How to Calculate Photon Energy from an Energy-Level Diagram

An electron moves from an energy level of \(-1.5\text{ eV}\) to one of \(-4.0\text{ eV}\). You subtract the initial energy from the final energy and get \(-2.5\text{ eV}\).

But has the atom emitted a photon with negative energy?

No. The atom has lost \(2.5\text{ eV}\), and the photon carries that energy away. The negative sign describes the atom’s energy change, not the photon’s energy.

To calculate photon energy from an energy-level diagram, find the magnitude of the difference between the initial and final energy levels:

\[ \boxed{E_{\text{photon}}=|E_f-E_i|} \]

Here, \(E_i\) is the initial energy level and \(E_f\) is the final energy level.

That is the central calculation. The important part is knowing how to subtract correctly, recognise whether a photon is absorbed or emitted, and check whether the transition is allowed.

01Read the energy-level diagram correctly

An energy-level diagram represents the allowed energy states of an electron in an atom.

Each horizontal line represents an allowed energy level. The vertical position represents its energy, with higher positions corresponding to higher energy.

For example, consider these three levels:

Notice that \(-1.0\text{ eV}\) is higher in energy than \(-5.0\text{ eV}\).

This can feel backwards at first. Remember that a more positive number represents a higher energy, even when both numbers are negative.

Why are some energy levels negative?

Energy is measured relative to a chosen reference point.

For many atomic energy-level diagrams, zero energy represents an electron that is completely free from the atom. Bound electron states have lower energies, so they are assigned negative values.

Think of sea level as a reference point. A location 5 metres below sea level has an altitude of \(-5\text{ m}\), but the distance between that location and one 2 metres below sea level is still \(3\text{ m}\).

The same idea applies to energy differences. The absolute values can be negative, but the energy transferred in a transition is determined by the difference between them.

One more detail: don’t assume that the vertical spacing in every diagram is drawn to scale. If numerical energy values are supplied, use those values rather than estimating distances between lines.

02Understand the difference between absorption and emission

An electron can move between energy levels in two directions.

The direction determines whether the atom absorbs or emits a photon.

Emission: the electron moves down

Suppose an electron starts at \(-1.0\text{ eV}\) and moves to \(-4.0\text{ eV}\).

The electron finishes at a lower energy level. The atom loses energy, which is released as a photon.

The atom’s energy change is:

\[ \Delta E_{\text{atom}}=E_f-E_i \]

\[ =-4.0-(-1.0) \]

\[ =-3.0\text{ eV} \]

The atom has lost \(3.0\text{ eV}\).

Therefore, the emitted photon has an energy of:

\[ E_{\text{photon}}=3.0\text{ eV} \]

Absorption: the electron moves up

Now reverse the transition.

An electron starts at \(-4.0\text{ eV}\) and absorbs a photon, moving to \(-1.0\text{ eV}\).

The atom gains energy:

\[ \Delta E_{\text{atom}}=E_f-E_i \]

\[ =-1.0-(-4.0) \]

\[ =+3.0\text{ eV} \]

The absorbed photon must therefore have an energy of \(3.0\text{ eV}\).

The two situations can be summarised:

The key distinction is that the atom’s energy change has a sign, but the photon energy is always positive.

For a single-photon transition:

\[ E_{\text{photon}}=|\Delta E_{\text{atom}}| \]

This is the safest equation to use when working from an energy-level diagram.

03Check whether the transition is allowed

There is another detail that can catch students out.

Just because two energy levels have a difference in energy does not automatically mean an electron can make that transition.

Some energy-level diagrams show arrows identifying allowed transitions. Others provide restrictions in the question.

In a more complete atomic model, selection rules determine which transitions are permitted. Some transitions are forbidden or have extremely low probabilities.

For HSC-style calculations, follow the information supplied in the question.

• If an arrow identifies an allowed transition, use that transition.
• If the question explicitly states that an electron moves between two levels, calculate that transition.
• If the question gives restrictions, do not assume that every pair of levels can be connected.

