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

Answer

C

Explanation

(c) According to formula =, `E=(hc)/(lambda)(v=(c)/(lambda))`
Energy E=hv
`3.03xx10^(-19)=(hc)/(lambda)`
`lambda=(6.63xx10^(-34)xx3.0xx10^(8))/(3.03xx10^(-19))`
`=6.56xx10^(-7) m`
`=6.56xx10^(-7)xx10^(9)nm`
=`6.56xx10^(2) nm`
`=656 nm`