How Standing Matter Waves Explain Allowed Electron Orbits

A hydrogen atom can emit a sharp red line of light at about 656 nm, but not every wavelength between red and violet. Why should an electron produce only certain colours if it can move around the nucleus?

The Bohr-de Broglie model gives an explanation: an electron has a matter wavelength, and only circular paths containing a whole number of those wavelengths can support a consistent wave pattern. These are the allowed electron orbits. Each has a particular energy, so an electron changing between them can emit or absorb only photons whose energies match the difference between two allowed levels.

The crucial idea is that the electron’s matter wave determines which states can exist, while the energy difference between states determines the wavelength of the light.

01Why did hydrogen’s spectrum need a new explanation?

Imagine looking directly at a glowing hydrogen discharge tube. You see a pinkish glow.

Now pass some of that light through a narrow slit and a diffraction grating. Instead of a continuous rainbow, you find separate bright lines, including a red line near 656 nm and a blue-green line near 486 nm.

Why are there gaps between the colours?

What the experiment actually measures

The apparatus works in a particular order:

1. A high-voltage supply is connected to electrodes at opposite ends of a low-pressure hydrogen discharge tube.
2. Moving electrons collide with hydrogen atoms, transferring energy and exciting some of them.
3. The excited atoms emit light as they return to lower-energy states.
4. Some light passes through a narrow slit. A lens can make the emerging light approximately parallel.
5. A prism or diffraction grating separates different wavelengths, producing bright lines at particular positions on a screen or detector.

Simply looking at the discharge tube cannot resolve those wavelengths. Adding the slit and dispersing element turns the glow into a measurable spectrum.

The prediction depends on the atomic model.

If electrons could continuously change their energies by arbitrary amounts, the model would not restrict their emitted light to particular energy differences.

If electrons could occupy only particular energies, photons would have to carry differences between those energies. A line spectrum would follow.

The separate hydrogen lines supported the second picture.

Bohr could explain the lines, but his model had a problem

Classical physics presented another difficulty.

An electron travelling in a circular orbit is accelerating because its velocity continually changes direction. According to classical electromagnetic theory, an accelerating charge emits radiation.

A classical orbiting electron should therefore lose energy and spiral towards the nucleus. Stable atoms would be difficult to explain.

In 1913, Niels Bohr proposed a different model. Electrons could occupy certain stable, or stationary, orbits without continuously radiating. Radiation occurred when an electron changed between these states.

This accounted for hydrogen’s spectral lines, but Bohr had to impose a rule restricting electron angular momentum.

Why those values, rather than others?

In 1924, Louis de Broglie proposed that matter has wave properties. Applying his idea to an electron moving around a nucleus gave a physical interpretation of Bohr’s restriction.

02How a standing matter wave selects an allowed orbit

Imagine drawing a repeating wave along a long strip of paper.

Now bend the strip into a circle and join its ends.

What happens if the strip contains exactly three wavelengths?

The beginning and end meet at the same stage of the wave. A crest joins a crest, and the pattern continues smoothly around the circle.

But suppose it contains two and a half wavelengths.

The beginning and end no longer have matching phases. One end might be at a crest while the other is at a trough.

For a consistent wave around a closed path, the wave must return to its starting point with the same phase.

That gives us the condition for an allowed circular matter wave:

\[
2\pi r=n\lambda_{\mathrm e}
\]

where:

• \(r\) is the radius of the circular orbit in metres (m).
• \(\lambda_{\mathrm e}\) is the electron’s de Broglie wavelength in metres (m).
• \(n\) is a positive integer: \(1,2,3,\ldots\).
• \(2\pi r\) is the circumference of the orbit.

The integer \(n\) tells us how many complete matter wavelengths fit around the circle.

Why whole wavelengths, not half-wavelengths?

A string fixed at both ends can have nodes at its endpoints. Consequently, its length can contain a whole number of half-wavelengths.

A circular path has no fixed endpoints. After one complete circuit, the wave must meet itself in the same phase.

That is why the circular condition is \(2\pi r=n\lambda_{\mathrm e}\), rather than a condition based on an arbitrary number of half-wavelengths.

