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

`E = (hc)/(lambda) rArr (E_(1))/(E_(2)) = (lambda_(1))/(lambda_(2)) rArr (3.32 xx 10^(-19))/(E_(2)) = (4000)/( 6000)`
`E_(2) = 4.98 xx 10^(-19) J = 3.1 eV`