Peak Wavelength vs Only Wavelength Emitted: How to Tell

A heated filament has a spectrum with its highest intensity at \(650\text{ nm}\). Does that mean it emits only red light at \(650\text{ nm}\)?

No. For a thermal source, the peak wavelength is simply the wavelength at which its continuous spectrum reaches maximum intensity. The source also emits radiation at many shorter and longer wavelengths.

A gas discharge tube behaves differently. It can emit light at specific wavelengths, producing separate spectral lines rather than a broad thermal spectrum.

The distinction is:

• Peak wavelength: the wavelength at which a continuous spectrum has its greatest spectral intensity.
• Only wavelength emitted: an idealised monochromatic source emits photons at one characteristic wavelength. A gas producing a single isolated emission line is an example of this model.

To distinguish them properly, you need to understand what a spectrum measures, what its peak represents, and how the area under its graph relates to total emitted power.

01Two different types of emission spectrum

Imagine comparing two light sources.

The first is a piece of hot metal. As its temperature increases, it emits electromagnetic radiation across a broad range of wavelengths. Its spectrum forms a smooth curve, with one wavelength producing the greatest spectral intensity.

The second is a low-pressure gas inside a discharge tube. Electrical energy excites its atoms, and electrons can move between specific energy levels. When electrons return to lower energy levels, they emit photons with particular energies.

Because photon energy is related to wavelength, these transitions produce characteristic emission lines.

What would you predict?

Suppose both sources produce a very tall peak at \(600\text{ nm}\).

Would that make them the same type of emitter?

No. The shape of the surrounding spectrum matters.

• The hot metal might have a broad curve that reaches its maximum at \(600\text{ nm}\).
• The gas might produce a narrow emission line centred at \(600\text{ nm}\), with very little emission at neighbouring wavelengths.

The numerical wavelength is the same, but its physical meaning is different.

Compare the spectra

One important qualification: real emission lines have a small but finite width. Their appearance depends partly on the physical conditions of the gas and the resolution of the measuring instrument.

An idealised single-wavelength source is a useful model, not a claim that every real source emits at one mathematically exact wavelength.

02How an experiment distinguishes the two

The distinction becomes much clearer when you measure the complete spectrum rather than looking at the source directly.

A spectrometer separates light into its component wavelengths and measures the intensity at each one.

Consider the following experimental arrangement.

Step 1: Prepare the light source

First, direct light from the source towards a narrow slit.

The slit produces a narrow beam, which passes through a collimating lens. The lens makes the light rays approximately parallel before they reach the diffraction grating.

The grating separates different wavelengths by diffracting them at different angles.

Step 2: Measure the different wavelengths

Place a detector where it can measure the separated light.

By moving the detector through different angles, you can measure the intensity at different wavelengths.

A calibration using known spectral lines allows each detector position to be associated with a wavelength.

The detector records the signal, giving you a set of measurements that can be plotted as spectral intensity against wavelength.

Step 3: Predict the results

If the source is a hot filament, you expect a broad distribution of intensities.

There should be a wavelength where the spectral intensity is greatest, but neighbouring wavelengths should also produce measurable signals.

If the source is a low-pressure gas discharge tube, you expect narrow peaks at particular wavelengths. Between these peaks, the measured intensity may be very small.

The important observation is not simply which wavelength produces the largest reading. It is whether the source produces a continuous distribution or separate spectral lines.

Why a screen alone is not enough

An early, simpler arrangement could use a prism or diffraction grating and a screen to display the spectrum.

This shows the colours or emission lines, but it does not provide a reliable quantitative comparison of their intensities. The human eye also responds differently to different wavelengths.

To improve the measurement, replace the screen with a calibrated detector.

The detector should have a known wavelength response, consistent measurement times, and fixed collection geometry. Its response must be corrected before comparing the actual spectral intensities.

This allows you to distinguish the shape of the spectrum from the limitations of the measuring equipment.

03Three quantities you must keep separate

When interpreting a spectral graph, three different measurements matter.

They are spectral intensity, peak wavelength, and integrated power.

Confusing them is one of the easiest ways to misinterpret a spectrum.

Spectral intensity: how strongly is each wavelength represented?

Spectral intensity describes how much radiation is emitted within a small wavelength interval.