The difference becomes clearer if you compare this model with progressive and standing waves.

The paper-strip analogy explains phase matching. Its limitation is that an electron’s matter wave is not a physical stripe wrapped around a tiny track. It represents quantum wave behaviour, and the circular picture is an approximation.

Connecting the wave to electron momentum

De Broglie’s relationship is:

\[
\lambda_{\mathrm e}=\frac{h}{p}
\]

For a non-relativistic electron:

\[
p=m_{\mathrm e}v
\]

Therefore:

\[
\lambda_{\mathrm e}=\frac{h}{m_{\mathrm e}v}
\]

Here:

• \(h=6.63\times10^{-34}\ \mathrm{J\,s}\) is Planck’s constant.
• \(p\) is the electron’s momentum in \(\mathrm{kg\,m\,s^{-1}}\).
• \(m_{\mathrm e}=9.11\times10^{-31}\ \mathrm{kg}\) is the electron’s mass.
• \(v\) is its speed in \(\mathrm{m\,s^{-1}}\).

Substitute the de Broglie relationship into the circular-wave condition:

\[
2\pi r=\frac{nh}{m_{\mathrm e}v}
\]

Rearranging gives:

\[
m_{\mathrm e}vr=\frac{nh}{2\pi}
\]

For circular motion, \(m_{\mathrm e}vr\) is the electron’s orbital angular momentum, \(L\).

Therefore:

\[
L=\frac{nh}{2\pi}
\]

This is Bohr’s angular momentum quantisation rule.

Bohr originally introduced the restriction as an assumption. De Broglie’s wave argument showed why such a restriction made physical sense: a wave on a closed path must have a consistent phase.

03Why the wave condition is not enough by itself

There’s a subtle problem.

You could draw a circle of almost any radius, choose an electron wavelength that fits around it, and satisfy the whole-wavelength condition.

Does that mean hydrogen has infinitely many different orbits with the same value of \(n\)?

No. The electron must also satisfy the physics of its interaction with the nucleus.

In the Bohr model, the attractive electrostatic force provides the centripetal force needed for circular motion:

\[
\frac{m_{\mathrm e}v^2}{r}
=
\frac{ke^2}{r^2}
\]

Here \(k\) is Coulomb’s constant, and \(e\) is the magnitude of the elementary electric charge.

The two conditions work together:

• The matter-wave condition restricts angular momentum.
• The electrostatic-force condition relates the electron’s speed to its orbital radius.

Together, for hydrogen in the simplified circular model, they give:

\[
r_n=n^2a_0
\]

where \(a_0=0.0529\ \mathrm{nm}\) is the Bohr radius, and \(1\ \mathrm{nm}=10^{-9}\ \mathrm{m}\).

They also give \(v_n\propto1/n\), meaning the electron’s modelled speed decreases as \(n\) increases.

This leads to specific allowed total energies:

\[
E_n=-\frac{13.6}{n^2}\ \mathrm{eV}
\]

An electronvolt (eV) is a unit of energy:

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

The energy is negative because we define the total energy of a separated electron and proton, infinitely far apart and at rest, as zero.

A bound electron has less energy than that reference state.

The first four hydrogen energy levels

These are rounded predictions of the idealised Bohr-de Broglie model, not measurements of literal electron tracks.

Notice that the radius and matter wavelength both increase as \(n\) increases.

However, the circumference increases in proportion to \(n^2\), while the matter wavelength increases in proportion to \(n\). So the number of wavelengths fitting around the orbit is still exactly \(n\).

Also notice the energy signs. The \(n=3\) state at −1.51 eV has more total energy than the \(n=2\) state at −3.40 eV. A number that is less negative is greater.

04How allowed orbits produce particular photon energies

We now have specific allowed energy levels.

How do they produce the red line at 656 nm?

An electron can change from one allowed energy state to another. If the atom releases energy as a photon, the photon carries the difference between the initial and final atomic energies.

The change in atomic energy is always:

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

where:

• \(E_i\) is the atom’s initial energy.
• \(E_f\) is its final energy.
• \(\Delta E_{\mathrm{atom}}\) is the signed energy change.

