How to Calculate an Electron’s de Broglie Wavelength

An electron gun is set to an accelerating voltage of 150 V. What is the wavelength of the electrons leaving the gun?

You don’t need to know their speed first. The accelerating voltage determines the electrons’ kinetic energy, which determines their momentum, which determines their de Broglie wavelength.

For an electron starting with negligible kinetic energy, the calculation is:

\[ \boxed{\lambda=\frac{h}{\sqrt{2m_e eV}} } \]

Or, using standard physical constants:

\[ \boxed{\lambda=\frac{1.227}{\sqrt{V}}\ \mathrm{nm}} \]

Here, \(V\) is the accelerating voltage entered in volts.

The important idea is that increasing the accelerating voltage makes the electrons move faster, increasing their momentum and decreasing their wavelength.

Let’s build the equation, work through the calculation, and then see how this relationship was tested experimentally.

01From accelerating voltage to de Broglie wavelength

There are three relationships connecting the accelerating voltage to the electron’s wavelength.

1. The accelerating voltage determines the electron’s change in kinetic energy.
2. The kinetic energy determines the electron’s momentum.
3. The momentum determines its de Broglie wavelength.

Understanding these connections makes the calculation much easier to remember.

Step 1: Calculate the kinetic energy gained

An electron has a negative electric charge.

When it moves from a negatively charged cathode towards a positively charged anode, the electric field accelerates it.

The work done by the electric field increases the electron’s kinetic energy.

The change in kinetic energy is:

\[ \Delta K=-q\Delta\phi \]

where:

• \(\Delta K\) is the change in kinetic energy, in joules.
• \(q\) is the electron’s charge, in coulombs.
• \(\Delta\phi\) is the change in electric potential, in volts.

The electron’s charge is \(q=-e\), where \(e\) is the positive magnitude of the elementary charge.

Therefore:

\[ \Delta K=e\Delta\phi \]

For an accelerating potential difference \(V\), the electron moves towards a higher electric potential, so:

\[ \boxed{\Delta K=eV} \]

If the electron starts approximately from rest, its initial kinetic energy is negligible. Its final kinetic energy is therefore:

\[ \boxed{K=eV} \]

This is the first important relationship.

Remember that an electron accelerated through 100 V gains 100 eV of kinetic energy.

An electron volt (eV) is a unit of energy, not voltage:

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

You can work in electron volts initially, but convert to joules when using the full momentum equation.

Step 2: Convert kinetic energy into momentum

The kinetic energy of a non-relativistic electron is:

\[ K=\frac{1}{2}m_ev^2 \]

Here:

• \(K\) is kinetic energy in joules.
• \(m_e\) is the electron’s mass.
• \(v\) is its speed in metres per second.

Momentum is:

\[ p=m_ev \]

Rearranging the momentum equation gives:

\[ v=\frac{p}{m_e} \]

Substitute this into the kinetic energy equation:

\[ K=\frac{1}{2}m_e \left(\frac{p}{m_e}\right)^2 \]

Simplifying:

\[ K=\frac{p^2}{2m_e} \]

Rearrange to find momentum:

\[ \boxed{p=\sqrt{2m_eK}} \]

We now have a way to calculate electron momentum directly from kinetic energy.

Step 3: Apply the de Broglie equation

The de Broglie relationship connects a particle’s momentum to its wavelength:

\[ \lambda=\frac{h}{p} \]

where:

• \(\lambda\) is the de Broglie wavelength, in metres.
• \(h\) is Planck’s constant.
• \(p\) is the magnitude of the electron’s momentum, in kg m/s.

Substituting our momentum equation:

\[ \lambda=\frac{h}{\sqrt{2m_eK}} \]

We already established that an electron starting from rest gains kinetic energy \(K=eV\).

Substitute this expression:

\[ \boxed{\lambda=\frac{h}{\sqrt{2m_e eV}}} \]

This is the equation you need to calculate an electron’s de Broglie wavelength from its accelerating voltage.

