The energy of a photon is given as, `Delta E`/atom `= 3.03 xx 10^(-19)J "atom"^(-1)` then, the wavelength `(lambda)` of the photon is

Options

A. `65.6 nm`
B. `656 nm`
C. `0.656 nm`
D. `6.56 nm`

Answer

B

Explanation

`Delta E =(hc)/(lambda)` or `lambda =(hc)/(Delta E)`
`= (6.62 xx 10^(-34) xx 3 xx 10^(8))/(3.08 xx 10^(-19)) = 656 nm`.