For emission, the electron moves to a lower-energy state:

\[
E_\gamma=E_i-E_f
\]

For absorption, the electron moves to a higher-energy state:

\[
E_\gamma=E_f-E_i
\]

Here \(E_\gamma\) is the photon’s energy. It is positive in both cases.

Keep the electron’s energy sign separate from the photon’s

A negative atomic energy change does not mean a negative-energy photon.

It means the atom has lost energy, which has been transferred to the emitted photon.

Connecting energy, frequency, and wavelength

Photon energy is related to its frequency by:

\[
E_\gamma=hf
\]

The frequency \(f\) is measured in hertz (Hz).

For light travelling through a vacuum:

\[
c=f\lambda_\gamma
\]

where \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\) is the speed of light, and \(\lambda_\gamma\) is the photon’s wavelength in metres.

Combining these relationships:

\[
E_\gamma=\frac{hc}{\lambda_\gamma}
\]

A larger photon energy means a higher frequency and a shorter wavelength.

Be careful: \(\lambda_\gamma\) is not the electron’s de Broglie wavelength \(\lambda_{\mathrm e}\). They describe different things.

05The contrasting case: The same red photon can be absorbed

Suppose a hydrogen electron starts at \(n=2\) instead of \(n=3\).

A photon with energy 1.89 eV arrives.

Will the electron emit another red photon, or can it absorb the incoming one?

It can absorb the photon and move from \(n=2\) to \(n=3\), provided the appropriate quantum transition is available.

The atomic energy change is now:

\[
\begin{aligned}
\Delta E_{\mathrm{atom}}
&=E_3-E_2\\
&=-1.51-(-3.40)\\
&=+1.89\ \mathrm{eV}
\end{aligned}
\]

The atom gains energy, but the incoming photon still has positive energy.

The forward and reverse processes involve the same energy gap:

Here is the less obvious part: a hydrogen atom initially in its ground state, \(n=1\), cannot absorb this 656 nm photon through an ordinary single-photon bound-state transition.

Its smallest excitation energy is:

\[
\begin{aligned}
E_2-E_1
&=-3.40-(-13.60)\\
&=10.20\ \mathrm{eV}
\end{aligned}
\]

A 1.89 eV photon does not provide that energy.

The fact that hydrogen can emit a particular wavelength does not mean hydrogen atoms in every initial state can absorb it.

The starting energy level matters.

A useful decision process

When interpreting an atomic transition, work in this order:

1. Identify the initial and final energy levels.
2. Calculate \(E_f-E_i\), preserving the signs of the energy levels.
3. If the result is negative, the atom loses energy through emission. If positive, the atom requires energy through absorption.
4. Use the positive photon energy to calculate frequency or wavelength.
5. Check that the proposed transition and initial population are physically possible.

In the simplified HSC energy-level treatment, we often consider transitions between levels labelled by \(n\). In the more complete quantum model, a matching energy difference is necessary but not always sufficient: selection rules involving other quantum numbers determine whether particular one-photon transitions can occur.

06What standing waves explain – and what the experiment actually established

It’s worth separating three claims.

Hydrogen spectroscopy established that the atom emits radiation at particular energies. This supports the existence of discrete atomic energy states.

Bohr’s model reproduced the hydrogen energy pattern by assuming stationary states and quantised angular momentum.

De Broglie’s proposal offered an explanation for the angular momentum restriction: the electron has wave properties, so a consistent state must satisfy a wave condition.

But observing spectral lines is not direct evidence that an electron wave physically wraps around a circular path.

That required a separate test of electron wave behaviour.

In 1927, the Davisson-Germer electron diffraction experiment provided evidence that electrons have wave properties through their scattering from a crystal. The experimental argument is developed in Electron diffraction evidence for de Broglie’s matter waves.

Electron diffraction supported de Broglie’s matter-wave hypothesis. It did not establish literal circular electron tracks inside atoms.

Where the circular-orbit model stops working

The Bohr-de Broglie picture is useful because you can see why wave conditions restrict states and why discrete energies produce spectral lines.

