How to Compare Absorption and Emission Spectra in HSC Physics

Imagine you pass white light through a cool sample of hydrogen gas, then compare its spectrum with the light produced by a hot hydrogen gas. One spectrum has dark lines. The other has bright lines.

The surprising result is that the lines appear at the same wavelengths.

**To compare the absorption and emission spectra of the same element, match their wavelengths. Each absorption line corresponds to an emission line produced by the reverse electron transition.** The electron moves between the same two energy levels in both cases, so the photon has the same energy, frequency, and wavelength.

The difference is the direction of the electron transition and whether the photon is absorbed or emitted.

## 1. The key to comparing the spectra

Start with two energy levels in an atom. Suppose an electron can occupy a lower energy level, \(E_1\), or a higher energy level, \(E_2\).

There are two possible transitions between them:

- **Absorption:** The electron gains energy and moves from \(E_1\) to \(E_2\). A photon is absorbed.
- **Emission:** The electron loses energy and moves from \(E_2\) to \(E_1\). A photon is emitted.

The energy difference is identical in both directions:

\[
\Delta E = E_2-E_1
\]

The photon energy is therefore the same magnitude for absorption and emission:

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

This is the central comparison rule. Reversing a transition changes the direction of energy transfer, not the size of the energy gap.

[IMAGE: An annotated atomic energy-level diagram showing two horizontal levels E1 and E2, with an upward arrow labelled absorption and a downward arrow labelled emission. Show a photon entering during absorption and a photon leaving during emission. Label both photon energies as the same magnitude, E2 minus E1. Keep the two transitions visually distinct but connect them to the same pair of levels.]

### Why do the lines occur at the same wavelength?

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

\[
E_{\text{photon}}=hf
\]

It is also related to wavelength by:

\[
E_{\text{photon}}=\frac{hc}{\lambda}
\]

Here:

- \(h\) is Planck's constant, \(6.63\times10^{-34}\ \text{J s}\).
- \(f\) is the photon frequency, measured in hertz (Hz).
- \(c\) is the speed of light in a vacuum, \(3.00\times10^8\ \text{m s}^{-1}\).
- \(\lambda\) is the photon wavelength, measured in metres (m).

For the same energy difference, the photon must have the same frequency and wavelength.

Therefore:

\[
\boxed{\lambda_{\text{absorption}}=
\lambda_{\text{emission}}}
\]

This assumes the same transition in the same element, with negligible differences from effects such as Doppler shifts or external magnetic fields.

### What changes between the two spectra?

The wavelength positions match, but the appearance of the spectra is different.

- An emission spectrum has bright lines against a dark background. The atoms release photons at particular wavelengths.
- An absorption spectrum has dark lines against a continuous bright background. The atoms remove photons at particular wavelengths from the incoming light.

[IMAGE: A side-by-side comparison of an emission spectrum and an absorption spectrum for the same hypothetical element. Show a horizontal visible-spectrum band in each. Put three narrow bright lines at exactly matching wavelength positions in the emission spectrum, and three narrow dark lines at those positions in the continuous rainbow of the absorption spectrum. Include a shared wavelength scale in nanometres and label the matching lines A, B and C.]

The easiest way to remember the comparison is:

| Feature | Emission | Absorption |
|---|---|---|
| Electron transition | Higher to lower energy | Lower to higher energy |
| Photon interaction | Photon emitted | Photon absorbed |
| Energy transfer | Atom loses energy | Atom gains energy |
| Spectrum | Bright lines | Dark lines |
| Wavelengths | Match the corresponding absorption lines | Match the corresponding emission lines |

For a more detailed explanation of how continuous, emission, and absorption spectra form, see [Continuous, Emission and Absorption Spectra: HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-continuous-emission-and-absorption-spectra/).

## 2. Compare the energy gaps, not the direction of the arrows

A common mistake is to give absorption and emission different photon energies because the electron travels in opposite directions.

The sign of the electron's energy change does reverse. The photon energy, however, is conventionally reported as a positive quantity.

Use this sign convention:

\[
\Delta E_{\text{atom}}=E_{\text{final}}-E_{\text{initial}}
\]

For absorption:

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

The atom gains energy, and the photon is absorbed.

For emission:

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

The atom loses energy, and the photon is emitted.