For a quantitative surface-emission graph, we can represent this using \(I_\lambda\), the spectral emission per unit surface area per unit wavelength.

For example, its units could be:

\[ \text{W m}^{-2}\text{ nm}^{-1} \]

A value of \(0.050\text{ W m}^{-2}\text{ nm}^{-1}\) near \(600\text{ nm}\) means that a wavelength interval of approximately \(1\text{ nm}\) around that point carries about \(0.050\text{ W m}^{-2}\), assuming the intensity is nearly constant across the interval.

The important point is that the vertical axis describes the distribution of radiation across wavelengths.

It does not directly give the total power emitted by the source.

Peak wavelength: where is the graph highest?

The peak wavelength, written as \(\lambda_{\max}\), is the horizontal coordinate corresponding to the greatest spectral intensity.

For a thermal spectrum:

\[ I_\lambda \text{ is maximum at } \lambda_{\max} \]

Notice that the peak wavelength is a position on the horizontal axis.

The peak intensity is a height on the vertical axis.

They are different quantities.

For a blackbody, the peak wavelength can be used to estimate temperature through Wien’s displacement law:

\[ \lambda_{\max}T=b \]

where:

• \(\lambda_{\max}\) is the peak wavelength in metres.
• \(T\) is absolute temperature in kelvin.
• \(b\) is Wien’s displacement constant, approximately \(2.90\times10^{-3}\text{ m K}\).

This relationship applies to the thermal continuum, not an isolated atomic emission line.

Integrated power: how much radiation is emitted altogether?

Now consider the entire area beneath the spectral intensity curve.

This area represents the total emitted power per unit surface area across the wavelength range being measured.

A tall, narrow peak can have a relatively small area.

A shorter, broader curve can have a much larger area.

That means the source with the greatest spectral intensity at one wavelength does not necessarily emit the greatest total power.

For a measured spectrum, the area can be estimated by dividing it into small sections and adding their areas.

For example, using trapezoids:

\[ R_{\text{band}}\approx \sum \frac{I_{\lambda_i}+I_{\lambda_{i+1}}}{2} \Delta\lambda \]

Here:

• \(R_{\text{band}}\) is the emitted power per unit surface area across the measured wavelength range, in \(\text{W m}^{-2}\).
• \(I_{\lambda_i}\) and \(I_{\lambda_{i+1}}\) are neighbouring spectral intensity measurements.
• \(\Delta\lambda\) is the wavelength interval between them, in nanometres.

If the emitting surface has area \(A_s\), the power emitted across that wavelength band is:

\[ P_{\text{band}}=A_sR_{\text{band}} \]

This calculation assumes the spectral intensity represents the source’s emitted power per unit surface area, rather than an uncorrected detector reading.

Keep the three ideas separate

A useful rule is:

Peak wavelength tells you where the spectrum is strongest. Area under the spectrum tells you how much radiation is emitted altogether.

05What happens when a source emits at specific wavelengths?

Now return to the gas discharge tube.

Unlike a hot solid, an excited gas can produce a line spectrum because its atoms have discrete allowed energy levels.

When an electron moves from a higher energy level to a lower one, it emits a photon.

The photon energy is:

\[ E=hf \]

Using the relationship between frequency and wavelength:

\[ c=f\lambda \]

we can also write:

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

where:

• \(E\) is photon energy in joules.
• \(h\) is Planck’s constant, \(6.63\times10^{-34}\text{ J s}\).
• \(c\) is the speed of light, \(3.00\times10^8\text{ m s}^{-1}\).
• \(\lambda\) is photon wavelength in metres.

Each allowed transition corresponds to a characteristic photon energy and therefore a characteristic wavelength.

Why this is different from a thermal peak

Suppose a gas produces a strong emission line at \(656\text{ nm}\).

That wavelength tells you about the energy difference between the relevant atomic energy levels.

It does not tell you the temperature of the gas using Wien’s displacement law.

The gas is not producing a broad blackbody distribution with a maximum at \(656\text{ nm}\). It is producing a characteristic emission line.

If several transitions occur, the gas may produce several lines.

The tallest line represents the greatest spectral intensity at one of those transitions, not the peak wavelength of a thermal continuum.

