How to Count Spectral Lines from Atomic Energy Levels

An atom is excited to its fourth energy level. As it returns towards the ground state, how many different emission lines could it produce?

In the simplified HSC Physics model, the maximum is six spectral lines, not three, provided every downward transition is allowed and all six energy differences are unique.

More generally, if an atom has \(N\) energy levels, including the ground state and the highest excited level, the maximum number of possible downward transitions is:

\[ \boxed{L=\frac{N(N-1)}{2}} \]

The important detail is that an electron can lose energy in stages. It does not have to fall directly from its highest energy level to the ground state. Every possible intermediate transition can produce another spectral line.

There is also a distinction between the number of possible transitions, the number of distinct wavelengths, and the number of photons emitted by one atom. Let’s work through each.

01Where the spectral lines come from

Before counting transitions, consider how scientists observe them.

A typical emission spectroscopy experiment uses a low-pressure gas discharge tube. The tube contains a gas between two electrodes connected to a high-voltage supply.

Electrons moving through the tube collide with gas atoms and transfer energy to them. Some atoms become excited, meaning their electrons occupy higher energy states.

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

The light travels out of the discharge tube and enters a narrow slit. A collimating lens produces a parallel beam, which then passes through a diffraction grating. The grating separates the light into different wavelengths, producing a pattern of bright lines on a screen or detector.

What should the experiment show?

Consider two possible models.

• If atoms could release any amount of energy, we would expect a continuous range of wavelengths.
• If atoms could only occupy specific energy states, we would expect particular energy differences and therefore separate spectral lines.

The observed emission spectrum contains discrete lines rather than every possible wavelength.

This supports the idea that atomic energy states are quantised.

The detector measures the wavelengths of emitted photons. The energy-level model explains those wavelengths as differences between allowed atomic energy states.

There is one limitation worth remembering: simply looking at the glowing tube is not enough to resolve individual wavelengths. The slit and diffraction grating improve the resolution, although very closely spaced or weak lines can still be difficult to distinguish.

Most importantly, a spectral line represents a photon energy. It does not represent an energy level by itself.

02Counting all possible transitions

Imagine an atom with four energy levels:

\[ E_1<E_2<E_3<E_4 \]

Here, \(E_1\) is the ground state, and \(E_4\) is the highest excited state.

An electron initially at \(E_4\) has three possible first transitions:

• \(E_4\rightarrow E_3\)
• \(E_4\rightarrow E_2\)
• \(E_4\rightarrow E_1\)

You might predict that there are only three possible spectral lines.

But that misses the transitions that can happen after the first drop.

If the electron falls to \(E_3\), it can continue to \(E_2\) or \(E_1\).

If it falls to \(E_2\), it can continue to \(E_1\).

The complete list is therefore:

The total number of possible transitions is:

\[ L=3+2+1=6 \]

Each transition can release a photon with a particular energy.

Deriving the general formula

The same pattern works for any number of energy levels.

Suppose the atom has \(N\) energy levels, numbered from 1 at the ground state to \(N\) at the highest excited state.

• The highest level has \(N-1\) possible lower levels.
• The next level has \(N-2\) possible lower levels.
• The next has \(N-3\).
• This continues until the second level, which has one possible lower level.

Therefore:

\[ L=(N-1)+(N-2)+\cdots+2+1 \]

Using the formula for the sum of consecutive integers:

\[ \boxed{L=\frac{N(N-1)}{2}} \]

This gives the following useful values:

A common terminology trap is the difference between an energy level and an excited state.

If the ground state is level 1, then the first excited state is level 2, the second excited state is level 3, and the third excited state is level 4.

So, if a question says an electron is excited to the third excited state, use \(N=4\):

\[ L=\frac{4(4-1)}{2}=6 \]

If a question instead calls it the \(n\)th excited state, with the ground state not included in that numbering, there are \(n+1\) levels in total. The maximum is then \(n(n+1)/2\).

03Connecting energy differences to spectral lines

Counting possible transitions is only part of the problem. To understand the actual spectral lines, we need to calculate the energy released during each transition.

Suppose an electron moves from an initial energy level \(E_i\) to a lower final energy level \(E_f\).

The change in the atom’s energy is:

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

Because the final energy is lower:

\[ \Delta E_{\text{atom}}<0 \]

The atom has lost energy. That energy is carried away by an emitted photon.