The photon energy is the magnitude of the atom's energy change:

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

This distinction prevents a second common error: treating a negative energy change as a negative photon energy.

### Worked example: Match an absorption line to an emission line

An electron in a hypothetical atom moves between energy levels of \(-5.0\ \text{eV}\) and \(-2.0\ \text{eV}\).

Calculate the photon energy for absorption and emission. Explain whether the two spectral lines have the same wavelength.

**Step 1**: Identify the initial and final energy levels.

For absorption, the electron moves from \(-5.0\ \text{eV}\) to \(-2.0\ \text{eV}\).

For emission, it moves in the opposite direction.

**Step 2**: Calculate the energy change for absorption.

\[
\begin{aligned}
\Delta E_{\text{atom}}
&=E_{\text{final}}-E_{\text{initial}}\\
&=(-2.0)-(-5.0)\\
&=+3.0\ \text{eV}
\end{aligned}
\]

The atom gains \(3.0\ \text{eV}\), so it absorbs a photon with energy \(3.0\ \text{eV}\).

**Step 3**: Calculate the energy change for emission.

\[
\begin{aligned}
\Delta E_{\text{atom}}
&=(-5.0)-(-2.0)\\
&=-3.0\ \text{eV}
\end{aligned}
\]

The atom loses \(3.0\ \text{eV}\), so it emits a photon with energy \(3.0\ \text{eV}\).

**Step 4**: Compare the photon energies.

\[
E_{\text{absorption}}=
E_{\text{emission}}=3.0\ \text{eV}
\]

The wavelengths are equal because both photons have the same energy.

**Conclusion:** The absorption and emission lines occur at the same wavelength. The negative sign for emission describes the atom's loss of energy, not a negative photon energy.

## 3. A practical method for comparing two spectra

In an experiment, you don't compare the brightness of the two spectra. You compare the positions of their lines along the wavelength scale.

Consider how the two spectra are produced.

For an emission spectrum, a low-pressure gas is excited, often using an electrical discharge. Light from the gas enters a narrow slit, passes through a diffraction grating or prism, and is separated into its component wavelengths.

For an absorption spectrum, a continuous light source shines through a sample of the element in gaseous form. The transmitted light then enters the spectrometer. Some wavelengths have reduced intensity because atoms in the gas have absorbed photons at those wavelengths.

[IMAGE: Two side-by-side experimental apparatus layouts. Upper layout: an electrically excited low-pressure gas discharge tube sending light through a narrow slit and diffraction grating into a detector, producing bright emission lines. Lower layout: a broad-spectrum lamp sending continuous light through a gas absorption cell, then through a narrow slit and diffraction grating into a detector, producing dark absorption lines on a continuous background. Label the direction of light travel and each apparatus component.]

The detector records light intensity against wavelength. An emission line appears as an intensity peak, while an absorption line appears as an intensity dip.

### A five-step comparison method

Use this method when an exam gives you two spectra, a table of wavelengths, or an energy-level diagram.

**Step 1: Put both spectra on the same wavelength scale.**

Check that wavelength increases in the same direction on both diagrams. Make sure you're comparing wavelengths in the same units, such as nanometres.

**Step 2: Match the line positions.**

Look for an absorption line and an emission line at the same wavelength.

Don't match lines based on their brightness, darkness, or apparent colour alone. Their horizontal positions matter.

**Step 3: Allow for measurement uncertainty.**

Experimental measurements aren't perfectly exact. If one line is measured at \(520.0\ \text{nm}\) and another at \(520.2\ \text{nm}\), they may represent the same underlying wavelength.

Check the instrument's resolution and measurement uncertainty before deciding that two lines are different.

**Step 4: Calculate the photon energy if required.**

Use:

\[
E_{\text{photon}}=\frac{hc}{\lambda}
\]

If the two wavelengths match, their photon energies match.

**Step 5: Check the electron transitions.**

Confirm that the absorption and emission transitions connect the same two allowed energy levels in opposite directions.

Also check whether the atoms occupy the appropriate initial energy states. An atom cannot absorb a photon through a transition if it isn't initially in the required lower state.

### Worked example: Compare two measured spectra

A hypothetical element, Element Q, produces the following emission and absorption wavelengths.