Also remember that the height of an emission line and its integrated power are different. A very narrow line can have a large maximum intensity but a relatively small total area.

06The tricky case: An emission line sitting on a thermal spectrum

So far, we’ve considered a thermal source and a gas discharge tube separately.

But what happens when both contribute to the same measured spectrum?

Imagine a hot surface viewed alongside an excited gas. The surface produces a broad thermal continuum, while the gas adds a narrow emission line.

The measured spectrum is the sum of the two contributions:

\[ I_{\lambda,\text{total}} = I_{\lambda,\text{thermal}} + I_{\lambda,\text{line}} \]

This creates an important complication.

The narrow emission line might be taller than the thermal continuum’s maximum.

If you simply select the highest point on the combined graph, you could identify the emission line rather than the thermal peak.

Which wavelength should you use?

If the question asks for the maximum intensity of the entire measured spectrum, you identify the highest point of the combined graph.

However, if the question asks for the temperature of the thermal emitter using Wien’s displacement law, you must identify the peak of the underlying thermal continuum.

These are not necessarily the same wavelength.

You need to separate the two contributions before applying Wien’s law.

There is a related situation with absorption lines. A cooler gas can absorb particular wavelengths from a continuous spectrum, creating narrow dips rather than additional peaks.

Those dips do not mean the thermal source emits only the wavelengths outside them. They indicate that particular wavelengths have been absorbed.

07A reliable method for interpreting spectral graphs

When you encounter a spectral graph in an HSC Physics question, use this sequence.

1. Read the axes.

Check whether the horizontal axis represents wavelength and whether the vertical axis represents spectral intensity per unit wavelength, total power, or detector counts.

The units matter.

2. Identify the shape.

Look for a broad, smooth thermal continuum or separate narrow emission lines.

A combination of both is also possible.

3. Find the correct peak.

For a thermal source, identify the maximum of the underlying continuous spectrum.

For a line spectrum, identify the individual line wavelengths and their relative intensities.

4. Decide what the question is asking for.

• Temperature: use the thermal continuum’s peak wavelength.
• Photon energy: use the wavelength of the relevant emission line.
• Total emitted power: integrate the spectral intensity across the relevant wavelength range.
• Strongest emission line: compare the maximum intensities of the individual lines.

5. Check whether the measuring equipment could affect your conclusion.

A low-resolution spectrometer might merge closely spaced lines into one apparent feature.

An uncorrected detector might also make some wavelengths appear stronger than they really are.

Do not assume every feature in a measured graph directly represents the source’s true spectrum.

Connecting peak wavelength to total power

Once you understand the difference between peak position and integrated emission, two important radiation laws become easier to distinguish.

Wien’s displacement law:

\[ \lambda_{\max}T=b \]

relates the peak wavelength of a blackbody spectrum to its temperature.

The Stefan-Boltzmann law is a useful extension:

\[ P=\sigma A_sT^4 \]

where:

• \(P\) is the total power emitted in watts.
• \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67\times10^{-8}\text{ W m}^{-2}\text{ K}^{-4}\).
• \(A_s\) is the emitting surface area in square metres.
• \(T\) is absolute temperature in kelvin.

This law applies to an ideal blackbody and describes its total emission across all wavelengths.

The distinction is simple:

• Wien’s law tells you where the spectrum peaks.
• Stefan-Boltzmann tells you the total power emitted.

Neither law should be applied directly to an isolated atomic emission line.

For a more detailed treatment of finding temperature from a thermal spectrum, see Wien’s Displacement Law: Finding Temperature from Peak Wavelength.

08Questions and solutions

Question 1

A heated metal object produces a continuous spectrum with the following measured values.

Determine the peak wavelength.

A student claims that the object emits only radiation at this wavelength.

Explain why the student’s conclusion is incorrect.

Solution 1

The peak wavelength is \(600\text{ nm}\), but the object emits radiation across a continuous range of wavelengths.

The peak wavelength is found by identifying the greatest spectral intensity in the table.

The maximum value is:

\[ I_{\lambda,\max}=5.8 \]

which occurs at:

\[ \boxed{\lambda_{\max}=600\text{ nm}} \]

However, the other wavelengths also have non-zero spectral intensities.

This means the object emits radiation at wavelengths both shorter and longer than \(600\text{ nm}\).

