Identification of a ion from the highest energy line in its emission spectrum?

This seems to be #"Li"^(2+)#. That agrees with the fact that it's a metal ion.

Well, since it has one electron, it is a hydrogen-like atom. This series is

#"H", "He"^(+), "Li"^(2+), "Be"^(3+)#, etc.

The ground-state energy for a hydrogen-like atom in units of electron-volts (to make it simpler to read) is given by:

#E_n = -"13.61 eV" cdot Z^2/n^2#

where #n# is the energy level and #Z# is the atomic number.

The emission line corresponds to the highest energy relaxation, and therefore corresponds to the #n=oo->1# transition (the transition between the ground state and the electron having entered the atom).

In other words, this frequency is based on its ionization energy.

Therefore, and you may recognize this as a form of the Rydberg equation,

#DeltaE_(oo1) = E_(oo) - E_1#

#= -"13.61 eV" cdot Z^2(1/oo^2 - 1/1^2)#

#= "13.61 eV" cdot Z^2#

Now, this transition releases photons, each of energy #E_"photon" = DeltaE# (per electron transition). Therefore,

#DeltaE_(oo1) = E_"photon" = hnu#

#= "13.61 eV" cdot Z^2#

where #h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant and #nu# is the frequency in #"s"^(-1)#.

You were given the frequency of #2.961 xx 10^16 "s"^(-1)#, so the energy gap is:

#DeltaE_(oo1) = (6.626 xx 10^(-34) "J"cdot"s")(2.961 xx 10^16 "s"^(-1))#

#= 1.962 xx 10^(-17) "J"#

In the same units as we used earlier, we need to convert to #"eV"#.

#1.962 xx 10^(-17) cancel"J" xx ("1 eV")/(1.602 xx 10^(-19) cancel"J") = "122.47 eV"#

Therefore:

#"122.47 eV" = "13.61 eV" cdot Z^2#

Solving for the atomic number,

#bbZ = sqrt("122.47 eV" cdot 1/"13.61 eV")#

#~~ bb3#

And so this is #color(blue)("Li"^(2+))#.