All measurements have an uncertainty of approximately \(\pm 0.3\ \text{nm}\).

| Line | Emission wavelength (nm) | Absorption wavelength (nm) |
|---|---:|---:|
| A | 421.0 | 421.2 |
| B | 520.0 | 519.8 |
| C | 612.0 | 612.3 |

Determine whether the spectra are consistent with the same atomic transitions. Calculate the frequency and photon energy associated with Line B.

**Step 1**: Compare the wavelength measurements.

Calculate the difference for each pair:

| Line | Wavelength difference (nm) |
|---|---:|
| A | 0.2 |
| B | 0.2 |
| C | 0.3 |

Each difference is smaller than the conservative combined measurement tolerance of \(0.6\ \text{nm}\).

The corresponding absorption and emission lines are therefore consistent with the same underlying wavelengths.

**Step 2**: Calculate the frequency of Line B.

Use the emission wavelength of \(520.0\ \text{nm}\).

Convert to metres:

\[
\lambda=520.0\times10^{-9}
=5.20\times10^{-7}\ \text{m}
\]

Using \(c=f\lambda\):

\[
\begin{aligned}
f&=\frac{c}{\lambda}\\
&=\frac{3.00\times10^8}
{5.20\times10^{-7}}\\
&=5.77\times10^{14}\ \text{Hz}
\end{aligned}
\]

**Step 3**: Calculate the photon energy.

Using \(E=hf\):

\[
\begin{aligned}
E&=(6.63\times10^{-34})
(5.77\times10^{14})\\
&=3.83\times10^{-19}\ \text{J}
\end{aligned}
\]

**Conclusion:** The two spectra have matching line positions within measurement uncertainty. Line B corresponds to a photon energy of \(3.83\times10^{-19}\ \text{J}\) and a frequency of \(5.77\times10^{14}\ \text{Hz}\).

The same calculation applies to the absorption line because its corresponding photon has the same energy.

## 4. The wavelength, frequency, and energy relationship

There is one more detail worth getting right: wavelength and photon energy change in opposite directions.

From:

\[
E_{\text{photon}}=\frac{hc}{\lambda}
\]

we can see that:

- A shorter wavelength means a higher photon energy.
- A longer wavelength means a lower photon energy.
- A higher frequency means a higher photon energy.
- A lower frequency means a lower photon energy.

[IMAGE: An annotated inverse relationship diagram showing three electromagnetic wavelengths, approximately 420 nm, 520 nm and 610 nm, alongside their corresponding photon energies. Show shorter wavelengths as higher-energy photons and longer wavelengths as lower-energy photons. Include a clear wavelength scale and an energy scale, with the inverse relationship visually apparent.]

This matters when interpreting spectral lines. Two lines from the same element can represent very different energy differences.

### A contrasting example: Same element, different transition

Suppose an emission spectrum contains a line at \(421\ \text{nm}\), while an absorption spectrum contains a line at \(612\ \text{nm}\).

A student might argue that both lines belong to the same element, so they must represent the same transition.

That reasoning is incorrect.

Calculate the photon energies:

For the \(421\ \text{nm}\) photon:

\[
\begin{aligned}
E_1&=\frac{hc}{\lambda}\\
&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{421\times10^{-9}}\\
&=4.72\times10^{-19}\ \text{J}
\end{aligned}
\]

For the \(612\ \text{nm}\) photon:

\[
\begin{aligned}
E_2&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{612\times10^{-9}}\\
&=3.25\times10^{-19}\ \text{J}
\end{aligned}
\]

The energy differences are not equal.

The two lines therefore cannot be the reverse transitions of the same pair of energy levels under the assumptions of this model.

They may still belong to the same element. That element simply has multiple allowed transitions with different energy differences.

**Remember:** Same element does not mean same wavelength for every line. Same transition in opposite directions means the same wavelength.

## 5. Why some lines appear in only one spectrum

There is an important qualification to the matching rule.

The energy levels of an atom determine which photon energies are possible, but not every possible energy gap necessarily produces an observable spectral line.

Two conditions matter:

1. The electron transition must be allowed.
2. The atom must occupy the appropriate initial energy state.

Quantum selection rules determine which transitions are allowed. You do not need to assume that every pair of energy levels can be connected by a transition.