However, modern quantum mechanics does not describe an electron as a tiny object repeatedly circling the nucleus along a definite path.

Schrödinger’s wave mechanics, developed in 1926, describes electrons using three-dimensional wavefunctions. Their spatial patterns are called atomic orbitals.

An orbital describes a probability distribution for finding the electron, rather than a definite trajectory.

The simplest hydrogen energy formula also ignores effects such as fine structure and external electric or magnetic fields. Multi-electron atoms need a more sophisticated treatment.

For the HSC-style hydrogen questions below, we’ll use the explicitly stated Bohr-de Broglie assumptions, so the calculations remain focused on allowed states, energy differences, and photon emission or absorption.

07Questions and solutions

Use the following rounded hydrogen energy levels unless a question specifies otherwise:

You may also use \(E_n=-13.6/n^2\ \mathrm{eV}\), \(hc=1240\ \mathrm{eV\,nm}\), and the constants defined earlier. Assume isolated hydrogen atoms, non-relativistic conditions, and negligible photon recoil. Treat the specified one-photon transitions as available.

Question 1

A hydrogen electron moves from \(n=2\) to \(n=1\).

Calculate:

a) The change in the atom’s energy, in eV.

b) The energy of the emitted photon, in joules.

c) The photon’s wavelength, in nanometres, and identify whether it is visible, infrared, or ultraviolet radiation.

Solution 1

The atom loses 10.20 eV, emits a photon carrying \(1.63\times10^{-18}\ \mathrm{J}\), and produces ultraviolet radiation with a wavelength of approximately 122 nm.

Part a: Atomic energy change

\[
\begin{aligned}
\Delta E_{\mathrm{atom}}
&=E_f-E_i\\
&=-13.60-(-3.40)\\
&=-10.20\ \mathrm{eV}
\end{aligned}
\]

The negative sign shows the atom loses energy.

Part b: Photon energy

The emitted photon receives this lost energy as a positive quantity:

\[
E_\gamma=10.20\ \mathrm{eV}
\]

Converting to joules:

\[
\begin{aligned}
E_\gamma&=10.20(1.602\times10^{-19})\\
&=1.63\times10^{-18}\ \mathrm{J}
\end{aligned}
\]

Part c: Wavelength

\[
\lambda_\gamma=\frac{hc}{E_\gamma}
\]

Using \(hc=1240\ \mathrm{eV\,nm}\):

\[
\begin{aligned}
\lambda_\gamma&=\frac{1240}{10.20}\\
&=121.6\ \mathrm{nm}\\
&\approx122\ \mathrm{nm}
\end{aligned}
\]

This is ultraviolet radiation.

A tempting mistake is to subtract the energy-level magnitudes in the wrong order and then call the photon’s energy negative. The atom’s energy change is negative; the emitted photon’s energy is positive.

Question 2

Two cells contain isolated hydrogen atoms.

• Cell A contains atoms initially in \(n=1\).
• Cell B contains atoms prepared in \(n=2\).

A 486 nm photon beam is directed through each cell in separate trials.

Assume single-photon interactions, negligible collisions, and available radiative transitions.

a) Determine which cell can absorb the 486 nm photons in a bound-to-bound transition, and identify the resulting energy level.

b) One atom that absorbs a photon in cell B subsequently returns to \(n=2\). During its return, it emits a 656 nm photon. Assuming its energy loss is entirely radiative, explain a possible sequence and calculate the wavelength of the other photon emitted.

c) A student argues that ground-state hydrogen must be able to absorb 486 nm because excited hydrogen can emit 486 nm. Explain the error.

Solution 2

Only cell B can absorb the 486 nm photon through the specified bound-to-bound transition, moving an electron from \(n=2\) to \(n=4\). A subsequent \(4\to3\to2\) cascade can produce a 656 nm photon and another photon of approximately 1880 nm.