For example, an energy difference of \(3.0\text{ eV}\) might exist between two levels, but if the direct transition is forbidden, you cannot claim that the atom emits a \(3.0\text{ eV}\) photon through that transition.

This distinction becomes particularly important when an electron can return to its original state through several smaller transitions.

04Convert photon energy into joules, frequency or wavelength

Energy-level diagrams commonly give energy values in electronvolts (eV).

However, some questions ask for photon energy in joules (J), frequency (Hz), or wavelength (m).

You need to know three relationships.

Converting electronvolts into joules

Use:

\[ 1\text{ eV}=1.602\times10^{-19}\text{ J} \]

Therefore:

\[ E(\text{J})=E(\text{eV})\times1.602\times10^{-19} \]

For example, a photon with an energy of \(2.5\text{ eV}\) has an energy in joules of:

\[ E=2.5\times1.602\times10^{-19} \]

\[ E=4.005\times10^{-19}\text{ J} \]

Always include the unit conversion. An answer in eV is not interchangeable with an answer in J.

Finding frequency

Photon energy and frequency are related by Planck’s equation:

\[ \boxed{E=hf} \]

where:

• \(E\) is photon energy in joules (J).
• \(h\) is Planck’s constant, \(6.626\times10^{-34}\text{ J s}\).
• \(f\) is frequency in hertz (Hz).

Rearranging:

\[ \boxed{f=\frac{E}{h}} \]

Finding wavelength

Photon energy and wavelength are related by:

\[ \boxed{E=\frac{hc}{\lambda}} \]

where:

• \(E\) is photon energy in joules (J).
• \(h\) is Planck’s constant.
• \(c\) is the speed of light, \(3.00\times10^8\text{ m s}^{-1}\).
• \(\lambda\) is wavelength in metres (m).

Rearranging:

\[ \boxed{\lambda=\frac{hc}{E}} \]

You can also use \(c=f\lambda\).

Important: Convert photon energy into joules before substituting into these equations. Also convert any wavelength given in nanometres into metres.

\[ 1\text{ nm}=1\times10^{-9}\text{ m} \]

A higher-energy photon has a higher frequency and a shorter wavelength.

05How experimental spectra connect to energy-level diagrams

Why do physicists use energy-level diagrams in the first place?

The diagrams provide a model for explaining the specific wavelengths of light emitted or absorbed by atoms.

Consider a standard emission spectroscopy experiment.

A low-pressure gas discharge tube contains a gas between two electrodes connected to a high-voltage supply. Electrons travelling through the gas collide with atoms, transferring energy to them.

Some electrons in the atoms move into higher energy states.

When these excited atoms return to lower energy states, they release photons.

The light travels out of the discharge tube, passes through a narrow slit and then through a collimating lens. The lens produces a parallel beam that enters a diffraction grating.

The grating separates the light into different wavelengths. A screen or detector records the resulting pattern of spectral lines.

What should the experiment show?

Consider two possible predictions.

If atoms could release any amount of energy, we would expect a continuous range of wavelengths.

However, if atoms have specific allowed energy states, only particular energy differences are possible. We would expect separate spectral lines.

The observed discrete lines support the idea that atomic energy levels are quantised.

There is an important distinction here:

• The spectrometer measures the wavelengths of light.
• The energy-level diagram models the allowed atomic energies.
• Physicists compare the measured photon energies with the differences between energy levels.

Simply looking at a glowing tube would not provide sufficiently precise wavelength measurements. The slit and diffraction grating help separate the different wavelengths so they can be measured individually.

The diagram is not a photograph of an electron’s physical path. It is a model that explains the observed energy changes.

06Worked examples

07Common mistakes to avoid

Before attempting exam questions, make sure you can recognise these mistakes.

One final distinction is particularly important.

If an electron drops directly from \(E_3\) to \(E_1\), it may emit one photon with energy equal to the full difference.

If it moves from \(E_3\) to \(E_2\), then from \(E_2\) to \(E_1\), it emits two photons.