The student has confused the wavelength of maximum intensity with the only wavelength emitted.

For a thermal source, the peak identifies the strongest part of a continuous distribution, not the boundaries of its emission.

Question 2

Two light sources are measured using the same calibrated spectrometer.

Their spectra can each be approximated by a triangular shape above a zero baseline.

• Source A produces a narrow emission line with a maximum spectral intensity of \(12\text{ W m}^{-2}\text{ nm}^{-1}\) and a full width of \(2.0\text{ nm}\).
• Source B produces a broad continuous spectrum with a maximum spectral intensity of \(1.0\text{ W m}^{-2}\text{ nm}^{-1}\) and a full width of \(100\text{ nm}\).

For each source:

1. Calculate the integrated irradiance across its measured spectral feature.
2. Identify which source has the greater peak spectral intensity.
3. Identify which source emits more energy across its measured wavelength range.

Explain why the source with the highest peak intensity does not necessarily emit the most total energy.

Solution 2

Source A has the greater peak spectral intensity, but Source B has the greater integrated irradiance.

The difference comes from the area beneath each spectrum.

Step 1: Calculate the integrated irradiance of Source A.

For a triangular spectrum:

\[ R=\frac{1}{2}\times\text{base}\times\text{height} \]

Substitute the values:

\[ R_A=\frac{1}{2}(2.0)(12) \]

\[ \boxed{R_A=12\text{ W m}^{-2}} \]

Step 2: Calculate the integrated irradiance of Source B.

\[ R_B=\frac{1}{2}(100)(1.0) \]

\[ \boxed{R_B=50\text{ W m}^{-2}} \]

Step 3: Compare the peak intensities.

Source A has:

\[ I_{\lambda,\max,A}=12 \]

Source B has:

\[ I_{\lambda,\max,B}=1.0 \]

Therefore, Source A has twelve times the peak spectral intensity of Source B.

Step 4: Compare the integrated emission.

\[ \frac{R_B}{R_A}=\frac{50}{12} \]

\[ \boxed{\frac{R_B}{R_A}=4.17} \]

Source B has approximately \(4.17\) times the integrated irradiance of Source A across the measured features.

The tempting mistake is to assume that the tallest peak must represent the greatest total emission.

But Source A’s peak is extremely narrow, while Source B emits over a much wider wavelength range.

The integrated area accounts for both the intensity and the width of the spectrum.

This is why peak height and total emitted power must be treated separately.

Question 3

A low-pressure gas discharge tube is known to emit two narrow spectral lines:

\[ \lambda_1=589.0\text{ nm} \]

\[ \lambda_2=589.6\text{ nm} \]

A spectrometer has a wavelength resolution of \(2.0\text{ nm}\).

The measured spectrum shows one broad feature centred at approximately \(589.3\text{ nm}\).

A student concludes that the gas emits only one wavelength, \(589.3\text{ nm}\), and that this must be its peak wavelength.

Assess this conclusion.

What improvement to the experimental arrangement would allow the two emission lines to be distinguished?

Solution 3

The student’s conclusion is incorrect. The single measured feature could be two unresolved emission lines rather than one wavelength emitted by the gas.

Step 1: Calculate the separation between the actual emission lines.

\[ \Delta\lambda=589.6-589.0 \]

\[ \boxed{\Delta\lambda=0.6\text{ nm}} \]

Step 2: Compare the separation with the instrument resolution.

The spectrometer has a resolution of \(2.0\text{ nm}\).

The two lines are separated by only \(0.6\text{ nm}\).

Because their separation is smaller than the instrument’s resolving capability, the detector cannot distinguish them as separate features.

Instead, their signals overlap and appear as one broader peak.

Step 3: Improve the resolution.

Use a spectrometer with a resolution substantially finer than \(0.6\text{ nm}\).

For example, a resolution of \(0.2\text{ nm}\) should allow the two lines to be distinguished under suitable measurement conditions.

Step 4: Interpret the result.

The measured centre at \(589.3\text{ nm}\) does not establish that the gas emits a single transition at that wavelength.

It is consistent with two closely spaced emission lines being merged by the instrument.

This question highlights an important experimental limitation: the appearance of a spectrum depends partly on the resolution of the measuring equipment.