Part a: Determine the photon energy and available excitation

The incoming photon has energy:

\[
\begin{aligned}
E_\gamma&=\frac{1240}{486}\\
&=2.55\ \mathrm{eV}
\end{aligned}
\]

For cell B:

\[
\begin{aligned}
E_4-E_2
&=-0.850-(-3.40)\\
&=2.55\ \mathrm{eV}
\end{aligned}
\]

The photon matches the \(n=2\to n=4\) excitation energy.

Cell A starts in \(n=1\). Its smallest excitation gap is 10.20 eV, so a 2.55 eV photon cannot excite its electron to another bound level.

The difference comes from the starting state, not the chemical identity of the atoms.

Part b: Account for the observed red photon

After absorbing the 486 nm photon, the atom is at \(n=4\).

A possible return sequence is:

\[
n=4\to n=3\to n=2
\]

The \(n=3\to n=2\) transition emits the familiar 656 nm photon, with energy approximately 1.89 eV.

The preceding \(n=4\to n=3\) transition releases:

\[
\begin{aligned}
E_\gamma
&=E_4-E_3\\
&=-0.850-(-1.51)\\
&=0.660\ \mathrm{eV}
\end{aligned}
\]

Its wavelength is:

\[
\begin{aligned}
\lambda_\gamma&=\frac{1240}{0.660}\\
&=1880\ \mathrm{nm}
\end{aligned}
\]

This second photon is infrared.

The energy check is:

\[
0.660+1.89=2.55\ \mathrm{eV}
\]

The two emitted photons together account for the energy absorbed.

It would be incorrect to conclude that only a 656 nm photon was emitted because that was the only one reported by the detector. The missing energy requires another transfer, which under the stated radiative-only assumption is the infrared photon.

Part c: Explain the misconception

A spectral line identifies an energy difference between states, not an absorption capability shared by every atom of the element.

An excited hydrogen atom can emit 486 nm during \(n=4\to n=2\).

A ground-state atom cannot absorb that photon through the reverse process because the reverse transition requires an electron already at \(n=2\).

The initial energy state is therefore essential when deciding whether a particular photon can be absorbed.

Question 3

A student proposes two circular paths for an electron in a hydrogen atom.

The student’s calculations give:

The student notices that both circumferences contain approximately two complete matter wavelengths.

They conclude that both paths must be allowed \(n=2\) orbits of hydrogen.

Evaluate the conclusion using:

\[
2\pi r=n\lambda_{\mathrm e}
\]

and:

\[
r_n=n^2a_0
\]

where \(a_0=0.0529\ \mathrm{nm}\).

Explain physically why satisfying the wavelength condition alone is insufficient.

Solution 3

Only path B is consistent with the \(n=2\) Bohr orbit of hydrogen. Both proposals approximately satisfy the matter-wave phase condition, but path A fails the electrostatic-force requirement that determines hydrogen’s permitted radius.

First, calculate the number of wavelengths around path A:

\[
\begin{aligned}
n_A&=\frac{2\pi r_A}{\lambda_A}\\
&=\frac{2\pi(0.159)}{0.500}\\
&\approx2.00
\end{aligned}
\]

Path A satisfies the circular phase condition to the precision of the supplied data.

Now check path B:

\[
\begin{aligned}
n_B&=\frac{2\pi(0.212)}{0.665}\\
&\approx2.00
\end{aligned}
\]

It also satisfies the phase condition.

So far, the student’s reasoning looks reasonable.

But the phase condition is only one requirement.

For hydrogen, the permitted Bohr radius at \(n=2\) is:

\[
\begin{aligned}
r_2&=n^2a_0\\
&=2^2(0.0529)\\
&=0.2116\ \mathrm{nm}\\
&\approx0.212\ \mathrm{nm}
\end{aligned}
\]

Path B agrees with this radius.

Path A, at 0.159 nm, does not.

Why can’t we simply adjust the electron wavelength to make path A work?

Because the electron’s wavelength determines its momentum and therefore its speed:

\[
v=\frac{h}{m_{\mathrm e}\lambda_{\mathrm e}}
\]

But for an electron moving in a circular Coulomb orbit, its speed and radius must also satisfy:

\[
\frac{m_{\mathrm e}v^2}{r}
=
\frac{ke^2}{r^2}
\]

Path A cannot satisfy both conditions simultaneously for hydrogen with the proposed values.