The total energy transferred can be identical, but the individual photon energies are different.

That distinction becomes important in the harder questions below.

08Questions and solutions

Question 1

An electron moves from an energy level of \(-0.9\text{ eV}\) to a lower energy level of \(-3.5\text{ eV}\).

Calculate:

(a) The energy of the emitted photon in eV.

(b) The energy of the emitted photon in J.

Use \(1\text{ eV}=1.602\times10^{-19}\text{ J}\).

Solution 1

The emitted photon has an energy of \(2.6\text{ eV}\), equivalent to \(4.17\times10^{-19}\text{ J}\).

(a) Calculate the energy in eV.

The initial energy is:

\[ E_i=-0.9\text{ eV} \]

The final energy is:

\[ E_f=-3.5\text{ eV} \]

Calculate the atom’s energy change:

\[ \Delta E_{\text{atom}}=E_f-E_i \]

\[ =-3.5-(-0.9) \]

\[ =-2.6\text{ eV} \]

Since the atom loses energy, the photon carries the positive magnitude:

\[ \boxed{E_{\text{photon}}=2.6\text{ eV}} \]

(b) Convert into joules.

\[ E=2.6\times1.602\times10^{-19} \]

\[ \boxed{E=4.17\times10^{-19}\text{ J}} \]

The negative sign belongs to the atom’s energy change, not the emitted photon.

Question 2

An atom has three energy levels:

\[ E_0=-5.0\text{ eV} \]

\[ E_1=-3.2\text{ eV} \]

\[ E_2=-1.0\text{ eV} \]

The atom is initially in its lowest energy state, \(E_0\).

A photon with an energy of \(2.2\text{ eV}\) strikes the atom.

(a) Calculate the energy required to move the electron from \(E_0\) to \(E_1\).

(b) Calculate the energy required to move the electron from \(E_0\) to \(E_2\).

(c) Can the atom absorb the incoming \(2.2\text{ eV}\) photon through a single transition from its initial state? Explain your reasoning.

Solution 2

The atom cannot absorb the \(2.2\text{ eV}\) photon through a single transition from \(E_0\). The photon energy does not match either available energy gap from the initial state.

(a) Calculate the energy required to move from \(E_0\) to \(E_1\).

\[ \Delta E=E_1-E_0 \]

\[ =-3.2-(-5.0) \]

\[ \boxed{\Delta E=1.8\text{ eV}} \]

The atom needs to absorb \(1.8\text{ eV}\) to reach \(E_1\).

(b) Calculate the energy required to move from \(E_0\) to \(E_2\).

\[ \Delta E=E_2-E_0 \]

\[ =-1.0-(-5.0) \]

\[ \boxed{\Delta E=4.0\text{ eV}} \]

The atom needs to absorb \(4.0\text{ eV}\) to reach \(E_2\).

(c) Determine whether the photon can be absorbed.

The incoming photon has an energy of \(2.2\text{ eV}\).

Neither available transition from \(E_0\) requires this amount of energy.

You might notice that:

\[ E_2-E_1=-1.0-(-3.2) \]

\[ =2.2\text{ eV} \]

This is exactly the incoming photon energy.

However, this transition starts at \(E_1\), not \(E_0\).

The atom is initially in \(E_0\), so it cannot make the \(E_1\rightarrow E_2\) transition unless it first reaches \(E_1\).

This is the important distinction: matching an energy gap is not enough. The electron must also start in the correct energy level.

Assuming no other absorption processes or energy levels are available, the photon is not absorbed.

Question 3

A spectrometer records an emission line with a wavelength of \(620\text{ nm}\).

An atom has the following energy levels:

Assume all downward transitions between the listed levels are allowed.

(a) Calculate the energy of the detected photon in joules.

(b) Convert the photon energy into eV.

(c) Identify the transition that could have produced the emission line.

(d) Explain why the transition from \(E_1\) to \(E_0\) cannot produce this particular spectral line.