The useful HSC calculation shortcut

Using the following physical constants:

The equation becomes:

\[ \boxed{\lambda=\frac{1.227}{\sqrt{V}}\ \mathrm{nm}} \]

Enter the numerical value of the accelerating voltage in volts.

For example:

Notice the pattern.

Increasing the accelerating voltage makes the wavelength smaller, but not in direct proportion to the voltage.

In fact:

\[ \lambda\propto\frac{1}{\sqrt{V}} \]

If you multiply the accelerating voltage by four, the wavelength halves.

02Worked examples

03How the relationship was tested: The Davisson-Germer experiment

Calculating an electron’s wavelength is useful, but how do we know that electrons actually behave like waves?

This question was investigated experimentally by Clinton Davisson and Lester Germer in the 1920s.

Their work provided important experimental evidence for de Broglie’s hypothesis that particles of matter have wave properties.

The problem they were investigating

Before this experiment, electrons were commonly studied through their behaviour as charged particles.

In 1924, Louis de Broglie proposed that matter should also have wave properties.

If this was correct, electrons should demonstrate phenomena associated with waves, including diffraction.

A crystal was an ideal material for testing this prediction.

Its atoms are arranged in a regular repeating structure. The regularly spaced atomic planes can act like a diffraction grating for waves with wavelengths comparable to their spacing.

Understanding the apparatus

The experiment was conducted inside an evacuated chamber.

The apparatus had four main components arranged in sequence.

1. An electron gun containing a heated filament released electrons from a cathode.
2. A positively charged anode accelerated the electrons towards the target.
3. A nickel crystal received the incoming electron beam.
4. A movable collector detected electrons scattered from the crystal at different angles.

The accelerating voltage could be adjusted to change the electrons’ kinetic energy and momentum.

The collector measured the intensity of scattered electrons. In practical terms, this meant measuring how many electrons entered the detector over time.

What should the experiment show?

There were two competing descriptions of the electrons.

A simple particle-only model would describe electrons as tiny charged particles scattering from the nickel surface.

However, if electrons behaved as waves, their interaction with the regularly spaced nickel atoms should produce diffraction.

For waves, constructive interference occurs when the path difference between waves scattered from neighbouring crystal planes satisfies:

\[ n\lambda=2d\sin\theta \]

This is Bragg’s law, where:

• \(n\) is the diffraction order, a positive integer.
• \(\lambda\) is the wavelength.
• \(d\) is the spacing between crystal planes.
• \(\theta\) is the angle between the incoming beam and the crystal planes.

At certain angles, scattered waves reinforce one another, producing a maximum in detected intensity.

At other angles, destructive interference reduces the intensity.

The important prediction was that the positions of these diffraction maxima should agree with the electron wavelength calculated from the accelerating voltage.

What did Davisson and Germer observe?

Their experiments revealed sharp maxima in the intensity of scattered electrons at particular angles and accelerating voltages.

This was not the smooth scattering pattern expected from a simple particle-only description.

The maxima were consistent with diffraction from the regular arrangement of atoms in the nickel crystal.

A particularly important historical detail is that the experiment was not originally designed to prove de Broglie’s hypothesis.

Davisson and Germer had been investigating electron scattering from nickel for another purpose. In 1925, an equipment accident caused their nickel target to become extremely heated and recrystallise into a few much larger crystals.

Their subsequent measurements began showing unexpected intensity peaks.

After Davisson learned about de Broglie’s hypothesis in 1926, the researchers recognised that these peaks could be evidence of electron diffraction.

They then deliberately investigated the effect using better-controlled crystal orientations, adjustable accelerating voltages, and measurements at different scattering angles.

The earlier target, containing many small randomly oriented crystals, made clear directional diffraction patterns difficult to resolve. Using a more ordered crystal allowed the researchers to investigate the diffraction maxima more systematically.

The vacuum was also important. It reduced unwanted collisions between the electrons and gas molecules before they reached the target.

The historical sequence is described in the American Physical Society’s account of the experiment. The original results were published in Davisson and Germer’s 1927 paper.

