Three of the strongest lines in the He + ion spectrum are observed at the following wavelength:

(1) 11.397 nm

(2) 54.030 nm

(3) 72.941 nm

Find the quantum numbers of the initial and final states for the transitions that give rise to these three lines. Do this by calculating, the wavelengths of lines that can originate from transitions involving any two of the four lowest levels. When a calculated wavelength matches an observed one, write down n-hi and n-lo for that line. Continue until you have assigned all three of the lines.

... ?

First let's calculate the energy involved in the transistions using Planck's formula E = hf. With given wavelength, we can replace f by c/lambda. Hence the formula will then become E = hc/lambda. Energy units will be in eV (electron volts). Just divide the energy units in J by electron charge to convert to eV.

(1) E1 = 108.79 eV

(2) E2 = 22.95 eV

(3) E3 = 17.00 eV

Since there are no external electric and magnetic fields, then we can ignore Stark and Zeeman Effects respectively. So, we only need to consider the n quantum number.

The energy of an electron in a nucleus of Z charge and at nth orbit is given by.

E_n = - 13.6*Z^2/n^2 eV

Let's calculate the energies of the electrons at different energy levels. For helium, Z = 2. Using the four lowest levels,

E_1 = - 54.40 eV

E_2 = - 13.60 eV

E_3 = - 6.04 eV

E_4 = - 3.40 eV

This time we need to calculate the transistion energies for every two pairs of energy levels. There are 4C2 = 6 possible transistion energies.

E_14 = E_4 - E_1 = 51.00 eV

E_13 = E_3 - E_1 = 48.36 eV

E_12 = E_2 - E_1 = 40.80 eV

E_24 = E_4 - E_2 = 10.20 eV

E_23 = E_3 - E_2 = 7.56 eV

E_34 = E_4 - E_3 = 2.64 eV

So it seems that no energy transition matches the change in energy for each jump. So the given wavelengths must be wrong.