Solve photoelectric, Compton-effect, and matter-wave problems

Ontario Grade 12 Physics

Revolutions in Modern Physics: Quantum Mechanics and Special Relativity

SPH4U F2.2 | Apply energy and momentum relationships to photons and particles

In SPH3U, you used energy conservation and described waves by properties such as wavelength and frequency. Those ideas also help solve problems involving light and particles at the atomic scale. A photon is a packet of light energy. An electron is a particle with mass and charge. In the photoelectric effect, light transfers energy to an electron. In the Compton effect, a photon and an electron exchange energy and momentum. The matter-wave model assigns a wavelength to a moving particle. These models are used here through algebraic relationships.

What you will learn

Use photon energy and the photoelectric energy relationship to find an electron’s maximum kinetic energy.
Calculate the wavelength change in a Compton interaction.
Use de Broglie’s relationship to find the wavelength of a moving particle.
Choose appropriate units, track direction where needed, and check whether answers are reasonable.

Set up the system and choose the model

A scalar has magnitude only. Energy, frequency, mass, and wavelength are scalars. A vector has both magnitude and direction. Momentum and velocity are vectors. In these problems, keep track of direction when particles or photons scatter. The photoelectric example below uses energy and speed magnitudes, so no direction is needed.

Define the physical system before calculating. For a photoelectric problem, the system can include the incoming light and an electron released from a material. For Compton scattering, include the incoming photon, the electron, and both outgoing particles. For a matter-wave problem, focus on the moving particle. Use the material or apparatus as the reference frame, meaning the frame from which positions and motion are described.

For scattering diagrams, choose the incoming photon’s direction as positive horizontal. Measure the photon’s scattering angle from that direction. An electron’s recoil direction can be described relative to the same axis. A direction is not needed when a question asks only for a wavelength or speed.

Identify the system, frame, known values, and unknown before choosing an equation.
Use SI units: joules for energy, metres for wavelength, seconds for time, kilograms for mass, and metres per second for speed.
A photon has energy and momentum but no rest mass. An electron has mass and can gain kinetic energy.

Photoelectric effect: energy delivered by light

The photoelectric effect occurs when light ejects electrons from a material. The work function is the minimum energy needed to remove an electron from that material. The maximum kinetic energy is the greatest kinetic energy among the emitted electrons. For a given material, light below the threshold frequency cannot eject electrons, even if its intensity is increased.

A photon’s energy depends on its frequency, not its intensity. The energy relationship says that the photon’s energy is used to overcome the work function; any remaining energy becomes the electron’s maximum kinetic energy. If the question gives a stopping potential, it is the potential difference that would just stop the fastest emitted electrons. The electrical energy change is related to the electron’s kinetic energy.

Use the SI values of Planck’s constant and the speed of light when needed:

h=6.626\times10^{-34}\ \mathrm{J\,s}

and

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

. The electron mass is

m_e=9.11\times10^{-31}\ \mathrm{kg}

, and the elementary charge is

e=1.602\times10^{-19}\ \mathrm{C}

.

hf=\phi+K_{\max}

Convert a work function given in electronvolts to joules before combining it with photon energy in joules.
An electronvolt is an energy unit: $1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$ .
A positive maximum kinetic energy means the photon has enough energy to eject electrons.

Compton effect: photon wavelength change

In the Compton effect, a photon scatters from an electron. The outgoing photon has a different direction and usually a longer wavelength than the incoming photon. The electron recoils. The change in photon wavelength depends on the scattering angle and the electron’s mass.

The angle is measured from the original photon direction. Wavelengths must use the same length unit. The Compton wavelength of the electron,

h/(m_ec)

, sets the scale of the shift. For a photon scattered straight ahead, the angle is zero and the shift is zero. At larger angles the shift increases.

A direction diagram is useful: draw the incoming photon along the positive horizontal axis, then draw the scattered photon at the stated angle. The formula gives a wavelength difference, not the outgoing wavelength itself. Add that difference to the incoming wavelength to find the outgoing wavelength.

