If the energy of a photon is given as,
`E = 3.03 xx10^(-19) J`, then, the wavelength `(lambda)` of the photon is

Options

A. 6.53 nm
B. 653 nm
C. 65.6 nm
D. 0.656 nm

Answer

B

Explanation

`E_("photon") (ev) =(3*03xx10^(-19))/(1*6xx10^(-19))=1*9 ev`
`lambda_("photon")(Å)=12408/(1*9)=6530 Å =653 nm`