For absorption, the atom must start in the lower state of the transition.

For emission, the atom must start in the higher state.

This explains why an element can have an emission line that does not appear in its absorption spectrum under particular experimental conditions.

It also explains why comparing the number of lines in two spectra isn't always a reliable test of whether they belong to the same element.

The next questions develop these ideas using energy-level diagrams and measured wavelengths.

## Questions and solutions

### Question 1

An element produces an absorption line and an emission line at \(480\ \text{nm}\).

Calculate the energy of each photon and explain how the electron transitions are related.

Use:

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

### Solution 1

Both photons have an energy of \(4.14\times10^{-19}\ \text{J}\), because they have the same wavelength.

**Step 1**: Convert the wavelength into metres.

\[
\lambda=480\times10^{-9}
=4.80\times10^{-7}\ \text{m}
\]

**Step 2**: Calculate the photon energy.

\[
\begin{aligned}
E_{\text{photon}}&=\frac{hc}{\lambda}\\
&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{4.80\times10^{-7}}\\
&=4.14\times10^{-19}\ \text{J}
\end{aligned}
\]

**Step 3**: Compare the electron transitions.

For absorption, the electron moves from a lower energy level to a higher one. The atom gains \(4.14\times10^{-19}\ \text{J}\).

For emission, the electron moves from the higher level to the lower one. The atom loses \(4.14\times10^{-19}\ \text{J}\).

Using the convention \(\Delta E_{\text{atom}}=E_{\text{final}}-E_{\text{initial}}\):

\[
\begin{aligned}
\Delta E_{\text{absorption}}&=+4.14\times10^{-19}\ \text{J}\\
\Delta E_{\text{emission}}&=-4.14\times10^{-19}\ \text{J}
\end{aligned}
\]

The photon energy remains positive in both cases.

The transitions are reverses of each other, so the spectral lines have the same wavelength.

### Question 2

An element has an emission line at \(450\ \text{nm}\) and an absorption line at \(600\ \text{nm}\).

A student claims that these must be reverse transitions because both lines belong to the same element.

Calculate the energy of each photon and determine whether the student's claim is correct. By what factor is the larger energy difference greater than the smaller one?

### Solution 2

The student's claim is incorrect. The two wavelengths correspond to different energy differences. The \(450\ \text{nm}\) photon has approximately \(1.33\) times the energy of the \(600\ \text{nm}\) photon.

**Step 1**: Calculate the energy of the \(450\ \text{nm}\) photon.

\[
\begin{aligned}
E_{450}&=\frac{hc}{\lambda}\\
&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{450\times10^{-9}}\\
&=4.42\times10^{-19}\ \text{J}
\end{aligned}
\]

**Step 2**: Calculate the energy of the \(600\ \text{nm}\) photon.

\[
\begin{aligned}
E_{600}&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{600\times10^{-9}}\\
&=3.32\times10^{-19}\ \text{J}
\end{aligned}
\]

**Step 3**: Calculate the energy ratio.

\[
\begin{aligned}
\frac{E_{450}}{E_{600}}
&=\frac{4.42\times10^{-19}}
{3.32\times10^{-19}}\\
&=1.33
\end{aligned}
\]

The \(450\ \text{nm}\) photon has approximately \(33\%\) more energy.

The tempting mistake is to assume that any two lines from the same element must match. However, the element has multiple allowed energy differences.

Only the absorption and emission lines associated with the same energy difference can be reverse transitions.

### Question 3

A hypothetical element has three energy levels:

\[
E_0=0,\qquad E_1=1.9\ \text{eV},
\qquad E_2=3.3\ \text{eV}
\]

Assume all transitions between these levels are allowed.

An absorption experiment uses a continuous source covering wavelengths from \(400\ \text{nm}\) to \(700\ \text{nm}\). Nearly all atoms initially occupy \(E_0\).

A separate emission experiment excites atoms to \(E_2\). The emission detector records wavelengths from \(400\ \text{nm}\) to \(900\ \text{nm}\).

Determine:

(a) Which absorption lines will be recorded.

(b) Which emission lines will be recorded.

(c) Which wavelengths appear in both spectra, and why do the remaining emission lines not appear in the absorption spectrum?

