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October 1, 2014

October 1, 2014

Posted by **Robin** on Tuesday, October 21, 2008 at 7:41pm.

The figure below represents part of the emission spectrum for a one-electron ion in the gas phase. All the lines result from electronic transmissions from excited states to the n = 3 state.

the graph is like this

A B

!! ! ! ! ! ! ! ! !

wavelength--->

pretend those are lines...

) What electronic transitions correspond to lines A and B?

A

n = 6 to n = 3

n = 4 to n = 2

n = 5 to n = 3

n = 4 to n = 1

B

n = 6 to n = 3

n = 4 to n = 2

n = 5 to n = 3

n = 4 to n = 1

- Chemistry AP -
**DrBob222**, Tuesday, October 21, 2008 at 8:22pmI can't tell from the drawing how the spectrum looks, particularly with regard to which is A and which is B.

But if you know the emitted energy starts at one of the higher levels and goes to n=3, there are only two answers that end with n = 3. Since E = h*frequency (also E = hc/wavelength so use the appropriate equation depending upon the units of your graph). The lines at the highest frequency (longest wavelength) are those that start at the nigher n. The lines at the lowest frequency (shortest wavelength) start at the lower n. Again, the way I have interpreted the drawing, the answer must be either 6 to 3 or 5 to 3.Please repost and clarify if I have misinterpreted.

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