A measured broad feature is not automatically proof of a continuous thermal spectrum.

Question 4

A heated surface produces a broad thermal continuum with a peak wavelength of \(680\text{ nm}\).

Its light passes through a cooler gas before reaching a detector.

The gas absorbs radiation at \(656\text{ nm}\), producing a narrow absorption feature.

The calibrated measurements give:

• Total continuum irradiance before absorption, across the measured wavelength band: \(180\text{ W m}^{-2}\).
• Integrated irradiance removed by the absorption feature: \(12\text{ W m}^{-2}\).

Assume the detector measures the transmitted light and collects negligible radiation re-emitted by the gas.

1. Calculate the percentage of the original band irradiance absorbed by the gas.
2. Explain why \(656\text{ nm}\) is not the thermal peak wavelength.
3. Identify the wavelength that should be used to estimate the heated surface’s temperature using Wien’s displacement law.

Solution 4

The gas absorbs approximately \(6.7\%\) of the original band irradiance, and the thermal peak wavelength remains \(680\text{ nm}\).

Step 1: Calculate the percentage of irradiance absorbed.

Use:

\[ \text{Percentage absorbed} = \frac{\text{absorbed irradiance}} {\text{original irradiance}} \times100 \]

Substitute:

\[ \text{Percentage absorbed} = \frac{12}{180}\times100 \]

\[ \boxed{\text{Percentage absorbed}=6.7\%} \]

The absorption feature therefore removes approximately \(6.7\%\) of the radiation that was originally present in the measured wavelength band.

Step 2: Interpret the absorption feature.

The gas absorbs photons with energies corresponding to particular electron transitions.

This reduces the measured intensity near \(656\text{ nm}\), producing a narrow dip in the spectrum.

It does not mean that the heated surface emits only wavelengths other than \(656\text{ nm}\).

The original thermal source still produces a continuous spectrum, but some of that radiation is absorbed before reaching the detector.

Step 3: Identify the correct peak wavelength.

The underlying thermal continuum reaches its maximum at:

\[ \boxed{\lambda_{\max}=680\text{ nm}} \]

This is the wavelength that should be used in Wien’s displacement law.

The \(656\text{ nm}\) feature is an absorption wavelength, not the thermal peak.

The important distinction is that a spectral feature can be caused by something happening to radiation after it leaves the original source.

You must identify the physical origin of the feature before interpreting it.

Question 5

A heated ceramic surface is observed through an excited gas.

A calibrated spectrometer records a combined spectrum consisting of thermal radiation from the ceramic and an additional emission line from the gas.

The following information is provided:

• The underlying thermal continuum has a smooth maximum at \(720\text{ nm}\).
• The thermal continuum has a spectral intensity of \(5.0\) arbitrary units per nanometre at \(720\text{ nm}\).
• At \(560\text{ nm}\), the thermal continuum has an intensity of \(3.0\) arbitrary units per nanometre.
• The gas produces a narrow emission line centred at \(560\text{ nm}\).
• The emission line has a triangular excess above the thermal continuum, with a maximum height of \(6.0\) arbitrary units per nanometre and a base width of \(1.0\text{ nm}\).
• The integrated area under the thermal continuum across the measured wavelength range is \(2000\) arbitrary units.
• The ceramic behaves approximately as an ideal blackbody.

Assume the detector response has been corrected and that the gas emission line is superimposed on, rather than replacing, the thermal continuum.

Two students make different claims.

Student A says:

“The highest point on the combined spectrum is at \(560\text{ nm}\), so that must be the peak wavelength to use in Wien’s displacement law.”

Student B says:

“The emission line has a very small area, so it can be ignored completely. It doesn’t affect anything important.”

Determine:

1. The wavelength of the maximum intensity in the combined measured spectrum.
2. The correct peak wavelength to use in Wien’s displacement law.
3. The integrated area of the additional emission line.
4. The percentage contribution of the emission line relative to the thermal continuum’s integrated area.
5. The approximate temperature of the ceramic surface.

Finally, assess both students’ arguments. Explain how a spectral feature can determine the highest point on a graph while contributing very little to the total emitted energy.

Solution 5

Student A identifies the highest point of the combined spectrum correctly but uses the wrong wavelength for Wien’s law. Student B correctly recognises that the emission line contributes very little integrated energy but incorrectly assumes it has no effect on the measured maximum.