[IMAGE: A question setup diagram with three horizontal atomic energy levels labelled E0 = 0 eV, E1 = 1.9 eV and E2 = 3.3 eV. Show the relative vertical positions accurately. Do not draw transition arrows or provide calculated wavelengths.]

### Solution 3

The absorption spectrum contains a line at approximately \(653\ \text{nm}\). The emission spectrum contains lines at approximately \(653\ \text{nm}\) and \(886\ \text{nm}\). The \(653\ \text{nm}\) line is shared.

**Step 1**: Determine the possible absorption transitions.

Nearly all atoms initially occupy \(E_0\).

Therefore, the relevant upward transitions are:

\[
E_0\rightarrow E_1
\]

and

\[
E_0\rightarrow E_2
\]

For the first transition:

\[
\Delta E=1.9\ \text{eV}
\]

Convert to joules:

\[
\Delta E=(1.9)(1.60\times10^{-19})
=3.04\times10^{-19}\ \text{J}
\]

Calculate the wavelength:

\[
\begin{aligned}
\lambda&=\frac{hc}{\Delta E}\\
&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{3.04\times10^{-19}}\\
&=6.54\times10^{-7}\ \text{m}\\
&=654\ \text{nm}
\end{aligned}
\]

For the second transition:

\[
\Delta E=3.3\ \text{eV}
=5.28\times10^{-19}\ \text{J}
\]

\[
\begin{aligned}
\lambda&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{5.28\times10^{-19}}\\
&=3.77\times10^{-7}\ \text{m}\\
&=377\ \text{nm}
\end{aligned}
\]

The \(377\ \text{nm}\) wavelength is outside the source's \(400\)-\(700\ \text{nm}\) range.

Therefore, only the approximately \(654\ \text{nm}\) absorption line is recorded.

**Step 2**: Determine the possible emission transitions.

Atoms initially occupy \(E_2\).

They can fall directly to \(E_0\):

\[
E_2\rightarrow E_0
\]

This releases \(3.3\ \text{eV}\), producing a photon at approximately \(377\ \text{nm}\).

Alternatively, they can fall to \(E_1\):

\[
E_2\rightarrow E_1
\]

The energy released is:

\[
\Delta E=3.3-1.9=1.4\ \text{eV}
\]

Converting to joules:

\[
\Delta E=2.24\times10^{-19}\ \text{J}
\]

The wavelength is:

\[
\begin{aligned}
\lambda&=\frac{hc}{\Delta E}\\
&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{2.24\times10^{-19}}\\
&=8.88\times10^{-7}\ \text{m}\\
&=888\ \text{nm}
\end{aligned}
\]

The electron can then fall from \(E_1\) to \(E_0\), producing the approximately \(654\ \text{nm}\) photon calculated earlier.

**Step 3**: Apply the detector's wavelength range.

The emission detector records \(400\)-\(900\ \text{nm}\).

- \(377\ \text{nm}\) is outside the range.
- \(888\ \text{nm}\) is inside the range.
- \(654\ \text{nm}\) is inside the range.

Therefore, the recorded emission lines are approximately \(654\ \text{nm}\) and \(888\ \text{nm}\).

**Step 4**: Explain the difference between the spectra.

The \(654\ \text{nm}\) line appears in both spectra because it corresponds to the reverse transitions \(E_0\rightarrow E_1\) and \(E_1\rightarrow E_0\).

The \(888\ \text{nm}\) emission line corresponds to \(E_2\rightarrow E_1\). Its reverse absorption transition would require an atom to start in \(E_1\).

However, nearly all atoms in the absorption experiment start in \(E_0\).

This is why the \(888\ \text{nm}\) line does not appear in that absorption spectrum.

### Question 4

A hypothetical atom has three energy levels:

\[
E_0=0,\qquad E_1=2.0\ \text{eV},
\qquad E_2=5.0\ \text{eV}
\]

The following transitions are allowed:

- \(E_0\leftrightarrow E_1\)
- \(E_1\leftrightarrow E_2\)

The direct transition \(E_0\leftrightarrow E_2\) is forbidden.

A gas sample is initially prepared with nearly all atoms in \(E_0\) for an absorption experiment. A separate sample is excited so that atoms initially occupy \(E_2\) for an emission experiment.