Use:

\[ h=6.626\times10^{-34}\text{ J s} \]

\[ c=3.00\times10^8\text{ m s}^{-1} \]

Solution 3

The detected photon has an energy of approximately \(3.21\times10^{-19}\text{ J}\), or \(2.00\text{ eV}\). It corresponds to the transition from \(E_3\) to \(E_2\).

(a) Calculate the photon energy in joules.

Convert the wavelength into metres:

\[ \lambda=620\times10^{-9}\text{ m} \]

Use:

\[ E=\frac{hc}{\lambda} \]

Substitute:

\[ E= \frac{(6.626\times10^{-34})(3.00\times10^8)} {620\times10^{-9}} \]

\[ \boxed{E=3.21\times10^{-19}\text{ J}} \]

(b) Convert into electronvolts.

\[ E(\text{eV})= \frac{3.21\times10^{-19}} {1.602\times10^{-19}} \]

\[ \boxed{E\approx2.00\text{ eV}} \]

(c) Identify the transition.

Compare the measured photon energy with the possible energy differences.

For the transition from \(E_3\) to \(E_2\):

\[ \Delta E=|-2.8-(-0.8)| \]

\[ =2.0\text{ eV} \]

This matches the measured photon energy.

Therefore:

\[ \boxed{E_3\rightarrow E_2} \]

is the transition responsible for the spectral line.

(d) Explain why \(E_1\rightarrow E_0\) is not responsible.

Calculate its energy difference:

\[ \Delta E=|-5.6-(-4.1)| \]

\[ =1.5\text{ eV} \]

This is different from the measured \(2.00\text{ eV}\).

Therefore, this transition would produce a photon with a different energy and wavelength.

The important experimental connection is that each spectral line corresponds to a particular photon energy, which can be matched against the allowed energy differences in the diagram.

Question 4

An atom contains three energy levels:

\[ E_3=-0.5\text{ eV} \]

\[ E_2=-2.0\text{ eV} \]

\[ E_1=-4.5\text{ eV} \]

An electron initially occupies \(E_3\) and eventually returns to \(E_1\).

There are two possible decay pathways:

• A direct transition from \(E_3\) to \(E_1\).
• A two-step transition from \(E_3\) to \(E_2\), followed by \(E_2\) to \(E_1\).

Assume all three transitions are allowed.

(a) Calculate the photon energy for the direct transition.

(b) Calculate the energy of each photon emitted during the two-step transition.

(c) Calculate the wavelength of each photon produced by all three individual transitions.

(d) Explain why the total energy released is the same for both pathways, even though the number and energies of the photons differ.

Solution 4

The direct transition produces one \(4.0\text{ eV}\) photon, while the two-step transition produces two photons with energies of \(1.5\text{ eV}\) and \(2.5\text{ eV}\). Both pathways release a total of \(4.0\text{ eV}\).

(a) Calculate the direct transition energy.

The electron moves from \(-0.5\text{ eV}\) to \(-4.5\text{ eV}\).

\[ E_{\text{photon}}= |-4.5-(-0.5)| \]

\[ \boxed{E_{\text{photon}}=4.0\text{ eV}} \]

Because the electron makes one direct transition, one photon carries the entire energy difference.

(b) Calculate the two photon energies in the cascade.

First transition: \(E_3\rightarrow E_2\).

\[ E_{\gamma 1}=|-2.0-(-0.5)| \]

\[ \boxed{E_{\gamma 1}=1.5\text{ eV}} \]

Second transition: \(E_2\rightarrow E_1\).

\[ E_{\gamma 2}=|-4.5-(-2.0)| \]

\[ \boxed{E_{\gamma 2}=2.5\text{ eV}} \]

The atom emits two separate photons.

(c) Calculate the wavelengths.

Start by converting each photon energy into joules.

