In this explainer, we will learn how to calculate the energy of the photon that is absorbed or released when an electron transitions from one atomic energy level to another. Show Recall the general structure of an atom, as shown by the diagram of a hydrogen atom below. The pink dot at the center, the nucleus, is made up of protons and neutrons. The blue dot around it is an electron. Electrons can occupy several different regions, called shells, as they orbit. These shells are numbered from the innermost to outermost, as seen in the diagram below. Shells are also called electron energy levels. This is because the shell an electron occupies determines how much energy it has. The least energetic electrons are in the innermost shell, shell 1, which is also called the ground state. As an electron moves further out from the center, increasing its shell number, it has more energy. An electron cannot be in between shells or energy levels. Shells and energy levels are only whole integers, the number of which is denoted by the principal quantum number, 𝑛. The lowest possible value for 𝑛 is 1, which is the innermost shell or energy level. The figure below demonstrates correct and incorrect electron placements. Each electron energy level corresponds to a specific amount of energy, and electrons must be in an 𝑛 energy level. This means each electron in a particular energy level has the same set amount of energy. For example, if we denote the first energy level as 𝐸, then all the electrons in the first energy level will have energy 𝐸. If the second energy level is 𝐸, then all electrons there will have energy 𝐸, and so on for all the energy levels. This is demonstrated in the figure below. Electron energy is very small, so electron volts (eV) are used instead of joules. The conversion between the two is 1=1.6×10.eVJ Every element has its own amount of energy for its shells, but this explainer will exclusively use the energy levels of hydrogen. The diagram below shows the energy levels, 𝑛, and corresponding electron binding energies of a hydrogen atom. The energy level is on the right, with the lowest possible level being the ground state of 1. We can see a large gap in between the first energy level and the others. This means that most of the energy of an electron comes from being in the first shell, with each subsequent level giving less and less energy. On the left axis, the energy is shown in negative electron volts. The reason it is negative is because it is not strictly showing how much energy the electron has, but rather how much energy it takes to remove it from that particular shell, the binding energy. This unusual convention is used because there is a ceiling to the amount of energy an electron can have, shown as an energy of 0 at the top of the graph. If the electron exceeds this level of energy (usually by absorbing the energy of a photon), it escapes the atom entirely, as seen in the diagram below. Electron in higher energy levels are more weakly bound to the atom, as they require less energy to transition out of the atom. Electrons can be expelled from an atom if they have a high enough energy, but electron movement occurs within the atom as well, transitioning between different energy levels. To transition upward in energy levels, an electron must gain energy, which it usually gains from absorbing a photon. To transition downward, an electron must lose energy, causing it to emit a photon. The exact amount of energy lost or gained depends on what levels the electron transitions between. For example, in the diagram below, an electron drops from the 2nd energy level, 𝐸, to the ground state 𝐸. In doing so, we see this energy escaping as a photon. Whenever an electron transitions downward, it releases a photon with the same energy as the difference in the energy levels. The exact difference in energy is given by the equation Δ𝐸=𝐸−𝐸. According to the graph above, the ground state, 𝐸, is −13.6 eV, and the 2nd energy level is −3.4 eV. Substituting in these values gives Δ𝐸=(−3.4)−(−13.6)−3.4−(−13.6)=10.2,eVeVeVeVeV so the energy of this photon is 10.2 eV. Let’s look at an example. The diagram shows the transition of an electron in a hydrogen atom from 𝑛=4 to 𝑛=2, emitting a photon as it does so.