The photon’s energy is therefore:

\[ \boxed{E_{\gamma}=E_i-E_f} \]

Notice the sign difference:

• The atom’s energy change is negative.
• The emitted photon’s energy is positive.

A photon does not have negative energy simply because the atom loses energy.

Frequency and wavelength

Photon energy is related to frequency by Planck’s equation:

\[ E_{\gamma}=hf \]

where:

• \(E_{\gamma}\) is photon energy in joules (J),
• \(h=6.63\times10^{-34}\ \text{J s}\) is Planck’s constant,
• \(f\) is frequency in hertz (Hz).

Frequency and wavelength are related by:

\[ c=f\lambda \]

where:

• \(c=3.00\times10^8\ \text{m s}^{-1}\) is the speed of light,
• \(\lambda\) is wavelength in metres (m).

Combining these equations gives:

\[ \boxed{E_{\gamma}=\frac{hc}{\lambda}} \]

This means:

• A larger energy difference produces a higher-frequency photon.
• A larger energy difference produces a shorter-wavelength photon.
• A smaller energy difference produces a lower-frequency photon with a longer wavelength.

What happens during absorption?

The reverse process is absorption.

When an atom absorbs a photon, its electron moves to a higher energy level. The atom gains energy, so:

\[ \Delta E_{\text{atom}}=E_f-E_i>0 \]

The absorbed photon’s energy is:

\[ E_{\gamma}=E_f-E_i>0 \]

For the same pair of energy levels, absorption and emission involve the same energy difference. They therefore have the same photon frequency and wavelength.

This is why the energy difference, rather than the direction of the transition, determines the spectral line.

06When should you use the formula?

The formula

\[ L=\frac{N(N-1)}{2} \]

is useful when:

• You know the number of energy levels.
• Every downward transition between the levels is allowed.
• You are counting possible transitions, and no two energy differences are equal.

There are two important limitations.

1. Some transitions may not be allowed

In more complete atomic models, quantum selection rules can prevent certain transitions.

If a question explicitly identifies which transitions are allowed, you must count only those transitions.

For example, if a four-level atom has six possible downward pairs but one transition is forbidden, there are only five allowed transitions.

An electron might still reach the lower level through a sequence of other allowed transitions. A forbidden direct transition does not automatically prevent a cascade through intermediate levels.

2. Different transitions can produce the same wavelength

As the previous example demonstrated, two different pairs of energy levels can have the same energy difference.

The correct number of distinct spectral lines is therefore the number of unique energy gaps among the allowed transitions.

A useful exam distinction is:

The maximum number of photons emitted by one atom in a complete downward cascade is also different from the total number of possible spectral lines.

For an atom with \(N\) levels, a cascade through every consecutive level produces at most \(N-1\) photons. A direct transition to the ground state produces just one photon, if that transition is allowed.

07A reliable method for HSC questions

When you see a question asking how many spectral lines can be produced after excitation to a particular level, use this process.

1. Identify the highest excited level.
2. Count the total number of energy levels, including the ground state.
3. Check whether all downward transitions are allowed.
4. Use \(N(N-1)/2\) to count possible transitions when all pairs are allowed.
5. If energy values are supplied, calculate the energy difference for every allowed transition.
6. Remove repeated energy differences when counting distinct spectral lines.
7. If asked for frequency or wavelength, use the positive photon energy and the equations \(E=hf\) and \(c=f\lambda\).

Keep the sign convention straight: the atom loses energy during emission, but the photon always carries positive energy.

08Questions and solutions

Question 1

An atom has four energy levels, labelled \(E_1\), \(E_2\), \(E_3\), and \(E_4\), with \(E_1\) being the ground state.

Assume all downward transitions are allowed and all energy differences are unique.

Calculate the maximum number of distinct emission spectral lines that can be produced when an electron is excited to \(E_4\).

Explain why counting only transitions directly from \(E_4\) gives an incomplete answer.

Solution 1

The maximum number of distinct emission lines is six.

There are four energy levels, so:

\[ N=4 \]

Using the transition-counting formula:

\[ L=\frac{N(N-1)}{2} \]

\[ L=\frac{4(4-1)}{2} \]

\[ \boxed{L=6} \]

The reason the answer is not three is that the electron can make further transitions after its initial drop.

The possible transitions are:

• From \(E_4\): three transitions.
• From \(E_3\): two transitions.
• From \(E_2\): one transition.