\Delta\lambda=\frac{h}{m_ec}(1-\cos\theta)

The photon’s wavelength increases by the calculated shift.
The angle belongs to the photon’s change in direction, not the electron’s recoil angle.
The wavelength shift is non-negative for scattering angles from zero to $180^\circ$ .

Matter waves: wavelength of a moving particle

The matter-wave model assigns a wavelength to a moving particle. This wavelength is called the de Broglie wavelength. A particle with greater momentum has a shorter de Broglie wavelength. Momentum is a vector, but the wavelength equation uses the magnitude of momentum.

For a non-relativistic particle, momentum magnitude is mass times speed. If an electron starts from rest and is accelerated through a potential difference, the electrical energy transferred becomes kinetic energy. That provides a way to find its speed or momentum before using the matter-wave relationship.

Keep the calculation in SI units. The resulting wavelength is in metres when Planck’s constant is in joule-seconds and momentum is in kilogram-metres per second. These relationships are used for the stated particle speeds and energies; do not apply the non-relativistic kinetic-energy relationship when a problem specifies a relativistic treatment.

\lambda=\frac{h}{p}

Use momentum magnitude when calculating a de Broglie wavelength.
For motion along a chosen axis, momentum direction follows velocity direction; wavelength itself has no direction.
Check whether the problem gives momentum directly or requires you to find it from speed or energy.

Worked example

Photoelectric effect: finding electron speed

Light of frequency

6.00\times10^{14}\ \mathrm{Hz}

shines on a material with a work function of

2.00\ \mathrm{eV}

. Find the maximum speed of an emitted electron. Treat the material as the reference frame and report speed as a magnitude.

Convert the work function
The photon energy and work function must use the same unit. Convert the given work function to joules using the electronvolt conversion.
$\phi=(2.00\ \mathrm{eV})(1.602\times10^{-19}\ \mathrm{J/eV})=3.20\times10^{-19}\ \mathrm{J}$
Find the photon energy
Use the photon relationship $E=hf$ . The frequency is already in inverse seconds, so the result is in joules.
$E=(6.626\times10^{-34}\ \mathrm{J\,s})(6.00\times10^{14}\ \mathrm{s^{-1}})=3.98\times10^{-19}\ \mathrm{J}$
Find maximum kinetic energy
Apply energy conservation for the photoelectric effect. The positive remainder is available as the electron’s maximum kinetic energy.
$K_{\max}=E-\phi=3.98\times10^{-19}\ \mathrm{J}-3.20\times10^{-19}\ \mathrm{J}=7.72\times10^{-20}\ \mathrm{J}$
Convert kinetic energy to speed
Use the non-relativistic kinetic-energy relationship and solve for speed. This energy is small compared with an electron’s rest energy, so this course-level relationship is suitable.
$v_{\max}=\sqrt{\frac{2K_{\max}}{m_e}}=\sqrt{\frac{2(7.72\times10^{-20}\ \mathrm{J})}{9.11\times10^{-31}\ \mathrm{kg}}}=4.12\times10^{5}\ \mathrm{m/s}$

Answer: The maximum electron speed is

4.12\times10^{5}\ \mathrm{m/s}

.

Check: The energy difference is positive, so emission is possible. The square root gives metres per second because joules per kilogram are equivalent to square metres per square second. The result is much less than the speed of light, consistent with the non-relativistic calculation.

Worked example

Compton effect: finding the scattered wavelength

A photon with an initial wavelength of

50.0\ \mathrm{pm}

scatters through

90.0^\circ

from an electron initially at rest. Find the outgoing photon wavelength. Use the incoming photon direction as the positive horizontal direction.