The correct thermal peak wavelength is \(720\text{ nm}\), giving a ceramic temperature of approximately \(4.0\times10^3\text{ K}\).

Step 1: Determine the maximum intensity of the combined spectrum.

At \(560\text{ nm}\), the thermal continuum contributes:

\[ I_{\lambda,\text{thermal}}=3.0 \]

The gas emission line adds an excess intensity of:

\[ I_{\lambda,\text{line}}=6.0 \]

Therefore, the combined intensity is:

\[ I_{\lambda,\text{total}}=3.0+6.0 \]

\[ I_{\lambda,\text{total}}=9.0 \]

At the thermal continuum maximum of \(720\text{ nm}\), the intensity is only \(5.0\).

Since \(9.0\) is greater than \(5.0\), the overall measured maximum occurs at:

\[ \boxed{\lambda_{\text{combined}}=560\text{ nm}} \]

Student A is correct about the highest point of the combined graph.

However, that point is produced by the additional emission line, not the thermal continuum.

Step 2: Identify the correct thermal peak wavelength.

Wien’s displacement law relates temperature to the maximum of the thermal spectrum.

The thermal continuum has its maximum at:

\[ \boxed{\lambda_{\max}=720\text{ nm}} \]

Therefore, this is the wavelength that must be used to calculate the ceramic’s temperature.

Using \(560\text{ nm}\) would incorrectly treat the gas emission line as the thermal maximum.

Step 3: Calculate the integrated area of the emission line.

The excess emission line is triangular.

Its base width is:

\[ \Delta\lambda=1.0\text{ nm} \]

Its maximum excess intensity is:

\[ I_{\text{excess}}=6.0 \]

Using the area of a triangle:

\[ A_{\text{line}}= \frac{1}{2}\times\text{base}\times\text{height} \]

Substitute:

\[ A_{\text{line}}= \frac{1}{2}(1.0)(6.0) \]

\[ \boxed{A_{\text{line}}=3.0\text{ arbitrary units}} \]

More precisely, the integrated area has units of arbitrary intensity units multiplied by nanometres.

Step 4: Calculate the percentage contribution of the line.

The thermal continuum has an integrated area of:

\[ A_{\text{thermal}}=2000 \]

The emission line contributes an additional area of:

\[ A_{\text{line}}=3.0 \]

Therefore:

\[ \text{Percentage contribution} = \frac{3.0}{2000}\times100 \]

\[ \boxed{\text{Percentage contribution}=0.15\%} \]

The emission line contributes only \(0.15\%\) of the thermal continuum’s integrated area across the measured range.

This is a very small contribution to the total energy represented by the spectrum.

However, its narrow shape gives it a large maximum intensity.

Step 5: Calculate the temperature of the ceramic.

Use Wien’s displacement law:

\[ \lambda_{\max}T=b \]

Rearrange:

\[ T=\frac{b}{\lambda_{\max}} \]

Convert the wavelength into metres:

\[ 720\text{ nm}=720\times10^{-9}\text{ m} \]

Substitute:

\[ T= \frac{2.90\times10^{-3}} {720\times10^{-9}} \]

\[ T=4.03\times10^3\text{ K} \]

Therefore:

\[ \boxed{T\approx4.0\times10^3\text{ K}} \]

Step 6: Assess the students’ reasoning.

Student A makes a subtle mistake.

The highest point of the combined spectrum really is at \(560\text{ nm}\). However, Wien’s law requires the peak wavelength of the thermal continuum, not necessarily the highest point of a spectrum containing additional emission features.

Student B also makes a mistake.

The emission line contributes very little integrated energy, but it still produces the highest local spectral intensity. A small area does not mean a feature cannot have a large height.

The two students have confused different properties of the same graph:

• Maximum spectral intensity depends on the greatest vertical height.
• Integrated power depends on the entire area beneath the curve.
• Thermal peak wavelength depends on the horizontal position of the maximum of the thermal continuum.

This is the central distinction to carry into HSC spectral analysis questions.

When a spectrum contains more than one physical contribution, identify what produces each feature before selecting a wavelength, calculating a temperature, or interpreting the total emitted power.