Therefore:

\[ L=3+2+1=6 \]

Counting only the three direct transitions from \(E_4\) ignores the photons that can be emitted during subsequent transitions.

Question 2

An electron in an atom initially occupies an energy level of \(-5.0\text{ eV}\).

It absorbs a photon and moves to an energy level of \(-1.0\text{ eV}\).

Calculate:

(a) The energy of the absorbed photon in joules.

(b) The frequency and wavelength of the photon.

(c) The signed change in the atom’s energy if the electron subsequently returns directly to its original energy level by emitting a photon.

Use:

• \(h=6.63\times10^{-34}\ \text{J s}\)
• \(c=3.00\times10^8\ \text{m s}^{-1}\)
• \(1\text{ eV}=1.60\times10^{-19}\text{ J}\)

Solution 2

The absorbed photon has an energy of \(4.0\text{ eV}\), equivalent to \(6.40\times10^{-19}\text{ J}\). Its frequency is approximately \(9.65\times10^{14}\text{ Hz}\), and its wavelength is approximately \(311\text{ nm}\).

(a) Calculate the absorbed photon energy.

The electron moves from:

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

to:

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

The atom’s energy change is:

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

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

\[ =+4.0\text{ eV} \]

The positive sign means the atom gains energy.

Therefore, the absorbed photon has energy:

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

Converting to joules:

\[ E_{\gamma}=4.0(1.60\times10^{-19}) \]

\[ \boxed{E_{\gamma}=6.40\times10^{-19}\text{ J}} \]

(b) Calculate the frequency and wavelength.

Using:

\[ E_{\gamma}=hf \]

Rearrange:

\[ f=\frac{E_{\gamma}}{h} \]

Substitute:

\[ f=\frac{6.40\times10^{-19}} {6.63\times10^{-34}} \]

\[ \boxed{f=9.65\times10^{14}\text{ Hz}} \]

Now use:

\[ \lambda=\frac{c}{f} \]

\[ \lambda=\frac{3.00\times10^8} {9.65\times10^{14}} \]

\[ \boxed{\lambda=3.11\times10^{-7}\text{ m}} \]

Therefore:

\[ \boxed{\lambda=311\text{ nm}} \]

(c) Determine the energy change during emission.

When the electron returns to its original level:

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

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

\[ \boxed{\Delta E_{\text{atom}}=-4.0\text{ eV}} \]

The atom loses 4.0 eV of energy.

The emitted photon carries:

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

It therefore has the same frequency and wavelength as the photon originally absorbed.

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

Question 3

An illustrative atom has five energy levels:

An electron is initially excited to \(E_5\).

It eventually reaches the ground state by emitting exactly two photons.

The first photon has twice the energy of the second photon.

Determine:

(a) The intermediate energy level through which the electron must pass.

(b) The energies of both photons.

(c) The wavelengths of both photons.

Solution 3

The electron must pass through \(E_3\), the 2.0 eV level. It emits a 4.0 eV photon first, followed by a 2.0 eV photon.

(a) Find the intermediate energy level.

Because the electron emits exactly two photons before reaching the ground state, it must follow this general pathway:

\[ E_5\rightarrow E_k\rightarrow E_1 \]

where \(E_k\) is the unknown intermediate energy.

The first photon has energy:

\[ E_{\gamma 1}=6.0-E_k \]

The second photon has energy:

\[ E_{\gamma 2}=E_k-0 \]

We are told that the first photon has twice the energy of the second:

\[ \frac{6.0-E_k}{E_k}=2 \]

Rearrange:

\[ 6.0-E_k=2E_k \]

\[ 6.0=3E_k \]

\[ \boxed{E_k=2.0\text{ eV}} \]

This corresponds to \(E_3\).

The required pathway is:

\[ \boxed{E_5\rightarrow E_3\rightarrow E_1} \]

(b) Calculate the photon energies.

The first transition is from 6.0 eV to 2.0 eV:

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

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

The second transition is from 2.0 eV to 0 eV:

\[ E_{\gamma 2}=2.0-0 \]

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

The energy ratio is:

\[ 4.0:2.0=2:1 \]

This agrees with the information in the question.

(c) Calculate both wavelengths.

