If we express the energy of a photon in `KeV` and the wavelength in angstroms , then energy of a photon can be calculated from the relation

Options

A. `E = 12.4 hv`
B. `E = 12.4 h//lambda`
C. `E = 12.4// lambda`
D. `E = hv`

Answer

C

Explanation

Energy of photon `E = (hc)/( lambda) (joules) = (hc)/( e lambda) ( eV)`
`rArr E( eV) = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(1.6 xx 10^(-19) xx lambda(Å)) = (12375)/(lambda (Å))`
`rArr E(keV) = (12.37)/( lambda (Å)) ~~ (12.4)/(lambda)`