What causes spectral lines?

Spectral lines are caused by the movement of the electrons between energy levels within an atom or ion.

Why do different elements have different spectral lines?

Different elements have different spectral lines because they have a different number of protons, therefore will have varying amount and arrangement of energy levels.

Why does neon have more spectral lines than hydrogen?

Neon has more spectral lines than hydrogen because neon has more protons and therefore energy levels, which will result in more transition of electrons between energy levels.

Why are spectral lines often referred to as atomic fingerprints?

Spectral lines often referred to as atomic fingerprints because each chemical element has a unique and constant set of spectral lines.

Spectral Lines: Types, Elements & Formula

Q: What are spectral lines?

Spectral lines are dark or colorful lines appearing in a uniform or continuous spectrum.

Types of Spectral Lines

In $17^{\text{th}}$ century, Isaac Newton passed a beam of light through a tiny slit and obtained the colors of the rainbow, as the white light split into its components. This is what we call a continuous spectrum, and it's displayed below.

Discrete Spectral Lines A continuous spectrum as all colors of rainbow StudySmarter Fig. 1 - A continuous spectrum made up of all wavelenghts of visible light.

Continuous spectrum contains all wavelengths of a light in a certain range.

Then, just over a hundred years later, Johann Balmer did a similar experiment, only using hydrogen. He passed current through a discharge tube: a tube filled with hydrogen gas held at low pressure, and as a result observed a bright pink glow. Now, passing this light through a prism or a diffraction grating, no longer resulted in a continuous spectrum, but rather a couple of colorful spectral lines as displayed in Figure $2$.

Spectral lines are dark or colorful lines appearing in a uniform or continuous spectrum.

Discrete Spectral Lines Four emission spectral lines of hydrogen - purple, blue, teal, and red. StudySmarter Fig. 2 - The visible emission spectral lines of hydrogen observed and analysed by Balmer.

There are two main types of discrete spectral lines: emission and absorption lines. We will define both of these now, however, the reason behind these explanations will become clearer later in the article.

Emission spectral line is caused by the transition of atoms from an excited state to a lower state.

An example of emission spectral lines are the hydrogen lines observed by Balmer. The reverse process of the emission is absorption, which causes absorption spectral lines to appear.

Absorption spectral line is caused by the transition of atoms from a ground state to an excited state.

The electrons absorb energy from their surroundings and therefore move to a higher state. Visually, it's represented by dark lines, over the continuous spectrum.

Discrete Spectral Lines An absorption spectrum of hydrogen, where four strips in purple, blue, teal, and red are missing from the continuous spectrum StudySmarter Fig. 3 - The visible absorption spectral lines of hydrogen.

If we compare the two spectrums (emission and absorption), it's obvious that the "missing" parts of the absorption spectrum perfectly align with the emission spectral lines, confirming that the two processes are inverse of one another.

Spectral Lines of Hydrogen

The spectral lines of Hydrogen were observed and analyzed by Balmer before the scientific community had a complete understanding about the structure of an atom. Only after Niels Bohr proposed his new atomic model in $1913$, the observations made by Balmer could be explained. Let's look at each of these events separately!

Balmer Equation

Balmer quickly figured out the exact wavelengths of each emission line in Figure $2$ ($410\, \mathrm{nm}$, $434\, \mathrm{nm}$, $486\, \mathrm{nm}$, and $656\, \mathrm{nm}$), and through trial and error came up with an equation connecting them all together. This relation is known as the Balmer equation or Balmer series and can be expressed mathematically as

$$\frac{1}{\lambda_n}=R \left (\frac{1}{2^2} - \frac{1}{n^2} \right ),$$

where $R$ is the Rydberg constant equal to $1.1\times10^7 \, \frac{1}{\mathrm{m}}$, and $n$ is an integer representing the principal quantum number.

The principal quantum number is a non-negative integer that indirectly portrays the size of the electron orbital and which discrete energy level an electron is located in.

The Balmer equation describes the wavelengths of transitions occurring between $n\geq3$ and $n=2$. In total, there are four transitions in the Balmer series, $n=6$ to $n=2$ corresponding to $410\, \mathrm{nm}$, $n=5$ to $n=2$ corresponding to $434\, \mathrm{nm}$, and so on.