For the direct transition:

\[ E=4.0\times1.602\times10^{-19} \]

\[ E=6.408\times10^{-19}\text{ J} \]

Use:

\[ \lambda=\frac{hc}{E} \]

\[ \lambda= \frac{(6.626\times10^{-34})(3.00\times10^8)} {6.408\times10^{-19}} \]

\[ \boxed{\lambda\approx310\text{ nm}} \]

For the first cascade photon:

\[ E=1.5\times1.602\times10^{-19} \]

\[ E=2.403\times10^{-19}\text{ J} \]

\[ \lambda= \frac{(6.626\times10^{-34})(3.00\times10^8)} {2.403\times10^{-19}} \]

\[ \boxed{\lambda\approx827\text{ nm}} \]

For the second cascade photon:

\[ E=2.5\times1.602\times10^{-19} \]

\[ E=4.005\times10^{-19}\text{ J} \]

\[ \lambda= \frac{(6.626\times10^{-34})(3.00\times10^8)} {4.005\times10^{-19}} \]

\[ \boxed{\lambda\approx496\text{ nm}} \]

The results are:

The direct photon has the shortest wavelength because it has the highest energy.

(d) Explain the conservation of energy.

For the direct transition:

\[ E_{\text{total}}=4.0\text{ eV} \]

For the two-step transition:

\[ E_{\text{total}}=1.5+2.5 \]

\[ E_{\text{total}}=4.0\text{ eV} \]

Both pathways transfer exactly the same total energy.

The tempting mistake is to assume that the two-step pathway also produces one \(4.0\text{ eV}\) photon.

It doesn’t. Each individual transition produces its own photon, and each photon carries only the energy lost during that particular step.

Energy is conserved, but the energy can be distributed across different numbers of photons.

Question 5

An atom has four allowed energy levels:

\[ E_A=-8.0\text{ eV} \]

\[ E_B=-6.0\text{ eV} \]

\[ E_C=-4.0\text{ eV} \]

\[ E_D=-2.0\text{ eV} \]

The allowed downward transitions are:

• \(D\rightarrow B\)
• \(D\rightarrow C\)
• \(C\rightarrow B\)
• \(C\rightarrow A\)
• \(B\rightarrow A\)

The direct transition \(D\rightarrow A\) is forbidden.

Every atom begins in state \(D\) and eventually returns to state \(A\). Assume that:

• Each allowed downward transition releases exactly one photon.
• No energy is lost through collisions or other non-radiative processes.
• All atoms complete their transitions.
• An ideal detector measures the total energy emitted by each atom but cannot initially distinguish individual photons.

(a) Calculate the total energy emitted by each atom as it moves from \(D\) to \(A\).

(b) Identify every possible complete transition pathway from \(D\) to \(A\). Calculate the energy of every photon emitted along each pathway.

(c) The detector measures \(6.0\text{ eV}\) of total emitted energy per atom. Can it determine how many photons were emitted? Explain why or why not.

(d) A second detector resolves the individual photons and finds that each atom emits exactly three photons, each with an energy of \(2.0\text{ eV}\).

Identify the transition pathway and explain why a single \(6.0\text{ eV}\) photon would be inconsistent with the supplied energy-level model.

Solution 5

The atom emits a total of \(6.0\text{ eV}\), but it can release that energy through either two photons or three photons. If exactly three photons of \(2.0\text{ eV}\) are detected, the unique pathway is \(D\rightarrow C\rightarrow B\rightarrow A\).

This question tests whether you can distinguish the total energy lost by an atom from the energy of each individual photon.

(a) Calculate the total energy emitted.

The initial energy is:

\[ E_i=-2.0\text{ eV} \]

The final energy is:

\[ E_f=-8.0\text{ eV} \]

The atom’s total energy change is:

\[ \Delta E_{\text{atom}}=E_f-E_i \]

\[ =-8.0-(-2.0) \]

\[ =-6.0\text{ eV} \]

The atom loses \(6.0\text{ eV}\).

Therefore, the total energy carried away by the photons is:

\[ \boxed{E_{\text{total}}=6.0\text{ eV}} \]

(b) Identify all possible pathways and photon energies.

We must follow the allowed transitions rather than assume that the electron can jump directly between any two levels.

There are three possible complete pathways.

