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[tex]_{1}[/tex] - 1/λ[tex]_{2}[/tex])
where:
ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*[tex]10^{-19}[/tex] J.
h = 4.135*[tex]10^{-15}[/tex] eVs
c = 3*[tex]10^{8}[/tex] m/s
λ[tex]_{1}[/tex] = 300 nm = 300*[tex]10^{-9}[/tex] m
λ[tex]_{2}[/tex] = 640 nm = 640*[tex]10^{-9}[/tex] m
Thus:
ΔE = 4.135*[tex]10^{-15}[/tex] eVs*3*[tex]10^{8}[/tex] m/s*([tex]\frac{1}{300*10^{-9}m } }-\frac{1}{640*10^{-9}m }[/tex])
ΔE = 4.135*[tex]10^{-15}[/tex]*3*[tex]10^{8}[/tex]*1.77*[tex]10^{6}[/tex] eV = 2.196 eV
