# Reason for same frequency of spectral line in infrared and visible region

I encountered this question while trying to understand the basics of spectroscopy and atomic structure.

Which electronic transition in Balmer series of hydrogen atom has same frequency as that of n=6 to n=4 transition in Helium positive ion?

I am aware that lines of Balmer series correspond to visible region of electromagnetic spectrum and that to Brackett series correspond to infrared region.

How is it that the frequency of a spectral line belonging to infrared region be same as that of line belonging to visible region knowing that frequency is characteristic property unique to an electromagnetic radiation?

• The Balmer series is said to be visible in case of Z=1 $\mathrm{n_1 = 2}$, The spectral lines are not in different regions. – Safdar Jul 29 at 11:13
• What do you mean by 'the spectral lines are not in different regions'? Spectral lines have different wavelengths and therefore should belong to different regions. – anushka verma Jul 29 at 11:40
• But that is if we consider them to be emitted from the same atom. Here you are asked to compare between two different atoms. – Safdar Jul 29 at 11:47

According to Bohr's theory of the structure of the hydrogen atom, the atom is defined as a system with a positively charged nucleus that has electrons revolving around it in discrete orbits of fixed size and energy and stated that the orbits were held stationary using electrostatic forces.

Upon calculation of the various parameters of the model, we get the energy of the $$\mathrm{n}^{\text{th}}$$ orbital in the hydrogen atom to be:

$$E=\frac{\mathrm{-13.6}}{\mathrm{n^2}} \mathrm{eV} \tag{1}$$

If we state that the atom to be considered were to be made mono-electronic and had an atomic number of Z, satisfying the conditions for the Bohr's model to work, we get that the energy of the $$\mathrm{n}^{\text{th}}$$ orbital to be:

$$E= -13.6 \cdot\frac{\mathrm{Z^2}}{\mathrm{n}^2} \mathrm{eV} \tag{2}$$

A spectral line is observed when the electron de-excites itself from a higher energy state to a lower energy state that is more stable. This is done by emitting a photon whose energy is the same as that of the difference of energy between the two levels. Applying $$\frac{\mathrm{hc}}{\lambda} =E$$, and $$\frac{\mathrm{c}}{\lambda} = f$$ we get

$$f = \frac{\mathrm{c}}{\lambda} = \mathrm{c} R_H\big[\mathrm{\frac{1}{n_1^2}-\frac{1}{n_2^2}\big]Z^2} \tag{3}$$

Here $$R_H$$ is the Rydberg constant whose value is $$\approx\pu{109,678 cm^-1}$$

In the question, you have been asked to figure out which line in the Balmer series($$\mathrm{n_1} = 2$$) of hydrogen would have the same frequency as that of the $$3^\text{rd}$$ line in the Brackett series($$\mathrm{n_1} = 4$$). I believe you would know how to solve the question and if not would now be able to via equating the frequencies of the two spectral lines.

Moving on to your doubt which is based on a misconception.

How is it that the frequency of a spectral line belonging to infrared region be same as that of line belonging to visible region knowing that frequency is characteristic property unique to an electromagnetic radiation?

The issue with this statement here is that you are assuming that the Balmer series is always visible for all atoms. This is only true for hydrogen as the the wavelengths provided for the different series are based on measurements taken on the hydrogen atom. That is, the Paschen series is infrared for hydrogen and this may change if the atom changes. As the atomic number increases, the frequency becomes greater as you can see from equation ($$3$$). This implies that at higher atomic numbers, there is a greater shift towards the UV region.

TL;DR: What you say is correct, frequency is a characteristic property unique for a certain electromagnetic radiation. However, the Balmer series is defined to be visible specifically for the hydrogen atom and not for any other mono-electronic atom/ion. As the atomic number becomes greater, a shift towards the UV region takes place.