.6 eV`

Answer

C

Explanation

Since, `E = (hc)/lamda implies E_1/E_2 = lamda_2/lamda_1`
`implies E_2 = (6000)/(4000) xx3.32 xx10^(-19) =3.1eV`