An externally constrained circular track could have different permitted wave patterns. Hydrogen does not provide an arbitrary track: its electrostatic interaction sets the relationship between radius and speed.

The trap is treating the matter-wave condition as a complete mechanical description rather than one condition within the model.

Question 4

A student investigates the red hydrogen emission line using a discharge tube, narrow slit, and diffraction grating.

The student measures the angle of the first-order bright maximum relative to the central maximum.

Two interchangeable gratings produce the following illustrative measurements:

For first-order diffraction:

\[
d\sin\theta=\lambda_\gamma
\]

where \(d\) is the spacing between adjacent grating lines, and \(\theta\) is the angle measured from the central maximum.

a) Calculate the wavelength measured using each grating.

b) Determine the approximate photon energy and identify the corresponding hydrogen transition from the supplied energy table.

c) A student claims the different angles show that the spectral lines are produced by the gratings, rather than hydrogen. Propose a control that would help test this explanation.

d) Explain whether this experiment directly establishes the existence of standing electron matter waves.

Solution 4

Both gratings give approximately 657 nm, consistent with hydrogen’s red \(n=3\to n=2\) emission line. The different angles arise from different grating spacings, not different atomic photon energies. The experiment supports quantised energy levels but does not directly establish electron matter waves.

Part a: Calculate the two wavelengths

A grating density of 500 lines/mm means the spacing is:

\[
\begin{aligned}
d_A&=\frac{10^{-3}}{500}\\
&=2.00\times10^{-6}\ \mathrm{m}
\end{aligned}
\]

For first-order diffraction:

\[
\begin{aligned}
\lambda_A&=d_A\sin19.2^\circ\\
&=(2.00\times10^{-6})\\
&\quad\times\sin19.2^\circ\\
&=6.58\times10^{-7}\ \mathrm{m}\\
&=658\ \mathrm{nm}
\end{aligned}
\]

For the 600 lines/mm grating:

\[
\begin{aligned}
d_B&=\frac{10^{-3}}{600}\\
&=1.67\times10^{-6}\ \mathrm{m}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
\lambda_B&=d_B\sin23.2^\circ\\
&\approx6.57\times10^{-7}\ \mathrm{m}\\
&=657\ \mathrm{nm}
\end{aligned}
\]

The measurements agree within the precision of the given angles.

Part b: Identify the photon and transition

Using approximately 657 nm:

\[
\begin{aligned}
E_\gamma&=\frac{1240}{657}\\
&=1.89\ \mathrm{eV}
\end{aligned}
\]

The energy difference between \(n=3\) and \(n=2\) is:

\[
\begin{aligned}
E_3-E_2
&=-1.51-(-3.40)\\
&=1.89\ \mathrm{eV}
\end{aligned}
\]

The measured wavelength is therefore consistent with the \(n=3\to n=2\) emission.

Part c: Design a control

Use the same grating with a broadband white-light source.

A broadband source should produce a continuous spread of wavelengths rather than the separated hydrogen lines.

The grating changes where different wavelengths appear. It does not create hydrogen’s specific atomic energy differences.

The original student’s argument confuses the instrument’s role in separating wavelengths with the atom’s role in producing them.

Part d: Interpret the evidence

The experiment supports the conclusion that hydrogen emits light at particular energies.

It does not distinguish every possible theoretical explanation for why those energies are quantised.

The de Broglie standing-wave interpretation requires the additional hypothesis that electrons possess matter wavelengths. Electron diffraction supplies independent experimental evidence for that wave behaviour.

Spectral lines and matter-wave diffraction support connected ideas, but they are not the same observation.

Question 5

An idealised hydrogen atom begins in an unknown allowed circular orbit, labelled \(n_i\).

It exchanges exactly one photon with a light field, then occupies another allowed state, labelled \(n_f\).