AnswerPart 1 The energy of the photon is the difference between the energy levels 4, 𝐸, and 2, 𝐸. This difference can be expressed using the equation Δ𝐸=𝐸−𝐸. We see from the graph that 𝐸 is −0.85 eV and that 𝐸 is −3.4 eV. Substituting these values into the equation gives Δ𝐸=(−0.85)−(−3.4)(−0.85)−(−3.4)=2.55.eVeVeVeVeV So, the energy of the photon to two decimal places is 2.55 eV. Part 2 We know the energy of the photon in electron volts from part 1, 2.55 eV. We now have to convert this into joules. Recall that the conversion ratio between electron volts and joules is 1=1.6×10,eVJ which means for every 1 eV, there are 1.6×10 J: 1.6×101.JeV Multiplying this relation by the energy the photon has, 2.55 eV, cancels the units of electron volts and gives us an answer in joules as follows: 1.6×101×2.55=4.08×10.JeVeVJ The energy in terms of joules to two decimal places is 4.08×10 J. When an electron transitions downward or upward, it emits or absorbs a photon respectively. Recall that this photon energy, 𝐸, can be expressed using the equation 𝐸=ℎ𝑓, where 𝑓 is the photon frequency and ℎ is Planck’s constant, 6.63×10 J⋅s. This energy can be related to the energy difference in electron energy levels in order to solve for the photon’s frequency. Let’s look at the example of an electron transition in a hydrogen atom from energy level 2, 𝐸, to the ground state 𝐸. The difference in energy in this case was determined to be 10.2 eV. This difference in energy between the electron energy levels is equal to the energy of the emitted photon: 𝐸=ℎ𝑓,𝐸=Δ𝐸=𝐸−𝐸. This means that by relating the energies together like this, we can determine the frequency of the photon. To do this, let’s put the energy of a photon equation in terms of its frequency, 𝑓. We can isolate the frequency in this equation by dividing both sides by Planck’s constant, ℎ, as follows: 𝐸=ℎ𝑓𝐸ℎ=ℎ𝑓ℎ, which causes ℎ to cancel on the right side, leaving behind only frequency: 𝐸ℎ=𝑓. Before going further, we need to convert Planck’s constant, ℎ, to electron volts so that the units match those given by the energy difference in the electron energy levels. There are 1.6×10 J in 1 eV: 11.6×10.eVJ So, multiplying this relation by Planck’s constant will cancel the units of joules, leaving behind electron voltseconds: 11.6×10×6.63×10⋅=4.144×10⋅.eVJJseVs So, Planck’s constant expressed in electron volts is 4.14×10 eV⋅s. We can now substitute in the values of energy and Planck’s constant into the equation that solves for frequency, 𝐸ℎ=𝑓. The energy of the photon is 10.2 eV, and Planck’s constant is 4.14×10 eV⋅s. This gives us 𝑓=(10.2)(4.14×10⋅).eVeVs The units of electron volts cancel, leaving behind only 1/s: (10.2)(4.14×10⋅)=2.46×101.eVeVss The units of 1/s are equivalent to hertz, Hz. This means that 2.46×101=2.46×10,sHz so the frequency of the photon is 2.46×10 Hz. Let’s look at an example. The diagram shows the transition of an electron in a hydrogen atom from 𝑛=3 to 𝑛=1, emitting a photon as it does so.
AnswerPart 1 The energy of the photon is the difference between the energy levels it transitions between. If the 3rd energy level is 𝐸 and the ground state is 𝐸, then we can express the energy difference as the equation Δ𝐸=𝐸−𝐸. The energy of 𝐸 is −1.51 eV and that of 𝐸 is −13.6 eV. Substituting these values into the equation gives (−1.51)−(−13.6)=12.09,eVeVeV so to two decimal places, the energy of the released photon, hereafter 𝐸, is 12.09 eV. Part 2 To find the frequency of this photon, we relate the photon’s frequency to its energy that was found in part 1. Recall that the equation for individual photon energy is 𝐸=ℎ𝑓, which can then be put in terms of frequency as 𝐸ℎ=𝑓. The value of 𝐸 from part 1 is 12.09 eV, and we are given Planck’s constant in electron volts as 4.14×10 eV⋅s. Substituting in these values into the equation gives us (12.09)(4.144×10⋅)=𝑓.eVeVs The units of electron volts cancel, leaving behind only 1/s: (12.09)(4.144×10⋅)=2.92×101.eVeVss The units of 1/s are the same as the units of hertz, Hz. So the answer becomes 2.92×101=2.92×10.sHz The frequency of this emitted photon to two decimal places is thus 2.92×10 Hz. Sometimes, it is more helpful to know the wavelength of a photon rather than its frequency. Recall that the equation for individual photon energy using wavelength is 𝐸=ℎ𝑐𝜆, where ℎ is Planck’s constant, 𝑐 is the speed of light, typically 3×10 m/s, and 𝜆 is the photon’s wavelength. In order to isolate the wavelength, 𝜆, we multiply both sides of the equation by 𝜆. This gives us 𝐸×𝜆=ℎ𝑐𝜆×𝜆, which cancels the 𝜆 on the right side 𝐸𝜆=ℎ𝑐. Now, we just need to divide both sides by the energy of the photon, 𝐸 as follows: 𝐸𝜆𝐸=ℎ𝑐𝐸. Canceling 𝐸 on the right yields 𝜆=ℎ𝑐𝐸. Let’s look at an example using this equation. The diagram shows the transition of an electron in a hydrogen atom from 𝑛=3 to 𝑛=2, emitting a photon as it does so.