Pathway 1: \(D\rightarrow B\rightarrow A\)

First transition:

\[ E_{\gamma 1}=|-6.0-(-2.0)| \]

\[ E_{\gamma 1}=4.0\text{ eV} \]

Second transition:

\[ E_{\gamma 2}=|-8.0-(-6.0)| \]

\[ E_{\gamma 2}=2.0\text{ eV} \]

Total:

\[ E_{\text{total}}=4.0+2.0 \]

\[ \boxed{E_{\text{total}}=6.0\text{ eV}} \]

This pathway produces two photons.

Pathway 2: \(D\rightarrow C\rightarrow A\)

First transition:

\[ E_{\gamma 1}=|-4.0-(-2.0)| \]

\[ E_{\gamma 1}=2.0\text{ eV} \]

Second transition:

\[ E_{\gamma 2}=|-8.0-(-4.0)| \]

\[ E_{\gamma 2}=4.0\text{ eV} \]

Total:

\[ E_{\text{total}}=2.0+4.0 \]

\[ \boxed{E_{\text{total}}=6.0\text{ eV}} \]

This pathway also produces two photons.

Pathway 3: \(D\rightarrow C\rightarrow B\rightarrow A\)

First transition:

\[ E_{\gamma 1}=|-4.0-(-2.0)| \]

\[ E_{\gamma 1}=2.0\text{ eV} \]

Second transition:

\[ E_{\gamma 2}=|-6.0-(-4.0)| \]

\[ E_{\gamma 2}=2.0\text{ eV} \]

Third transition:

\[ E_{\gamma 3}=|-8.0-(-6.0)| \]

\[ E_{\gamma 3}=2.0\text{ eV} \]

Total:

\[ E_{\text{total}}=2.0+2.0+2.0 \]

\[ \boxed{E_{\text{total}}=6.0\text{ eV}} \]

This pathway produces three photons.

The possible outcomes are:

(c) Can the detector determine the number of photons?

No. Measuring the total emitted energy does not reveal how many photons carried that energy.

All three pathways release \(6.0\text{ eV}\), but two pathways produce two photons and the other produces three.

This is a consequence of conservation of energy. The total energy transferred is fixed by the initial and final states, but the intermediate transitions determine how that energy is divided between photons.

A detector that measures only the total energy cannot distinguish these pathways.

(d) Identify the pathway from the individual photon measurements.

The atom emits exactly three photons, each with an energy of \(2.0\text{ eV}\).

Only one of the allowed pathways produces this result:

\[ \boxed{D\rightarrow C\rightarrow B\rightarrow A} \]

Each step releases \(2.0\text{ eV}\), giving three photons:

\[ E_{\text{total}}=2.0+2.0+2.0 \]

\[ \boxed{E_{\text{total}}=6.0\text{ eV}} \]

Now consider the tempting alternative.

The overall difference between \(D\) and \(A\) is \(6.0\text{ eV}\). You might therefore predict that the atom emits one photon with an energy of \(6.0\text{ eV}\).

However, the direct transition \(D\rightarrow A\) is explicitly forbidden.

An energy difference exists, but the transition cannot occur under the supplied model. The atom must instead follow one of the allowed pathways through intermediate levels.

If an experiment actually detected one \(6.0\text{ eV}\) photon from this atom, the result would contradict the stated model. The direct transition would need to be reconsidered, or the model would need to account for another process.

This is the important lesson: the energy difference determines the photon energy only when the corresponding transition is allowed and occurs as a single-photon transition.

09Where to go next

Once you can calculate photon energies from individual transitions, the next useful skill is working out how many different spectral lines an excited atom can produce.

Remember that different transitions can sometimes have the same energy difference, while one electron can also return to its ground state through several intermediate levels.

These ideas connect directly to How to Count Spectral Lines from Atomic Energy Levels.

For now, keep one calculation at the centre of your working:

\[ \boxed{E_{\text{photon}}=|E_f-E_i|} \]

Calculate the difference for each allowed transition separately, keep the photon energy positive, and only convert units when the question requires it.