Analysis using the Bohr-de Broglie model gives the following relationships between the final and initial states:

\[
\frac{r_f}{r_i}=\frac94
\]

and:

\[
\frac{\lambda_{\mathrm e,f}}
{\lambda_{\mathrm e,i}}=\frac32
\]

A spectrometer identifies the exchanged photon as having a wavelength of approximately 656 nm. The measurement does not reveal whether the photon entered or left the atom.

A student makes two claims:

1. The two ratios uniquely identify the initial and final quantum numbers.
2. Because the electron’s matter wavelength increases, its momentum and kinetic energy must decrease. Therefore, the atom must have emitted the photon.

Using:

\[
2\pi r=n\lambda_{\mathrm e}
\]

and:

\[
E_n=-\frac{13.6}{n^2}\ \mathrm{eV}
\]

answer the following.

a) Determine the relationship between \(n_f\) and \(n_i\). Explain whether the two measured ratios uniquely identify the states.

b) Use the photon wavelength to determine the actual initial and final levels.

c) Decide whether the photon was absorbed or emitted.

d) Calculate the changes in the electron’s kinetic energy and electrostatic potential energy. Explain how these changes are consistent with the direction of photon energy transfer.

For part d, use the circular Coulomb-orbit model:

\[
\frac{m_{\mathrm e}v^2}{r}
=
\frac{ke^2}{r^2}
\]

with electrostatic potential energy:

\[
U=-\frac{ke^2}{r}
\]

Assume non-relativistic hydrogen, no collisions or other energy transfers, and negligible recoil.

Solution 5

The atom moves from \(n=2\) to \(n=3\) by absorbing a 656 nm photon. Its total energy increases by approximately 1.89 eV, even though the electron’s kinetic energy decreases by 1.89 eV. The electrostatic potential energy increases by approximately 3.78 eV, more than compensating for the kinetic energy decrease.

The two student claims fail for different reasons.

Part a: Use the standing-wave condition

The starting condition is:

\[
2\pi r=n\lambda_{\mathrm e}
\]

For the initial and final states:

\[
\begin{aligned}
2\pi r_i&=n_i\lambda_{\mathrm e,i}\\
2\pi r_f&=n_f\lambda_{\mathrm e,f}
\end{aligned}
\]

Divide the second expression by the first:

\[
\frac{r_f}{r_i}
=
\frac{n_f}{n_i}
\frac{\lambda_{\mathrm e,f}}
{\lambda_{\mathrm e,i}}
\]

Substitute the given ratios:

\[
\frac94=\frac{n_f}{n_i}\frac32
\]

Therefore:

\[
\frac{n_f}{n_i}=\frac32
\]

That does not uniquely identify the two levels.

Possible integer pairs include:

\[
2\to3,\qquad4\to6,\qquad6\to9
\]

All satisfy the same ratio.

This defeats the first claim. A ratio can establish a relationship between states without identifying their absolute quantum numbers.

Part b: Use the photon energy to identify the states

The observed photon has energy:

\[
\begin{aligned}
E_\gamma&=\frac{1240}{656}\\
&=1.89\ \mathrm{eV}
\end{aligned}
\]

Because \(n_f=3n_i/2\), the atomic energy change is:

\[
\Delta E
=
13.6\left(
\frac{1}{n_i^2}
–
\frac{1}{n_f^2}
\right)
\]

Substitute the quantum-number ratio:

\[
\begin{aligned}
\Delta E
&=13.6\left(
\frac1{n_i^2}
-\frac4{9n_i^2}
\right)\\
&=\frac{7.56}{n_i^2}\ \mathrm{eV}
\end{aligned}
\]

For the smallest possible initial integer level, \(n_i=2\):

\[
\begin{aligned}
\Delta E&=\frac{7.56}{2^2}\\
&=1.89\ \mathrm{eV}
\end{aligned}
\]

This matches the photon.

For the next possible initial level, \(n_i=4\):

\[
\begin{aligned}
\Delta E&=\frac{7.56}{4^2}\\
&=0.4725\ \mathrm{eV}
\end{aligned}
\]

That is too small.

All larger permitted initial levels give still smaller energy gaps because \(n_i^2\) increases.

