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# finding the intital excited state

Submitted by ckoepsel on Thu, 11/22/2007 - 20:45

Consider an electron for a hydrogen atom in an excited state. The maximum wavelength of electromagnetic radiation that can completely remove (ionize) the electron  from the H atom is 1460 nm. What is the initial excited state for the electron (n = ?)?

I think i should use the equation: E = -2.178 X 10-18 J (Z2/n2)

but i am not sure how to find Z or n. Should Z be 1? Or do I need to use another equation because i need to be finding n?

OK, using Bohr's equation for the hydrogen atom (Z = 1) will give you the energy at a specific energy level (n). To determine the energy level from which the ionized electron left (ninitial), we'll use a slightly modified version of the Bohr equation.

We know that the change in energy is related to the wavelength of the photon emitted. I believe Planck's equation will do that nicely.
?E = hc
?

Now the energy needed for ionization will be equal to the energy of the final energy level minus the energy of the initial energy level.

Using Bohr's equation we get

?E = [-2.178 X 10-18 J (12/nfinal2)] - [-2.178 X 10-18 J (12/ninitial2)]

let's "do the tighten-up" and we get

?E = -2.178 X 10-18 J |  1    -     1    |
| nfinal2  ninitial2|

solve for ninitial and your done.

Check your text where they discuss the Bohr Model and electron energies.

Good luck.

[Hint: nfinal = ?, so 1/nfinal2 = ?]