For the first photon:

\[ E_{\gamma 1}=4.0(1.60\times10^{-19}) \]

\[ =6.40\times10^{-19}\text{ J} \]

Using:

\[ \lambda_1=\frac{hc}{E_{\gamma 1}} \]

\[ \lambda_1= \frac{(6.63\times10^{-34})(3.00\times10^8)} {6.40\times10^{-19}} \]

\[ \boxed{\lambda_1=311\text{ nm}} \]

For the second photon:

\[ E_{\gamma 2}=2.0(1.60\times10^{-19}) \]

\[ =3.20\times10^{-19}\text{ J} \]

Therefore:

\[ \lambda_2= \frac{(6.63\times10^{-34})(3.00\times10^8)} {3.20\times10^{-19}} \]

\[ \boxed{\lambda_2=622\text{ nm}} \]

Notice that the photon with twice the energy has half the wavelength.

This question also highlights an important distinction: one atom following this particular cascade emits two photons, even though other possible transitions may exist in the atom.

Question 4

An atom has five energy levels with the following energies:

The only allowed downward transitions are:

• \(E_5\rightarrow E_4\)
• \(E_5\rightarrow E_3\)
• \(E_4\rightarrow E_2\)
• \(E_3\rightarrow E_2\)
• \(E_2\rightarrow E_1\)

An ensemble of atoms is repeatedly excited to \(E_5\).

(a) Determine the number of distinct emission spectral lines that can be produced.

(b) Identify all possible complete cascade pathways from \(E_5\) to \(E_1\).

(c) Can a photon with energy 7 eV be emitted directly during a transition from \(E_5\) to \(E_1\)? Explain.

Solution 4

The atom can produce four distinct spectral lines, and there are two possible complete cascade pathways from \(E_5\) to \(E_1\).

(a) Calculate the number of distinct lines.

First, calculate the photon energy for each allowed transition.

The unique photon energies are:

\[ 1,\ 2,\ 3,\ 4\text{ eV} \]

Therefore:

\[ \boxed{4\text{ distinct spectral lines}} \]

There are five allowed transitions but only four different energy gaps.

The two transitions:

\[ E_5\rightarrow E_4 \]

and

\[ E_4\rightarrow E_2 \]

both release 3 eV.

They therefore produce the same spectral line.

(b) Identify the complete cascade pathways.

The first possible pathway is:

\[ E_5\rightarrow E_4\rightarrow E_2\rightarrow E_1 \]

The emitted photon energies are:

\[ 3\text{ eV},\ 3\text{ eV},\ 1\text{ eV} \]

This cascade produces three photons.

The second possible pathway is:

\[ E_5\rightarrow E_3\rightarrow E_2\rightarrow E_1 \]

The emitted photon energies are:

\[ 4\text{ eV},\ 2\text{ eV},\ 1\text{ eV} \]

This cascade also produces three photons.

These are the only complete pathways because the question specifies which transitions are allowed.

(c) Determine whether a direct 7 eV photon can be emitted.

No, a direct 7 eV photon cannot be emitted.

A direct transition would require:

\[ E_5\rightarrow E_1 \]

The energy difference would be:

\[ E_{\gamma}=7-0=7\text{ eV} \]

However, this transition is not included in the list of allowed transitions.

The electron must instead reach the ground state through the specified intermediate levels.

This illustrates why you cannot automatically use the \(N(N-1)/2\) formula when a question explicitly restricts the allowed transitions.

Question 5

An atom has four non-degenerate energy levels:

\[ E_1=0 \]

\[ E_2=\Delta \]

\[ E_3=3\Delta \]

\[ E_4=4\Delta \]

where \(\Delta\) is a positive energy.

Assume every downward transition is allowed.

A large ensemble of identical atoms is repeatedly excited to \(E_4\), and the emitted light is analysed using a spectrometer.

(a) Calculate the number of possible downward transitions and the number of distinct spectral line positions.

(b) Explain why the spectrometer records fewer distinct lines than the number of possible transitions. Identify which transitions produce identical photon energies.

(c) List every possible complete cascade pathway from \(E_4\) to \(E_1\). State the number of photons emitted in each pathway.

(d) Now suppose an external influence raises the energy of \(E_3\) slightly, changing it to:

\[ E_3=3\Delta+\delta \]

where:

\[ 0<\delta<\frac{\Delta}{2} \]

All other energy levels remain unchanged, and every downward transition is still allowed.

Calculate the six possible photon energy differences after this change. Determine the new number of distinct spectral lines.

Explain why changing just one energy level can increase the number of observed line positions without creating any new allowed transitions.

Solution 5

Initially, there are six possible downward transitions but only four distinct spectral line positions.

