Ahydrogen atom in the ground state absorbs a photon of wavelength 93.8 nm. what energy level does the electron reach? this excited atom then emits a photon of wavelength 1094.0 nm. what energy level does the electron fall to?

So you can use the Bohr formula En = -13.6*(1/n2^2 - 1/n1^2) eV where En is in electron volts. Energy of a photon is Ep = 1.24/wl eV where wl is wavelength in microns So 93.8 nm = 0.0938 um so Ep = 13.22 ev. Now atom radiates a photon wl' = 7.459 um loosing an energy E' of E' = 1.24/7.459 = 0.166 eV SO the atom has a net gain in energy of Ep - E' =13.053 eV TO find th eenergy level, set Ep - E' = En 13.053 = -13.6 (1/n2^2 - 1/n1^2) and set n1 = 1 = ground state. then solve for n2 -13.055/13.6 = 1/n2^2 - 1 --> 1 - 13.055/13.6 = 1/n2^2 0.040193 = 1/n2^2 ---> n2 = sqrt(1/0.040193) = 4.98 ~ 5 So n final is 5

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