Assume the detectors cover wavelengths from \(400\ \text{nm}\) to \(700\ \text{nm}\).

(a) Calculate the wavelengths of the allowed transitions.

(b) Identify the lines expected in each spectrum.

(c) A student predicts an additional line at approximately \(248\ \text{nm}\), based on the total energy difference between \(E_0\) and \(E_2\). Explain why this prediction is incorrect.

Use \(1\ \text{eV}=1.60\times10^{-19}\ \text{J}\).

[IMAGE: An energy-level diagram for Question 4 with E0 at 0 eV, E1 at 2.0 eV and E2 at 5.0 eV. Show a connector between E0 and E1 and another between E1 and E2, indicating the permitted level pairs without arrowheads. Show no direct connector between E0 and E2. Do not include wavelengths or transition directions.]

### Solution 4

The absorption spectrum contains a line at \(620\ \text{nm}\). The emission spectrum contains lines at approximately \(413\ \text{nm}\) and \(620\ \text{nm}\). There is no direct \(248\ \text{nm}\) transition because \(E_0\leftrightarrow E_2\) is forbidden.

**Step 1**: Calculate the wavelength for \(E_0\leftrightarrow E_1\).

The energy difference is:

\[
\Delta E=2.0\ \text{eV}
\]

Convert to joules:

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

Therefore:

\[
\begin{aligned}
\lambda&=\frac{hc}{\Delta E}\\
&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{3.20\times10^{-19}}\\
&=6.22\times10^{-7}\ \text{m}\\
&=622\ \text{nm}
\end{aligned}
\]

**Step 2**: Calculate the wavelength for \(E_1\leftrightarrow E_2\).

The energy difference is:

\[
\Delta E=5.0-2.0=3.0\ \text{eV}
\]

Convert to joules:

\[
\Delta E=4.80\times10^{-19}\ \text{J}
\]

Therefore:

\[
\begin{aligned}
\lambda&=\frac{hc}{\Delta E}\\
&=\frac{(6.63\times10^{-34})(3.00\times10^8)}
{4.80\times10^{-19}}\\
&=4.14\times10^{-7}\ \text{m}\\
&=414\ \text{nm}
\end{aligned}
\]

Both wavelengths are within the detector's range.

**Step 3**: Determine the absorption spectrum.

Nearly all atoms start in \(E_0\).

The only allowed upward transition from this state is:

\[
E_0\rightarrow E_1
\]

Therefore, the absorption spectrum contains a line at approximately \(622\ \text{nm}\).

The atom cannot directly absorb a photon to move from \(E_0\) to \(E_2\), because that transition is forbidden under the stated conditions.

**Step 4**: Determine the emission spectrum.

Atoms initially at \(E_2\) can undergo:

\[
E_2\rightarrow E_1
\]

producing a photon at approximately \(414\ \text{nm}\).

They can then undergo:

\[
E_1\rightarrow E_0
\]

producing a photon at approximately \(622\ \text{nm}\).

Therefore, the emission spectrum contains both wavelengths.

**Step 5**: Explain why the \(248\ \text{nm}\) prediction is wrong.

The student has calculated the total energy difference:

\[
E_2-E_0=5.0\ \text{eV}
\]

This would correspond to a wavelength of approximately \(248\ \text{nm}\).

However, calculating an energy difference does not prove that the corresponding transition is allowed.

The atom cannot make the direct \(E_0\leftrightarrow E_2\) transition under the conditions specified.

Instead, it moves through the intermediate level \(E_1\), releasing energy in two separate photons.

This distinction is important: an energy gap can exist mathematically without producing an observable spectral line.

### Question 5

An element has four allowed energy levels:

\[
E_0=0,\qquad E_1=1.5\ \text{eV}
\]

\[
E_2=3.0\ \text{eV},\qquad E_3=4.2\ \text{eV}
\]

The allowed transitions are:

\[
E_0\leftrightarrow E_1
\]

\[
E_1\leftrightarrow E_2
\]

\[
E_2\leftrightarrow E_3
\]

\[
E_0\leftrightarrow E_2
\]

\[
E_1\leftrightarrow E_3
\]

The direct transition \(E_0\leftrightarrow E_3\) is forbidden.