A numerical check: The 54 V experiment

Consider electrons accelerated through 54 V.

Their predicted de Broglie wavelength is:

\[ \lambda=\frac{1.227}{\sqrt{54}} \]

\[ \boxed{\lambda=0.167\ \mathrm{nm}} \]

Independent analysis of the nickel diffraction geometry gives a wavelength of approximately 0.165 nm.

The agreement is very close.

This matters because the wavelength calculated from the accelerating voltage agrees with the wavelength inferred independently from the diffraction pattern.

The electrons were therefore behaving in the way predicted by the matter-wave model.

The experiment did not mean that electrons were simply tiny water waves. Rather, it demonstrated that their behaviour includes wave-like properties that can be measured through diffraction.

04Don’t confuse electron wavelengths with photon wavelengths

There is an important distinction between calculating an electron’s de Broglie wavelength and calculating the wavelength of a photon emitted during an atomic transition.

Both calculations involve Planck’s constant, but they use different relationships.

Electron wavelength

For an accelerated electron:

\[ \boxed{\lambda_{\mathrm{dB}}=\frac{h}{p}} \]

Its wavelength depends on its momentum.

You first calculate its kinetic energy from the accelerating voltage, then use that energy to find momentum.

Photon wavelength

For a photon:

\[ E_\gamma=hf \]

And because electromagnetic waves travel at the speed of light:

\[ c=f\lambda_\gamma \]

Combining these equations:

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

Here:

• \(E_\gamma\) is photon energy.
• \(f\) is photon frequency.
• \(\lambda_\gamma\) is photon wavelength.
• \(c\) is the speed of light.

The photon wavelength is determined by its energy, not by the momentum of an accelerated electron.

What happens during an atomic transition?

Electrons bound inside atoms occupy discrete energy levels.

They cannot absorb or emit arbitrary amounts of energy through transitions between these levels. Only allowed transitions can occur.

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

Define the change in the atom’s energy as:

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

For absorption:

\[ E_f>E_i \]

Therefore:

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

The atom absorbs a photon with energy:

\[ E_\gamma=E_f-E_i \]

For emission:

\[ E_f<E_i \]

Therefore:

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

The atom loses energy, but the emitted photon still has positive energy:

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

Its frequency and wavelength are also positive.

\[ f=\frac{E_\gamma}{h}>0 \]

\[ \lambda_\gamma=\frac{hc}{E_\gamma}>0 \]

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

05Common mistakes to avoid

One final limitation: the shortcut equation assumes non-relativistic electrons, for which \(K=p^2/(2m_e)\) is valid. This is an excellent approximation for ordinary HSC-style accelerating voltages. At electron energies in the hundreds of keV, relativistic momentum must be considered.

For more background on the underlying momentum-wavelength relationship, see de Broglie Matter Waves: HSC Physics Explained Clearly.

06Questions and solutions

The following questions progress from a straightforward voltage calculation to more demanding applications involving electric potential, diffraction, experimental evidence, and atomic energy levels.

Question 1

An electron starts from rest and is accelerated through a potential difference of 250 V.

Calculate its de Broglie wavelength in picometres.

Use:

• \(h=6.626\times10^{-34}\ \mathrm{J\,s}\)
• \(m_e=9.11\times10^{-31}\ \mathrm{kg}\)
• \(e=1.602\times10^{-19}\ \mathrm{C}\)

Solution 1

The electron’s de Broglie wavelength is approximately 77.6 pm.

Because the electron starts from rest, its kinetic energy is entirely due to the accelerating voltage.

