Shortest wavelength for Li2+ and Be3+ in same series

[StudentofPhysics]

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?

 

[JeffKoch]

Assuming an equation like (2) is valid, does the answer depend on the particular values of n1 and n2?

 

[StudentofPhysics]

ok, i misunderstood what nf should be.

I figured it out now by solving for it by the first given wavelength then applying that number as the nf for the second.

Thank you.