Hello chemists!
I have used the following equation E = R (1/ni2 - 1/nf2) followed by wavelength = ch/E to calculate the visible wavelengths.
I know that at n=3 the wavelength will be 663 nm, which is red and visible.
I know that at n=4 the wavelength will be 486 nm, which is violet and visible.
I know that at n=5 the wavelength will be 434 nm, which is violet as well and visible.
I know that at n=6 the wavelength will be 412 nm, which is still visible
I know that at n=7 the wavelength will be 397 nm, which is no longer visible.
HOWEVER, I now realize that this is the case for any element in the Balmer series, and am having a blonde moment as to how I can relate this to Li and then to Li2+
Can anyone steer me on the right direction as to how I would then calculate the longest and shortest wavelengths of Li and Li2+ or would I just use the nm calculations from above in relations to it.
For example the shortest wavelength seen in the visible spectrum of Li 2+ would be 412 nm, and the longest wavelength seen in the visible spectrum of Li2+ would be 663 nm.
Any direction or explanation would be so welcomed!
Thank you so much
From: ..somewhat lost in chemistry....