Step 1: Calculate the kinetic energy.

\[ K=eV \]

\[ K=(1.602\times10^{-19})(250) \]

\[ K=4.005\times10^{-17}\ \mathrm{J} \]

Step 2: Calculate the momentum.

\[ p=\sqrt{2m_eK} \]

Substitute:

\[ p=\sqrt{ 2(9.11\times10^{-31}) (4.005\times10^{-17}) } \]

\[ p=8.54\times10^{-24}\ \mathrm{kg\,m/s} \]

Step 3: Calculate the wavelength.

\[ \lambda=\frac{h}{p} \]

\[ \lambda= \frac{6.626\times10^{-34}} {8.54\times10^{-24}} \]

\[ \lambda=7.76\times10^{-11}\ \mathrm{m} \]

Convert to picometres:

\[ 1\ \mathrm{pm}=10^{-12}\ \mathrm{m} \]

Therefore:

\[ \boxed{\lambda=77.6\ \mathrm{pm}} \]

The result is physically reasonable. An electron accelerated through a few hundred volts has a wavelength on the scale of atomic dimensions, making it suitable for investigating crystal structures through diffraction.

Question 2

An electron travels from point A to point B.

At A, it has a kinetic energy of 200 eV. Point B is at an electric potential 80 V lower than point A.

Assuming the electron reaches B without any other interactions:

(a) Calculate its kinetic energy at B.

(b) Calculate its de Broglie wavelength at B.

(c) Explain whether its wavelength increases or decreases during the journey.

Solution 2

The electron reaches B with a kinetic energy of 120 eV and a de Broglie wavelength of approximately 112 pm. Its wavelength increases during the journey.

(a) Final kinetic energy

Step 1: Identify the potential difference.

Since B is at a lower electric potential:

\[ \Delta\phi=-80\ \mathrm{V} \]

Step 2: Calculate the kinetic energy change.

For an electron:

\[ \Delta K=e\Delta\phi \]

Therefore:

\[ \Delta K=-80\ \mathrm{eV} \]

The electron loses 80 eV of kinetic energy.

Step 3: Calculate its final kinetic energy.

\[ K_f=K_i+\Delta K \]

\[ K_f=200-80 \]

\[ \boxed{K_f=120\ \mathrm{eV}} \]

(b) Final de Broglie wavelength

Use the shortcut equation:

\[ \lambda=\frac{1.227}{\sqrt{K_f}}\ \mathrm{nm} \]

Substitute:

\[ \lambda=\frac{1.227}{\sqrt{120}} \]

\[ \boxed{\lambda=0.112\ \mathrm{nm}=112\ \mathrm{pm}} \]

(c) Explain the wavelength change

The electron’s wavelength increases.

Its kinetic energy decreases from 200 eV to 120 eV, so its momentum also decreases.

Since:

\[ \lambda=\frac{h}{p} \]

a decrease in momentum produces an increase in wavelength.

The important trap here is treating the 80 V potential difference as an additional 80 eV of kinetic energy. The electron is moving towards a lower electric potential, so it is being retarded, not accelerated.

Question 3

An electron beam is accelerated from rest and directed towards a crystalline material.

At an accelerating voltage of 100 V, a first-order diffraction maximum occurs at a Bragg angle of \(30^\circ\).

The same crystal is then used, but the accelerating voltage is increased to 400 V.

Assuming the same crystal planes and diffraction order are involved, calculate the new Bragg angle.

Explain why the new angle is not exactly half the original angle.

Solution 3

The new Bragg angle is approximately 14.5°, not exactly 15°.

The key is that wavelength changes with the square root of accelerating voltage, while Bragg’s law involves the sine of the diffraction angle.

Step 1: Determine how the wavelength changes.

The de Broglie relationship gives:

\[ \lambda\propto\frac{1}{\sqrt V} \]

The voltage increases from 100 V to 400 V.

Therefore:

\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{100}{400}} \]

\[ \frac{\lambda_2}{\lambda_1} =\frac{1}{2} \]

The wavelength halves.

Step 2: Apply Bragg’s law.

\[ n\lambda=2d\sin\theta \]

The crystal spacing \(d\) and diffraction order \(n\) remain unchanged.

