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

Answer

C

Explanation

The energy of photon,
`E = (hc)/(lambda) = 3.03 xx 10^(-19)`
or `lambda = (6.626 xx 10^(-34) xx 3xx 10^(8))/(3.03 xx 10^(-19))`
or `lambda = (19.878)/(3.03) xx 10^(-7) = 6.56 xx 10^(-7) m = 656` nm