After raising \(E_3\), all six transitions have different energy differences, producing six distinct spectral lines.

(a) Calculate the number of possible transitions and distinct lines.

There are four energy levels.

The maximum number of possible downward transitions is:

\[ L=\frac{N(N-1)}{2} \]

\[ L=\frac{4(4-1)}{2} \]

\[ \boxed{L=6} \]

Now calculate the energy differences.

The unique photon energies are:

\[ \Delta,\ 2\Delta,\ 3\Delta,\ 4\Delta \]

Therefore:

\[ \boxed{4\text{ distinct spectral lines}} \]

(b) Explain the repeated spectral lines.

Two pairs of transitions produce identical photon energies.

The first pair is:

\[ E_2\rightarrow E_1 \]

and:

\[ E_4\rightarrow E_3 \]

Both release:

\[ E_{\gamma}=\Delta \]

The second pair is:

\[ E_3\rightarrow E_1 \]

and:

\[ E_4\rightarrow E_2 \]

Both release:

\[ E_{\gamma}=3\Delta \]

Since photon energy determines frequency and wavelength, each pair produces a single spectral line position.

This explains why six possible transitions result in only four distinct lines.

The tempting assumption is that every transition must produce its own spectral line. That is only true when all the energy differences are unique.

(c) List every possible complete cascade pathway.

There are four possible pathways from \(E_4\) to \(E_1\).

The first is a direct transition:

\[ E_4\rightarrow E_1 \]

It releases:

\[ 4\Delta \]

This produces one photon.

The second pathway is:

\[ E_4\rightarrow E_2\rightarrow E_1 \]

It releases:

\[ 3\Delta,\ \Delta \]

This produces two photons.

The third pathway is:

\[ E_4\rightarrow E_3\rightarrow E_1 \]

It releases:

\[ \Delta,\ 3\Delta \]

This also produces two photons.

The fourth pathway is:

\[ E_4\rightarrow E_3\rightarrow E_2\rightarrow E_1 \]

It releases:

\[ \Delta,\ 2\Delta,\ \Delta \]

This produces three photons.

The possible photon counts from one complete cascade are therefore:

An individual atom does not emit all six possible transitions during one decay. It follows one of these pathways.

(d) Determine the effect of raising \(E_3\).

The new energy levels are:

\[ E_1=0 \]

\[ E_2=\Delta \]

\[ E_3=3\Delta+\delta \]

\[ E_4=4\Delta \]

We must recalculate every transition involving \(E_3\), while remembering that the other energy levels have not changed.

The six photon energy differences are now:

The distinct energy differences are:

\[ \Delta-\delta \]

\[ \Delta \]

\[ 2\Delta+\delta \]

\[ 3\Delta \]

\[ 3\Delta+\delta \]

\[ 4\Delta \]

Because:

\[ 0<\delta<\frac{\Delta}{2} \]

these six values are all different.

For example:

\[ \frac{\Delta}{2}<\Delta-\delta<\Delta \]

and:

\[ 2\Delta<2\Delta+\delta<2.5\Delta \]

Similarly:

\[ 3\Delta<3\Delta+\delta<3.5\Delta \]

None of these values coincide with another transition’s energy difference.

Therefore:

\[ \boxed{6\text{ distinct spectral lines}} \]

The subtle point is that changing one energy level affects three different transitions:

• \(E_3\rightarrow E_1\)
• \(E_3\rightarrow E_2\)
• \(E_4\rightarrow E_3\)

The other three transitions remain unchanged.

Initially, two pairs of transitions had identical energy differences. Raising \(E_3\) separates those pairs, so the previously overlapping spectral lines become distinguishable.

However, the number of allowed transitions has not changed. There were six before the adjustment, and there are still six afterwards.

Only their energy differences have changed.

This is why counting transitions and counting distinct spectral lines are related, but not always identical, problems.

The key idea to carry into other spectroscopy questions is that each allowed transition corresponds to an energy difference, and each unique energy difference corresponds to a spectral line position.

For a broader application, see Identifying Elements with Spectral Lines: HSC Physics, which explores how these wavelengths can identify elements.

A particularly useful special case is the Balmer series, where hydrogen electrons make transitions that finish at \(n=2\). See The Balmer Series: How Hydrogen Produces Spectral Lines.

The distinction to remember in an exam is simple: count allowed transitions first, check for repeated energy differences second, and only then state the number of distinct spectral lines.