Therefore:

\[ \lambda\propto\sin\theta \]

So:

\[ \frac{\sin\theta_2}{\sin\theta_1} = \frac{\lambda_2}{\lambda_1} \]

Substitute:

\[ \sin\theta_2 = \frac{1}{2}\sin30^\circ \]

\[ \sin\theta_2=0.5\times0.5 \]

\[ \sin\theta_2=0.25 \]

Step 3: Calculate the new angle.

\[ \theta_2=\sin^{-1}(0.25) \]

\[ \boxed{\theta_2=14.5^\circ} \]

Why isn’t the answer 15°?

Because the relationship is between wavelength and the sine of the angle, not the angle itself.

The approximation \(\sin\theta\approx\theta\) only works for sufficiently small angles when the angle is expressed in radians.

Here, using the inverse sine gives the correct result.

This question combines two relationships:

• de Broglie wavelength depends on accelerating voltage.
• Bragg diffraction depends on wavelength and crystal geometry.

You need both to predict the movement of a diffraction maximum.

Question 4

Davisson and Germer investigated electron diffraction using a nickel crystal.

In one experimental arrangement, electrons were accelerated through 54 V.

The wavelength calculated from the accelerating voltage was compared with an independently determined wavelength of 0.165 nm, obtained from the crystal’s diffraction geometry.

(a) Calculate the de Broglie wavelength predicted by the accelerating voltage.

(b) Calculate the percentage difference between the predicted wavelength and the independently determined value, using 0.165 nm as the reference value.

(c) Explain why the agreement provides evidence for the wave nature of electrons.

(d) Explain why using a single-crystal target and an evacuated chamber improved the experiment.

Solution 4

The accelerating voltage predicts a wavelength of approximately 0.167 nm, which differs from the independently determined value by only 1.2%.

This agreement supports the interpretation that electrons undergo diffraction as predicted by their de Broglie wavelength.

(a) Calculate the predicted wavelength

Use:

\[ \lambda=\frac{1.227}{\sqrt V} \]

Substitute:

\[ \lambda=\frac{1.227}{\sqrt{54}} \]

\[ \boxed{\lambda=0.167\ \mathrm{nm}} \]

(b) Calculate the percentage difference

The percentage difference relative to the reference value is:

\[ \%\text{ difference} = \frac{|\text{predicted}-\text{reference}|} {\text{reference}}\times100 \]

Substitute:

\[ = \frac{|0.167-0.165|}{0.165} \times100 \]

\[ \boxed{1.2\%} \]

The two values are very close.

(c) Explain the experimental significance

The accelerating voltage provides a prediction of the electron’s wavelength using de Broglie’s equation.

The diffraction pattern provides an independent way of estimating that wavelength using the known crystal spacing and Bragg’s law.

The close agreement supports the wave model because the electrons produce constructive interference at angles consistent with their predicted wavelength.

A simple classical particle-only description does not account for this diffraction pattern.

The important point is that the experiment compares two independent methods of determining wavelength, rather than simply assuming the de Broglie equation must be correct.

(d) Explain the apparatus improvements

A single-crystal target has a regular, known arrangement of atomic planes.

This makes it easier to identify directional diffraction maxima. A target containing many randomly oriented crystal grains can average out these patterns and make them harder to distinguish.

An evacuated chamber reduces collisions between electrons and gas molecules before they reach the nickel target.

These collisions could change electron momentum and interfere with the measurements.

Together, these improvements make the diffraction pattern easier to observe and interpret.

Question 5

An electron beam is accelerated from rest through 120 V before entering a chamber containing atoms of an illustrative model gas.

Each atom has three allowed energy levels:

\[ E_0=0\ \mathrm{eV} \]

\[ E_1=30\ \mathrm{eV} \]

\[ E_2=80\ \mathrm{eV} \]

Initially, every atom is in the ground state \(E_0\).

The allowed upward excitations from the ground state are:

\[ E_0\rightarrow E_1 \]

and

\[ E_0\rightarrow E_2 \]

The following downward transitions are also allowed:

\[ E_2\rightarrow E_1 \]

\[ E_2\rightarrow E_0 \]

\[ E_1\rightarrow E_0 \]

Assume the following:

• Each incident electron causes at most one excitation.
• The electron transfers exactly the energy required for that excitation.
• No other energy losses or ionisation occur.
• Each excited atom eventually returns to its ground state.
• All emitted photons are detected.