Two experiments are performed:

- In the absorption experiment, nearly all atoms initially occupy \(E_0\).
- In the emission experiment, atoms are excited to \(E_3\) and can decay through any of the allowed downward transitions.

Assume the spectrometer can detect wavelengths from \(400\ \text{nm}\) to \(1100\ \text{nm}\).

(a) Calculate all distinct wavelengths expected in the absorption spectrum.

(b) Determine all possible emission wavelengths, including those produced during cascades through intermediate levels.

(c) Compare the two spectra.

(d) A student counts five allowed downward transitions and predicts five distinct emission lines. Explain why this prediction is incorrect.

(e) One emission line matches an absorption line at approximately \(827\ \text{nm}\). Can you conclude that every photon producing this emission line comes from the reverse of the corresponding absorption transition? Justify your answer.

[IMAGE: A detailed energy-level setup for Question 5. Show four horizontal levels at E0 = 0 eV, E1 = 1.5 eV, E2 = 3.0 eV and E3 = 4.2 eV, with vertical spacing proportional to energy. Connect the five permitted level pairs using neutral, undirected connectors: E0-E1, E1-E2, E2-E3, E0-E2 and E1-E3. Do not draw transition arrows, calculate wavelengths, or identify matching spectral lines.]

### Solution 5

The absorption spectrum contains two distinct wavelengths, approximately \(413\ \text{nm}\) and \(827\ \text{nm}\).

The emission spectrum contains four distinct wavelengths, approximately \(459\ \text{nm}\), \(413\ \text{nm}\), \(827\ \text{nm}\), and \(1033\ \text{nm}\).

The key complication is that two different transitions have the same energy difference. They therefore produce photons with the same wavelength, so five allowed transitions produce only four distinct spectral lines.

**Step 1**: Calculate the absorption wavelengths.

Nearly all atoms initially occupy \(E_0\).

The allowed upward transitions from this state are:

\[
E_0\rightarrow E_1
\]

and

\[
E_0\rightarrow E_2
\]

For the first transition:

\[
\Delta E=1.5\ \text{eV}
\]

Use the convenient conversion:

\[
\lambda(\text{nm})\approx
\frac{1240}{E(\text{eV})}
\]

This follows from \(E=hc/\lambda\) and the conversion between joules and electronvolts.

Therefore:

\[
\lambda=\frac{1240}{1.5}
=827\ \text{nm}
\]

For the second transition:

\[
\Delta E=3.0\ \text{eV}
\]

\[
\lambda=\frac{1240}{3.0}
=413\ \text{nm}
\]

The direct transition \(E_0\rightarrow E_3\) is forbidden, so there is no absorption line corresponding to that energy gap.

The absorption spectrum therefore contains:

\[
\boxed{413\ \text{nm},\quad827\ \text{nm}}
\]

**Step 2**: Calculate the possible emission wavelengths.

Atoms initially occupy \(E_3\).

There are two possible first transitions.

For:

\[
E_3\rightarrow E_2
\]

the energy released is:

\[
\Delta E=4.2-3.0=1.2\ \text{eV}
\]

Therefore:

\[
\lambda=\frac{1240}{1.2}
=1033\ \text{nm}
\]

For:

\[
E_3\rightarrow E_1
\]

the energy released is:

\[
\Delta E=4.2-1.5=2.7\ \text{eV}
\]

Therefore:

\[
\lambda=\frac{1240}{2.7}
=459\ \text{nm}
\]

Both are within the detector's wavelength range.

**Step 3**: Follow the possible cascades.

An electron that reaches \(E_2\) can make either of two further transitions.

For:

\[
E_2\rightarrow E_1
\]

\[
\Delta E=3.0-1.5=1.5\ \text{eV}
\]

Therefore:

\[
\lambda=\frac{1240}{1.5}
=827\ \text{nm}
\]

For:

\[
E_2\rightarrow E_0
\]

\[
\Delta E=3.0-0=3.0\ \text{eV}
\]

Therefore:

\[
\lambda=\frac{1240}{3.0}
=413\ \text{nm}
\]

An electron that reaches \(E_1\) can then fall to \(E_0\):

\[
E_1\rightarrow E_0
\]

The energy released is:

\[
\Delta E=1.5\ \text{eV}
\]