Therefore the only matching pair is:

\[
n_i=2,\qquad n_f=3
\]

This is the first major reasoning hinge: the geometric and matter-wave ratios leave multiple possibilities, but the measured photon energy removes the ambiguity.

Part c: Determine the direction of energy transfer

The initial total energy is:

\[
E_2=-3.40\ \mathrm{eV}
\]

The final total energy is:

\[
E_3=-1.51\ \mathrm{eV}
\]

Therefore:

\[
\begin{aligned}
\Delta E_{\mathrm{atom}}
&=-1.51-(-3.40)\\
&=+1.89\ \mathrm{eV}
\end{aligned}
\]

The atom gains total energy.

It must have absorbed the photon.

The second student’s conclusion is incorrect, even though their statement that the electron loses momentum and kinetic energy is correct within this model.

Part d: Account for kinetic and potential energy

The key is to distinguish kinetic energy from total mechanical energy.

The electron’s kinetic energy is:

\[
K=\frac12m_{\mathrm e}v^2
\]

The electrostatic attraction supplies the centripetal force:

\[
\frac{m_{\mathrm e}v^2}{r}
=
\frac{ke^2}{r^2}
\]

Multiplying by \(r/2\) gives:

\[
\frac12m_{\mathrm e}v^2
=
\frac{ke^2}{2r}
\]

Therefore:

\[
K=\frac{ke^2}{2r}
\]

The electrostatic potential energy is:

\[
U=-\frac{ke^2}{r}
\]

So:

\[
U=-2K
\]

The total energy is:

\[
\begin{aligned}
E&=K+U\\
&=K-2K\\
&=-K
\end{aligned}
\]

This gives two useful relationships for the circular Coulomb-orbit model:

\[
K=-E
\]

and:

\[
U=2E
\]

Now apply them to both states.

At \(n=2\):

\[
\begin{aligned}
K_2&=+3.40\ \mathrm{eV}\\
U_2&=-6.80\ \mathrm{eV}
\end{aligned}
\]

At \(n=3\):

\[
\begin{aligned}
K_3&=+1.51\ \mathrm{eV}\\
U_3&=-3.02\ \mathrm{eV}
\end{aligned}
\]

The kinetic energy change is:

\[
\begin{aligned}
\Delta K&=K_3-K_2\\
&=1.51-3.40\\
&=-1.89\ \mathrm{eV}
\end{aligned}
\]

The potential energy change is:

\[
\begin{aligned}
\Delta U&=U_3-U_2\\
&=-3.02-(-6.80)\\
&=+3.78\ \mathrm{eV}
\end{aligned}
\]

Add the changes:

\[
\begin{aligned}
\Delta E&=\Delta K+\Delta U\\
&=-1.89+3.78\\
&=+1.89\ \mathrm{eV}
\end{aligned}
\]

Energy is conserved.

The electron moves to a larger modelled orbit, where it travels more slowly and has a longer de Broglie wavelength. Its kinetic energy decreases.

But its electrostatic potential energy becomes much less negative.

That potential-energy increase is greater than the kinetic-energy decrease, so the atom’s total energy increases.

The photon is absorbed, not emitted.

This is the deeper reason the second claim fails: gaining total energy does not necessarily mean gaining kinetic energy.

The same principle appears in a familiar gravitational example. A spacecraft moving into a higher circular orbit can end up travelling more slowly despite having gained total mechanical energy.

The analogy works because both systems involve attractive inverse-square forces. It stops at the quantum restriction: a spacecraft can occupy a continuous range of classical orbits, whereas an atomic electron has discrete quantum states.

08Where this leads next

Standing matter waves explain why the Bohr-de Broglie model restricts electrons to particular states. Electrostatic attraction helps determine the energies of those states, and the differences between them determine the photons an atom can absorb or emit.

The next useful question is what happens when an excited atom has several possible routes back to a lower energy state.

A single initial state might produce one photon through a direct transition or several photons through intermediate states. The total energy released is fixed by the initial and final states, but the individual photon energies depend on the path.

That’s the connection between allowed electron states and the pattern of spectral lines observed from a collection of excited atoms.