Answer the following:

(a) Calculate the initial kinetic energy and de Broglie wavelength of the incoming electrons.

(b) Determine every possible outgoing electron kinetic energy and calculate the corresponding de Broglie wavelengths.

(c) An experiment detects a 50 eV photon followed by a 30 eV photon. Explain how these photons can arise from a single excitation event.

(d) Calculate the frequency and wavelength of the 50 eV and 30 eV photons.

(e) Another student argues that a 70 eV outgoing electron should also be possible because the difference between \(E_2\) and \(E_1\) is 50 eV.

Explain precisely why this prediction is incorrect, despite the fact that a 50 eV transition is allowed.

Solution 5

Only two outgoing electron kinetic energies are possible: 90 eV and 40 eV.

Their corresponding de Broglie wavelengths are approximately 129 pm and 194 pm, respectively.

The key distinction is that the electron initially encounters atoms in the ground state. It cannot directly excite an atom through an energy gap that is only available between two already-excited levels.

(a) Initial electron energy and wavelength

The electrons start from rest and are accelerated through 120 V.

Their kinetic energy is:

\[ K_i=eV \]

Therefore:

\[ \boxed{K_i=120\ \mathrm{eV}} \]

Calculate the initial de Broglie wavelength:

\[ \lambda_i=\frac{1.227}{\sqrt{120}} \]

\[ \boxed{\lambda_i=0.112\ \mathrm{nm}=112\ \mathrm{pm}} \]

(b) Possible outgoing electron energies

The atoms initially occupy \(E_0\).

There are only two permitted excitations from this state.

The first is:

\[ E_0\rightarrow E_1 \]

The energy transferred is:

\[ \Delta E=30-0=30\ \mathrm{eV} \]

The electron loses 30 eV.

Its remaining kinetic energy is:

\[ K_f=120-30 \]

\[ \boxed{K_f=90\ \mathrm{eV}} \]

Its de Broglie wavelength is:

\[ \lambda=\frac{1.227}{\sqrt{90}} \]

\[ \boxed{\lambda=0.129\ \mathrm{nm}=129\ \mathrm{pm}} \]

The second possible excitation is:

\[ E_0\rightarrow E_2 \]

The energy transferred is:

\[ \Delta E=80-0=80\ \mathrm{eV} \]

The electron loses 80 eV.

Its remaining kinetic energy is:

\[ K_f=120-80 \]

\[ \boxed{K_f=40\ \mathrm{eV}} \]

Its de Broglie wavelength is:

\[ \lambda=\frac{1.227}{\sqrt{40}} \]

\[ \boxed{\lambda=0.194\ \mathrm{nm}=194\ \mathrm{pm}} \]

We can summarise the results:

There are no other outgoing energies under the stated assumptions.

(c) Explain the 50 eV and 30 eV photons

The two photons arise from a cascade of atomic transitions following the initial excitation to \(E_2\).

First, the incident electron transfers 80 eV to the atom:

\[ E_0\rightarrow E_2 \]

The electron therefore leaves with 40 eV.

The excited atom can then undergo the following two transitions.

First:

\[ E_2\rightarrow E_1 \]

The atom loses:

\[ 80-30=50\ \mathrm{eV} \]

It emits a 50 eV photon.

However, the atom is not yet in its ground state. It still occupies \(E_1\).

It then undergoes:

\[ E_1\rightarrow E_0 \]

The atom loses another:

\[ 30-0=30\ \mathrm{eV} \]

It emits a second photon with energy 30 eV.

Therefore, the sequence is:

\[ 80\ \mathrm{eV} \rightarrow 50\ \mathrm{eV}+30\ \mathrm{eV} \]

The total energy emitted is:

\[ 50+30=80\ \mathrm{eV} \]

This accounts for the original energy transferred by the incoming electron.

