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I believe you would use the equation for the energy level of an H- atom, but use it for the initial and final n. In this case, you would be solving for n(initial). Set this difference equal to the equation for the energy of a photon, since it takes energy to change from one energy level to another.
The Rydberg equation is 1/lambda = R((1/n^2(initial))-(1/n^2(final)), where R is Rydberg's constant. You have the frequency, which means you can find the wavelength using c=(wavelength)(frequency). Once you have the wavelength, you know the wavelength and the final electron state, which is 4. Since you have both of these numbers, you can solve for n-initial, which after you do the work is 6.
The change in energy of the electron is equal to energy of the light emitted. Having the frequency of the light allows you to calculate its energy (E=hv). You also know the energy of n=4 because E(n)=-hR/n^2. Also note that ∆E=-E(emitted light)=Ef-Ei = E(n=?) - E(n=4). With this information, you can find the other n value. Dr. Lavelle wants you to avoid using the Rydberg equation.
Use the equation for the energy level of an H-atom. (1/lambda=R((1/n^2 (initial)-(1/n^2(final))). Since we have the frequency, the lambda can be calculated. As well plug in n=4 for the final energy level and solve for the initial energy level.
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