A couple of years later, Johannes Rydberg generalized the Balmer equation to account for all transitions, rather than just the visible spectrum. The modified version is as follows

$$\frac{1}{\lambda_n}=R \left (\frac{1}{n_1^2} - \frac{1}{n_2^2} \right ),$$

where all terms remain the same as for the Balmer series, except $n_1$ and $n_2$ correspond to two different energy levels, and $n_2 > n_1$.

Considering Balmer derived this expression directly from observing hydrogen, it does not explain its emission spectrum fully, only the part corresponding to the visible spectrum. There are five other sets of series, named after the people who discovered them: Lyman, Paschen, Brackett, Pfund, and Humphreys series, where the Lyman series is for the ultraviolet region, and the rest are for the infrared region. The main difference between these series is which level the transition occurs from. As mentioned earlier, in the Balmer series, the electron jumps to or from $n=2$, meanwhile in the Lyman series, for instance, it happens from the ground state $n=1$.

Although the Balmer equation was and remains correct purely based on observations, physicists at the time of its discovery couldn't explain the meaning of these emission lines. That's where Niels Bohr and his model come in.

Bohr Model

Bohr, just like many physicists of the time, wasn't satisfied with the atomic theory. It was a known fact that an atom consists of positively charged protons and negatively charged electrons, however, why the atom didn't collapse remained a mystery. So did the spectral lines we discussed earlier, as nobody could explain the consistent emission of certain wavelengths of light by certain elements. Bohr found a way to explain both of these phenomena by suggesting a new model for the hydrogen model in particular.

What made his model stand out from the previous versions of the atomic model, was the application of the early quantum theory. He used Max Planck and Albert Einstein's ideas about the quantization of energy, stating that in the quantum model the emission and absorption of electromagnetic radiation occurs in discrete energy packets called photons.

A photon is a particle of electromagnetic radiation.

The energy of a photon can be calculated as follows

$$ E = h f$$

where, $E$ is the energy of the photon measured in joules ($\mathrm{J}$), $h$ is the Planck’s constant ($h=6.626\times10^{-34} \, \mathrm{J}\cdot\mathrm{s}$), and $f$ is the frequency of the light wave in $\mathrm{Hz}$.

Keeping that in mind, Bohr hypothesized that the potential energy of an atom is also quantized, meaning that an electron can only possess a certain amount of energy depending on its location relative to the nucleus. Depending on the element and the number of protons in its atom, there is a discrete number of energy levels at fixed distances from the nucleus. When an electron occupies a certain energy level, it doesn't radiate any energy. However, if an electron absorbs or emits a photon of very specific wavelength, it can move between these energy levels, as visualized in Figure $4$.

Discrete Spectral Lines The energy levels of an atom, where an electron is transitioning between the levels and emitting energy StudySmarter

Fig. 4 - Transitioning of electrons between energy levels results in the emission of energy.

The Bohr model not only explained the reasoning behind the spectral lines experimentally observed by Balmer, but also provided clarification to the Rydberg constant, which initially was just a fitting parameter.

Bohr managed to obtain the same value for the Rydberg constant as Balmer, only in his case it came as a result of other physics constants, rather than just a constant of proportionality. The full expression used to calculate the Rydberg constant is

$$ R_\infty = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c}$$

where the infinity symbol indicates a heavier atom, and the newly introduced constants are the rest mass of an electron $m_e$, the elementary charge $e$, and the permittivity of free space $\epsilon_0$.

Spectral Lines Formula

When an electron absorbs energy, it becomes excited and therefore moves to a higher energy level with a larger radius. The electron remains in this state for a short period of time and then returns to the lower energy level, simultaneously emitting a photon. This photon will have a very specific wavelength, as all orbits are a set distance from the nucleus and will require a constant amount of energy when moving between two specific energy levels.

We can calculate the amount of energy required to complete this transition by subtracting the energy of the two energy levels; it is also equal to the energy of the photon. Mathematically, that can be expressed as follows

$$ E_{\text{emitted photon}}=\left | \Delta E \right |=E_j-E_i$$

where $i$ and $j$ are the principal quantum numbers, referring to the specific energy levels ($n=i$ is a higher energy level and $n=j$ is a lower one).

The first five energy levels of a hydrogen atom are visible in figure below.

Discrete Spectral Lines The first five energy levels of a hydrogen atom starting at the ground state, with increasing energy as the energy level increases StudySmarter Fig. 5 - The first five energy levels of a hydrogen atom.

If the atom is in the $n=4$ level, how much photon energy will it require returning to its ground state? If a photon of energy $11 \, \mathrm{eV}$ was fired at this atom, what would happen?