There is another possible decay route:

\[ E_2\rightarrow E_0 \]

In this case, the atom emits a single photon with energy 80 eV.

Both decay routes are allowed by the question.

(d) Calculate the photon frequencies and wavelengths

Use:

\[ E_\gamma=hf \]

Therefore:

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

Also:

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

For convenience, use:

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

and:

\[ c=3.00\times10^8\ \mathrm{m/s} \]

For the 50 eV photon:

Convert its energy into joules:

\[ E_\gamma=50(1.602\times10^{-19}) \]

\[ E_\gamma=8.01\times10^{-18}\ \mathrm{J} \]

Calculate its frequency:

\[ f=\frac{8.01\times10^{-18}} {6.626\times10^{-34}} \]

\[ \boxed{f=1.21\times10^{16}\ \mathrm{Hz}} \]

Calculate its wavelength:

\[ \lambda_\gamma=\frac{c}{f} \]

\[ \lambda_\gamma= \frac{3.00\times10^8} {1.21\times10^{16}} \]

\[ \boxed{\lambda_\gamma=24.8\ \mathrm{nm}} \]

For the 30 eV photon:

\[ E_\gamma=30(1.602\times10^{-19}) \]

\[ E_\gamma=4.806\times10^{-18}\ \mathrm{J} \]

Calculate its frequency:

\[ f=\frac{4.806\times10^{-18}} {6.626\times10^{-34}} \]

\[ \boxed{f=7.25\times10^{15}\ \mathrm{Hz}} \]

Calculate its wavelength:

\[ \lambda_\gamma=\frac{3.00\times10^8} {7.25\times10^{15}} \]

\[ \boxed{\lambda_\gamma=41.4\ \mathrm{nm}} \]

The results are:

All photon energies, frequencies, and wavelengths are positive.

For the 50 eV transition, the atom’s energy change is actually:

\[ \Delta E_{\mathrm{atom}}=30-80=-50\ \mathrm{eV} \]

But the emitted photon has:

\[ E_\gamma=+50\ \mathrm{eV} \]

The signs describe different things.

(e) Why a 70 eV outgoing electron is impossible

The prediction is incorrect because the atom initially occupies the ground state.

Although the energy difference between \(E_2\) and \(E_1\) is 50 eV, this does not mean that a ground-state atom can absorb 50 eV and reach another allowed energy level.

The 50 eV gap exists between two excited states:

\[ E_2-E_1=80-30=50\ \mathrm{eV} \]

An atom starting at \(E_0\) cannot use this gap as its initial excitation energy.

Its only allowed upward transitions are:

\[ E_0\rightarrow E_1 \]

and:

\[ E_0\rightarrow E_2 \]

These require 30 eV and 80 eV, respectively.

A 70 eV outgoing electron would mean that the incoming electron lost:

\[ 120-70=50\ \mathrm{eV} \]

That would require a 50 eV excitation from the ground state, which is not allowed.

The subtle point is that the 50 eV transition becomes possible only after the atom has already absorbed 80 eV and reached \(E_2\).

It can then emit 50 eV while falling to \(E_1\), followed by another 30 eV when it reaches the ground state.

This is why you must track the atom’s initial state, its allowed transitions, and its subsequent decay rather than simply listing every possible energy difference between pairs of levels.

The outgoing electron’s de Broglie wavelength depends on the kinetic energy remaining after the collision. The emitted photons have their own wavelengths and frequencies, determined by the atomic energy differences.

07Where to go next

Once you can calculate an electron’s de Broglie wavelength from its accelerating voltage, the next step is understanding how that wavelength produces an observable diffraction pattern.

Bragg’s law connects the electron wavelength to the spacing between crystal planes and predicts the angles at which diffraction maxima occur.

For a closer look at that experimental evidence, continue with Electron diffraction evidence for de Broglie’s matter waves.

That connection completes the chain: accelerating voltage determines kinetic energy, kinetic energy determines momentum, momentum determines wavelength, and wavelength determines the diffraction pattern.