Therefore:

\[
\lambda=827\ \text{nm}
\]

We have now identified five possible downward transitions:

| Downward transition | Energy released (eV) | Wavelength (nm) |
|---|---:|---:|
| \(E_3\rightarrow E_2\) | 1.2 | 1033 |
| \(E_3\rightarrow E_1\) | 2.7 | 459 |
| \(E_2\rightarrow E_1\) | 1.5 | 827 |
| \(E_2\rightarrow E_0\) | 3.0 | 413 |
| \(E_1\rightarrow E_0\) | 1.5 | 827 |

**Step 4**: Explain why there are only four distinct emission lines.

The student has correctly counted five allowed downward transitions but has incorrectly assumed that every transition must produce a different wavelength.

Notice that:

\[
E_2-E_1=1.5\ \text{eV}
\]

and

\[
E_1-E_0=1.5\ \text{eV}
\]

These are two different transitions with exactly the same energy difference.

Since photon energy determines wavelength, both transitions produce photons at \(827\ \text{nm}\).

The spectrometer records one line position at \(827\ \text{nm}\), even though photons at that wavelength can come from two different transitions.

The emission spectrum therefore has four distinct line positions, not five.

**Step 5**: Determine what the matching \(827\ \text{nm}\) line actually tells us.

The absorption line at \(827\ \text{nm}\) comes from:

\[
E_0\rightarrow E_1
\]

Its exact reverse is:

\[
E_1\rightarrow E_0
\]

However, the emission spectrum also contains photons at \(827\ \text{nm}\) from:

\[
E_2\rightarrow E_1
\]

Both transitions release \(1.5\ \text{eV}\).

Therefore, matching wavelengths establish a matching energy difference, but they do not always identify a unique pair of energy levels.

This is the subtle distinction. The reverse of a particular transition always has the same photon energy, but two different transitions can independently have equal energy differences.

**Step 6**: Compare the complete spectra.

| Spectrum | Distinct wavelengths |
|---|---|
| Absorption | 413 nm, 827 nm |
| Emission | 413 nm, 459 nm, 827 nm, 1033 nm |

The \(413\ \text{nm}\) absorption line corresponds to the reverse of the \(E_2\rightarrow E_0\) emission transition.

The \(827\ \text{nm}\) absorption line corresponds to the reverse of \(E_1\rightarrow E_0\), but the emission line also receives contributions from \(E_2\rightarrow E_1\).

The \(459\ \text{nm}\) and \(1033\ \text{nm}\) emission lines have no corresponding absorption lines in this experiment because their reverse transitions require atoms to initially occupy \(E_1\) or \(E_2\). Nearly all absorption atoms start in \(E_0\).

This example shows why comparing complete energy-level structures is more reliable than simply counting lines or matching one wavelength.

## 6. What to remember in an HSC response

When comparing absorption and emission spectra, keep these relationships clear:

| What you're comparing | Correct interpretation |
|---|---|
| Same transition, opposite directions | Same photon energy and wavelength |
| Absorption electron transition | Atom gains energy |
| Emission electron transition | Atom loses energy |
| Photon energy | Positive in both processes |
| Shorter wavelength | Higher frequency and photon energy |
| Longer wavelength | Lower frequency and photon energy |
| Missing absorption line | The required lower state may not be populated |
| Different numbers of lines | May result from different populations, allowed transitions, or overlapping wavelengths |
| Same wavelength from two transitions | Possible if the energy differences are equal |

For the calculation, remember:

\[
\boxed{E_{\text{photon}}=hf=\frac{hc}{\lambda}}
\]

And for the electron:

\[
\boxed{\Delta E_{\text{atom}}
=E_{\text{final}}-E_{\text{initial}}}
\]

The first equation gives the positive photon energy. The second tells you whether the atom gains or loses energy.

One final experimental qualification: real spectral lines may be broadened or shifted by the motion of the gas, collisions, external fields, and the resolution of the spectrometer. The comparison method assumes that these effects are negligible or have been accounted for.

Once you can match absorption and emission lines using energy differences, the next useful step is learning how to use a complete spectral pattern to identify an unknown element. That requires looking beyond individual wavelengths and understanding how the full set of allowed atomic transitions creates a characteristic spectrum.