An electron is excited from the n=1 ground state to the n=3 state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true.

A. It takes more energy to ionize (completely remove) the electron from n=3 than from the ground state.

B. The electron is farther from the nucleus on average in the n=3 state than the n=1 state.

C. The wavelength of light emmitted if the electron drops from n=3 to n=2 will be shorter than the wavelength of light emitted if the electron falls from n=3 to n=1.

D. The wavelength of light emitted when the electron returns to the ground state from n=3 will be the same as the wavelength of light absorbed to go from n=1 to n=3.

E. For n=3, the electron is in the first excited state.

I got really confused in this chapter so a good explanation would help me very much. I appreciate you taking your time to help.

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  1. hint: a, c, e are false.

    note on c. energy is proportional to (1/2^2-1/3^2) or proportional to 1/4-1/9
    which is about .16
    in the second case, energy prop to (1/1-1/9)=about .9
    so energy is higher in the second case, so wavelength is shorter.

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  2. You need to re-read the chapter because all I can do is repeat much of what your text has. But here is a summary.
    You have 1 proton in the nucleus. The single electron usually occupies the first shell which we will call n = 1. If energy is added, and there are various ways to do that, the electron can absorb that energy and move to n = 2 or n = 3 or n = 4 or n = infinity (where is is the ionized H atom or H^+). When it is in n = 1 it is said to be in the ground state. When in any other orbit (shell), it is an excited H atom and when the electron has been completely removed (that's when the e is at infinity from the proton) it is not only excited but is said to be ionized as well. within a few nanoseconds of becoming an excited atom the electron can return to the ground state by various routes. If it starts as an excited atom with the e in n = 4, it can go from 4 to 3 to 2 to 1 in 3 steps; it can go from 4 to 2 to 1 or it can go from 4 to 1. Each of those transitions emits energy and the amount of energy emitted by going from 2 to 1 is exactly the same energy it took to get it from 1 to 2 in the first place. Likewise, moving from n = 5 to n = 1 requires the same energy as getting from 1 to 5. The wavelength of the light is proportional to the energy. More energy means shorter wavelengths.
    Hope this helps. I'll be glad to answer follow up questions but this short summary should allow you to answer the question above.

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  3. This helped a lot! Thank you. The following are my answers and reasons.

    A. False because n=3 has energy so only need to add a bit. N=1 or ground state requires more energy to ionize because it is smaller (the beginning)

    B. True because the electron that occupies the first she'll is n=1 so then when you add energy it gets farther from the nucleus.

    C. False because the wavelengths of light are proportional to energy and more energy means shorter wavelengths. So going 1-3 takes more energy so then the wavelength would be shorter.

    D. True because it takes the same amount of energy to go back (3-1) as it does to get to 1-3.

    E. False because n=1 is the ground state so any other orbit is exited. n=2 is the nest one after n=1

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  4. ok?

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