a ground state hydrogen atom absorbs a photon of light having a wavelength of 92.57 nm. it then gives off a photon having a wavelength of 388.8 nm. what is the final state of the hydrogen atom? values for physical constants can be found in the chempendix.

According to the Bohr model, electrons orbit the nucleus at constant energies. Higher energy levels are present in orbits that are farther from the nucleus. Light is released as energy when electrons drop to a lower energy level.

How do we fix?

1) With electron energy levels specified by the formula -k/n2, where k is a constant and n is the primary quantum number of the specific level the electron is occupying (usually, ground state, unless perturbed).

2) Assume that the electron in this problem is being moved between the three numbered energy levels; otherwise, you would need to use the additional data that E1 = -13.6 eV. However, since you were unaware of this, the energy levels of the electron must have been clearly defined.

3) The variations in the energies of those energy levels are represented by the wavelengths of the absorbed and emitted light. The n1 level has energy 1/n12, and the n2 level has energy 1/n22, assuming the ground state has reference energy of 1/12 (omit the minus sign for convenience throughout).

4) Now, simply translate the two energy changes into ratios: The ratio is reversed in relation to the corresponding energy level differences for 388.8/94.57nm since the energies are reciprocal to the wavelengths = 3.87142 = (1-1/n1^2)/(1/n1^2 - 1/n2^2 ) . All that remains is more algebraic labor; first, create the common denominators, and then eliminate all of the (n1 + n2) common denominators. Thus, n22*(n12 -1) / (n22 - n12) = 3.87142 is obtained.

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