# Determination of initial excited state via luminescence

A one electron species initially in the some excited state ($n_i$) is irradiated with a light of wavelength 121 nm when the electron is promoted to a further higher orbit ($n_f$). In returning back to the ground state, it gives an emission spectrum containing 15 bright lines. Out of these 15 lines, 9 lines were found to have wavelengths smaller than 121 nm while 5 lines have wavelengths greater than 121 nm. What is the value of $n_i$?

I got the value of $n_f$ using this:

$$\frac{(n_f-1+1)(n_f-1)}{2}=15$$ $$\therefore n_f=6$$

We do not know the atomic number($Z$), nor $E_{ground state}$, nor $n_i$. What is the approach we should use to solve this problem?

• You haven't yet used the information that the light was 121 nm in wavelength. That is a key consideration. Commented Jul 17, 2016 at 15:22
• @CurtF. I tried using it by substituting in Wave number equation, but it has 2 unknowns - $n_i$ and $Z$. Commented Jul 17, 2016 at 16:16

Sometimes formulae are more hindrance than help, and perhaps there are some red herrings in your problem.

A one electron species initially in some excited state $n_i$ is irradiated with a light of wavelength 121 nm when the electron is promoted to a further higher orbit $n_f$. In returning back to the ground state, it gives an emission spectrum containing 15 bright lines. Out of these 15 lines, 9 lines were found to have wavelengths smaller than 121 nm while 5 lines have wavelengths greater than 121 nm. What is the value of $n_i$?

There being fifteen bright lines suggest that $n_f = 6$, since $15 = 5 + 4 + 3 + 2 + 1$, there being $5$ transitions from the $n = 6$ state to $n = 5,\,4,\,3,\,2,\,1$; $4$ transitions from the $n = 5$ state, and so on. This is the basis for the formula you've employed.

Long wavelengths correspond to low energy; we see that the transition from $n_i$ to $n = 6$ is the sixth lowest in energy. $E \propto n_i^{-2}-n_f^{-2}$, so we pull out our handy-dandy calculator and work out: $$6 \to 5: 0.012,\,5 \to 4: 0.023,\,6 \to 4: 0.035,\, 4 \to 3: 0.049,\, 5 \to 3: 0.071,\, 6 \to 3: 0.083.$$ Thus $n_i = 3$. We did not need to use the exact wavelength of light used for irradiation and we did not need to know the identity of the species, because this problem is only concerned with the relative energy differences between energy levels.

• I am sorry but could u explain the sixth lowest in energy part in more detail? Commented Jul 18, 2016 at 14:44
• Well, five lines (in the emission spectrum) have wavelengths greater than 121 nm, and the transition from $n_i$ to $n_f$ has a wavelength of 121 nm. Since the transition from $n_i$ to $n_f$ has the sixth longest wavelength, it is the sixth lowest in energy. Commented Jul 18, 2016 at 14:46
• So we actually have to calculate the energy of all 15 wavelengths like 6 to 5, 6 to 4 etc and then rank it and find the answer? Commented Jul 18, 2016 at 14:52
• Well, you can be smart about it and: (i) graph everything, in which case you simply need to measure distances with a ruler, or (ii) note that $E_{2\to 1} < E_{3\to 1}, E_{4\to 1}, ...$ and $E_{3\to 2} < E_{4\to 2}, E_{5\to 2}, ...$, and simply compare $E_{2\to 1}$ to $E_{3\to 2}$ and $E_{6\to 3}$. These are actually the only three energies we need compute, since we know that $E_{6\to 3}$ is at least greater than $E_{6\to 4},\,E_{6\to 5},\,E_{5\to 3},\,E_{5\to 4}$, and $E_{4\to 3}$. Commented Jul 18, 2016 at 15:00
• I postulated that those were the six lowest energy transitions based on the fact that $E \propto n^{-2}$, and I felt that my having obtained an answer that worked was enough justification for my postulate. I think the easiest way to see that my postulate is acceptable is to look up an energy level diagram for hydrogen and compare distances between energy levels. Commented Jul 18, 2016 at 15:08