Solution

We can calculate the amount of energy required by simply subtracting the energy of the two energy levels.

\begin{align} \left | \Delta E \right | & =E_j-E_i \\ E_{4\rightarrow1}&=E_4-E_1 \\ E_{4\rightarrow1}&= (-0.85 \, \mathrm{eV}) - (-13.6 \, \mathrm{eV}) = 12.8 \, \mathrm{eV}. \end{align}

In order for a photon to be emitted or absorbed, it has to have an identical energy to the difference between any two energy levels. We need to check all the potential options:

\begin{align} E_{2\rightarrow1}&= (-3.39 \, \mathrm{eV}) - (-13.6 \, \mathrm{eV}) = 10.2 \, \mathrm{eV} \\ E_{3\rightarrow1}&= (-1.51 \, \mathrm{eV}) - (-13.6 \, \mathrm{eV}) = 12.1 \, \mathrm{eV} \\ E_{5\rightarrow1}&= (-0.54 \, \mathrm{eV}) - (-13.6 \, \mathrm{eV}) = 13.1 \, \mathrm{eV}. \end{align}

None of these answers match the photon energy exactly; therefore it won't have any effect on this particular atom.

This energy formula can then be used to calculate the wavelength of any photon emitted or absorbed by the electron.

Spectral Lines Wavelength

To find the expression for the wavelength, we need to consider the Planck relation

$$ E = h f.$$

By reexpressing the frequency in terms of wavelength

$$ f =\frac{c}{\lambda}$$

we get the following equation:

\begin{align} E&=\frac{hc}{\lambda} \\ \lambda &=\frac{hc}{E}. \end{align}

Here, $E$ is the energy of a photon, $h$ is the Planck’s constant, $c$ is the speed of light ($c=3\times10^8 \, \frac{\mathrm{m}}{\mathrm{s}}$), and $\lambda$ is the wavelength of the light in meters ($\mathrm{m}$).

Finally, we can plug in the expression for the momentum of a photon

$$ p = \frac{E}{c} $$

and obtain

$$ \lambda = \frac{h}{p} $$

where $p$ is the momentum of a particle measured in kilogram meter per second $\left ( \frac{\mathrm{kg \cdot m}}{\mathrm{s}} \right )$. Let's look at an example problem for finding the wavelength of a photon!

Using the same diagram of the hydrogen atom as in the previous example (Figure $5$), calculate the wavelengths of the emitted photons for all transitions between $n=3$ and the ground state. Are any of these in the visible spectrum of light?

Solution

Considering we know the energy emitted by the photons, we can calculate the wavelength using the equation we derived earlier

$$ \lambda =\frac{hc}{E},$$

where $h$ and $c$ are constant values ($h=6.626\times10^{-34} \, \mathrm{J}\cdot\mathrm{s}$ and $c=3\times10^8 \, \frac{\mathrm{m}}{\mathrm{s}}$).

Firstly, we need to convert the Planck constant from $\mathrm{J}\cdot\mathrm{s}$ to $\mathrm{eV}$. $1 \, \mathrm{eV}$ is equal to $1.60\times10^{-19} \, \mathrm{J}$, so

\begin{align} h &= \left (6.626\times10^{-34} \, \cancel{\mathrm{J}}\cdot\mathrm{s}\right ) \times \frac{1 \, \mathrm{eV}}{1.60\times10^{-19} \, \cancel{\mathrm{J}}} \\h&= 4.14\times10^{-15}\, \mathrm{eV}. \end{align}

Now, we find the energy of the photons

$$ \left | \Delta E \right | =E_j-E_i $$

at each transition, and get the following values:

\begin{align} E_{2\rightarrow1}&= (-3.39 \, \mathrm{eV}) - (-13.6 \, \mathrm{eV}) = 10.2 \, \mathrm{eV}, \\ E_{3\rightarrow1}&= (-1.51 \, \mathrm{eV}) - (-13.6 \, \mathrm{eV}) = 12.1 \, \mathrm{eV}, \\ E_{3\rightarrow2}&= (-1.51 \, \mathrm{eV}) - (-3.39 \, \mathrm{eV}) = 1.88 \, \mathrm{eV}. \end{align}

