- #1
StudentofPhysics
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1. Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of lines in the lithium spectrum, the shortest wavelength is 1222.2 nm. For the same series of lines in the beryllium spectrum, what is the shortest wavelength?
2. 1/ lamba = 1.097 x 10^7 (z^2) (1/nf^2 - 1/ni^2)
3. My only question is what I use for the n's? Since its the smallest I assume ni = 0. What about nf?
2. 1/ lamba = 1.097 x 10^7 (z^2) (1/nf^2 - 1/ni^2)
3. My only question is what I use for the n's? Since its the smallest I assume ni = 0. What about nf?