AnswerPart 1 The energy of the photon is the difference between the energy levels it transitions between. If the 3rd energy level is 𝐸 and the 2nd energy level is 𝐸, then we can express the difference as the equation Δ𝐸=𝐸−𝐸. The value of 𝐸 is −1.51 eV and that of 𝐸 is −3.4 eV. Substituting these values into the equation yields Δ𝐸=(−1.51)−(−3.4)(−1.51)−(−3.4)=1.89,eVeVeVeVeV so the energy of the photon is 1.89 eV. Part 2 To find this photon’s wavelength, we relate the wavelength to its energy that was found in part 1. Recall that the equation relating the energy of a photon to its wavelength is 𝐸=ℎ𝑐𝜆, which can then be put in terms of wavelength as 𝜆=ℎ𝑐𝐸. We know the energy from part 1, 𝐸, is 1.89 eV. The other values are 𝑐, the speed of light, which is 3×10 m/s, and Planck’s constant, ℎ, which is 4.14×10 eV⋅s. Substituting these values into the equation gives 𝜆=4.14×10⋅3×10/(1.89).eVsmseV The units of electron volts cancel, as we divide the Planck’s constant in the numerator by the energy in the denominator as follows: 𝜆=2.19×103×10.sms When we multiply these last two terms together, the units of seconds cancel, leaving behind only metres as follows: 2.19×103×10=6.57×10.smsm Now, we just have to put our answer in terms of nanometres. There are 10 metres in one nanometre, as seen in the relation 110,nmm which we can then multiply by the wavelength in metres to give it in terms of nanometres as follows 110×6.57×10=657.nmmmnm The wavelength to the nearest nanometre is 657 nm. In order for an electron to transition upward, it must absorb a photon. The energy of the absorbed photon determines to which energy level they transition to, with higher photon energies meaning higher energy levels. If the energy of the photon is very high, the electron can be expelled from the atom entirely, causing the atom to be ionized. This is shown in the diagram below. In order for this to happen, the energy of the photon must exceed the maximum binding energy of the electron shells. This maximum binding energy is found at the ground state, or electron energy level 1. So, for hydrogen, a photon must have 13.6 or more electron volts of energy to make an electron transition out of the atom. Let’s look at an example of a photon being absorbed by an electron. The diagram shows the binding energy of each energy level of a hydrogen atom. If an electron is in the ground state, what wavelength of photon must it absorb in order for the hydrogen atom to become completely ionized? Use a value of 4.14×10 eV⋅s for the Planck constant. Give your answer to one decimal place. AnswerThe energy needed for the hydrogen atom to become completely ionized means the energy required for the electron to leave completely. Since this electron is in ground state, the electron needs to absorb a photon with an energy, 𝐸, of at least 13.6 eV. Recall that the equation that relates wavelength to photon energy is 𝐸=ℎ𝑐𝜆, which can then be put into terms of wavelength as 𝜆=ℎ𝑐𝐸. We know the photon energy needed to expel the electron, 13.6 eV, Planck’s constant 4.14×10 eV⋅s, and the speed of light, 3×10 m/s. Substituting these values into the equation gives 𝜆=4.14×10⋅3×10/(13.6),eVsmseV so when we divide the Planck’s constant in the numerator by the energy in the denominator, the units of electron volts cancel to leave behind seconds as follows: 𝜆=3.05×103×10.sms Multiplying these last two terms together cancels the units of seconds, leaving just metres, which gives 3.05×103×10=9.13×10.smsm We now need to put this answer in terms of nanometres. There are 10 metres in one nanometre. This is given by the relation 110,nmm which we can then multiply by the wavelength in metres to give it in terms of nanometres as 110×9.13×10=91.3.nmmmnm So, the wavelength required of a photon to completely ionize this hydrogen atom is 91.3 nm. The transitions between the electron energy levels are grouped into series. When an electron transitions downward to a particular energy level, it has a special series name. The names of these series are given in the table below.
The series name is regardless of the energy level it transitions from. Electrons transitioning down to the 2nd energy level from the 3rd, 4th, or 5th energy levels would all be in the Balmer series. The figure below shows Lyman, Balmer, and Paschen series transitions for electrons in a hydrogen atom. Let’s look at an example. The diagram shows four possible transitions that an electron can make between the energy levels of a hydrogen atom. What name is given to this series of transitions? AnswerIt does not matter where the electrons are transitioning from, only where they end. The name of these series and the energy level they transition down to is given by the table below.
Consulting this table, we see that when an electron transitions down to the 2nd energy level, it does so to the Balmer series. All of the electrons in this graph are transitioning to the 2nd energy level. The answer is the Balmer series. Let’s summarize what we have learned in this explainer.