Finally, plug in our obtained values and the constants into the wavelength equation and obtain the following results:

\begin{align} \lambda_{2\rightarrow1}&=\frac{\left (4.14\times10^{-15} \, \mathrm{eV} \cdot \mathrm{s} \right )\left (3\times10^8 \, \frac{\mathrm{m}}{\mathrm{s}}\right )}{\left ( 10.2 \, \mathrm{eV}\right )} \\ \lambda_{2\rightarrow1}&=1.22\times10^{-7} \, \mathrm{m} = 122 \, \mathrm{nm}, \end{align}

\begin{align} \lambda_{3\rightarrow1}&=\frac{\left (4.14\times10^{-15} \, \mathrm{eV} \cdot \mathrm{s} \right )\left (3\times10^8 \, \frac{\mathrm{m}}{\mathrm{s}}\right )}{\left ( 12.1 \, \mathrm{eV}\right )} \\ \lambda_{3\rightarrow1}&=1.03\times10^{-7} \, \mathrm{m} = 103 \, \mathrm{nm}, \end{align}

\begin{align} \lambda_{3\rightarrow2}&=\frac{\left (4.14\times10^{-15} \, \mathrm{eV} \cdot \mathrm{s} \right )\left (3\times10^8 \, \frac{\mathrm{m}}{\mathrm{s}}\right )}{\left ( 1.88 \, \mathrm{eV}\right )} \\ \lambda_{3\rightarrow2}&=6.61\times10^{-7} \, \mathrm{m} = 661 \, \mathrm{nm}. \end{align}

The ultraviolet region is roughly $100$ to $400 \, \mathrm{nm}$, which means $\lambda_{2\rightarrow1}$ and $\lambda_{3\rightarrow1}$ are both UV light. The final transition of $\lambda_{3\rightarrow2}$ corresponds with red light in the visible light spectrum.

Spectral Lines of Elements

Based on the number of protons in the nucleus of an atom, we can determine the number of electrons in a neutral atom. Considering different elements have a different number of protons, they will have varying energy levels. The varying amount and arrangement of energy levels in an atom lead to the distinctive properties of each chemical element. So, just like hydrogen, each element will have a unique set of spectral lines which will be identical every time. Spectral lines can be used to very accurately recognize the chemical composition of any material.

This is particularly useful when analyzing distant stars, as the spectral lines can not only tell us the chemical makeup of a star, but also the temperature and density of said elements. Examples of some celestial objects and their spectral lines can be seen in Figure $6$ below.

Discrete Spectral Lines Spectral lines of ten different stars and nebulae including our sun StudySmarter Fig. 6 - Spectral lines of various stars and nebulae.

Spectral Lines - Key takeaways

Emission spectral line is caused by the transition of atoms from an excited state to a lower state.
Absorption spectral line is caused by the transition of atoms from a ground state to an excited state.
Bohr hypothesized that the potential energy of an atom is quantized, meaning that an electron can only possess a certain amount of energy depending on its location relative to the nucleus.
An electron can only possess a certain amount of energy depending on its location relative to the nucleus.
The amount of energy required to complete a transition between two energy levels is found by subtracting the energy of the two energy levels; it is also equal to the energy of the photon.
To calculate the wavelength of any photon emitted or absorbed by the electron, we use $\lambda = \frac{h}{p}$.
Spectral lines can be used to very accurately recognize the chemical composition of any material, including many celestial objects.

References

Fig. 1 - Continuous spectrum (https://commons.wikimedia.org/wiki/File:Spectral_lines_continuous.png) by Sassospicco (https://it.wikipedia.org/wiki/Utente:Sassospicco) is licensed by Public Domain.
Fig. 2 - Visible lines in the hydrogen emission spectrum (https://commons.wikimedia.org/wiki/File:Emission_spectrum-H.png) by Merikanto is licensed by Public Domain.
Fig. 3 - Absorption lines (https://commons.wikimedia.org/wiki/File:Spectral-lines-absorption.svg) by Stkl is licensed by Public Domain.
Fig. 4 - Transition of electrons in Bohr model, StudySmarter Originals.
Fig. 5 - The first five energy levels of a hydrogen atom, StudySmarter Originals.
Fig. 6 - Spectra of the fixed stars and nebulae compared with the sun-spectrum and other spectra (https://commons.wikimedia.org/wiki/File:NIEdot362.jpg) by Dodd, Mead and Company (https://commons.wikimedia.org/wiki/Category:Dodd,_Mead_and_Company) is licensed by Public Domain.

Spectral Lines

StudySmarter Editorial Team

Types of Spectral Lines