# Calculate the wavelength of the photon absorbed to promote an electron from n=2 to n=4? n refers to the energy level E4=-1.3612x10^-19J E2=-5.445x10^-19J c=3.00x10^8 m/s h=6.6x10^-34 J/s

we can let delate E = Efinal - Einitial = Efinal(1/nf^2) - Einitial(1/ni^2)

or in our problem delta E = E(4) - E(2) = E4(1/4^2) - E2(1/2^2)

now we supply the values of E int he equation

delta E = [(-1.3612x10^-19J)(1/16)] - [(-5.445x10^-19)(1/4)]

delta E = 1.2762x10^-19 J

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we can let delate E = Efinal - Einitial = Efinal(1/nf^2) - Einitial(1/ni^2)

or in our problem delta E = E(4) - E(2) = E4(1/4^2) - E2(1/2^2)

now we supply the values of E int he equation

delta E = [(-1.3612x10^-19J)(1/16)] - [(-5.445x10^-19)(1/4)]

delta E = 1.2762x10^-19 J

now we have

delta E = hc/wavelength

therefore:

wavelength = hc/delta E

by supplying the given and our value of delta E, we now have:

delta E = (6.626x10^-34J-s)(3.00x10^8m/s)/(1.2762x10^-19 J)

delta E = 1.5576x10^-6 m = 1558 nm

hope this helps :)

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