Question

An atom (not a hydrogen atom) absorbs a photon whose associated wavelength is 300 nm and then immediately emits photon whose associated wavelength is 640 nm. How much net energy is absorbed by the atom in this process?

Answer 1

Answer:

The net energy is 2.196 eV

Explanation:

Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.

Using:

ΔE = h*c*(1/λ$_{1}$ - 1/λ$_{2}$)

where:

ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*$10^{-19}$ J.

h = 4.135*$10^{-15}$ eVs

c = 3*$10^{8}$ m/s

λ$_{1}$ = 300 nm = 300*$10^{-9}$ m

λ$_{2}$ = 640 nm = 640*$10^{-9}$ m

Thus:

ΔE = 4.135*$10^{-15}$ eVs*3*$10^{8}$ m/s*($\frac{1}{300*10^{-9}m } }-\frac{1}{640*10^{-9}m }$)

ΔE = 4.135*$10^{-15}$*3*$10^{8}$*1.77*$10^{6}$ eV = 2.196 eV