Calculate the wavelength shift
The photon’s scattering angle is measured from its incoming direction. At $90.0^\circ$ , the cosine is zero, so the shift equals the electron Compton wavelength.
$\Delta\lambda=\frac{(6.626\times10^{-34}\ \mathrm{J\,s})}{(9.11\times10^{-31}\ \mathrm{kg})(3.00\times10^8\ \mathrm{m/s})}(1-\cos90.0^\circ)=2.43\times10^{-12}\ \mathrm{m}$
Find the outgoing wavelength
Convert the shift to picometres, then add it to the initial wavelength. The scattered photon has the longer wavelength.
$\lambda_f=50.0\ \mathrm{pm}+2.43\ \mathrm{pm}=52.4\ \mathrm{pm}$

Answer: The outgoing photon wavelength is

52.4\ \mathrm{pm}

.

Check: The shift has units of length because

h/(m_ec)

has units of metres. The outgoing wavelength is greater than the incoming wavelength, as expected for this scattering process.

Worked example

Matter waves: electron accelerated from rest

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

150\ \mathrm{V}

. Find its de Broglie wavelength. Use the apparatus as the reference frame and treat the electron as non-relativistic.

Find the kinetic energy
For an electron accelerated from rest, the gained kinetic energy equals the charge magnitude times the potential difference. The result is expressed in joules.
$K=eV=(1.602\times10^{-19}\ \mathrm{C})(150\ \mathrm{J/C})=2.40\times10^{-17}\ \mathrm{J}$
Find the momentum magnitude
Combine $K=p^2/(2m_e)$ with the kinetic energy just found. Take the positive root because the equation asks for momentum magnitude.
$p=\sqrt{2m_eK}=\sqrt{2(9.11\times10^{-31}\ \mathrm{kg})(2.40\times10^{-17}\ \mathrm{J})}=6.61\times10^{-24}\ \mathrm{kg\,m/s}$
Calculate the wavelength
Use the de Broglie relationship. Since the question asks for wavelength, report a positive length; the electron’s direction does not change that magnitude.
$\lambda=\frac{h}{p}=\frac{6.626\times10^{-34}\ \mathrm{J\,s}}{6.61\times10^{-24}\ \mathrm{kg\,m/s}}=1.00\times10^{-10}\ \mathrm{m}$

Answer: The electron’s de Broglie wavelength is

1.00\times10^{-10}\ \mathrm{m}

, or

0.100\ \mathrm{nm}

.

Check: The units reduce to metres. A moving electron has a finite wavelength, and greater momentum would give a shorter wavelength. The energy is low enough for the stated non-relativistic model.

Common mistakes and how to avoid them

Using light intensity to calculate the energy of one photon.

Correction: Use frequency in

E=hf

to find the energy of one photon. Intensity is not a substitute for frequency in that relationship.

Subtracting the Compton wavelength shift from the initial photon wavelength.

Correction: For the stated Compton scattering model, add the shift to the initial wavelength to find the scattered photon’s wavelength.

Using a particle’s speed in place of momentum in the de Broglie equation.

Correction: Use momentum magnitude. If only speed is given, first calculate momentum from mass and speed.

Mixing electronvolts and joules in one energy subtraction.

Correction: Convert all energy terms to the same unit before applying energy conservation.

Lesson summary

For photoelectric problems, find photon energy, subtract the work function, and use the remaining energy as maximum electron kinetic energy.
For Compton problems, calculate the wavelength shift from the scattering angle, then add it to the initial wavelength.
For matter-wave problems, find momentum magnitude and use the de Broglie relationship.
Keep units consistent, report sensible significant figures, and check signs and physical reasonableness.

Check your understanding

Question 1

A photon has frequency

5.00\times10^{14}\ \mathrm{Hz}

. What is its energy?

$3.31\times10^{-19}\ \mathrm{J}$
$1.33\times10^{-19}\ \mathrm{J}$
$9.94\times10^{-20}\ \mathrm{J}$
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Show answer and explanation

3.31\times10^{-19}\ \mathrm{J}

Use

E=hf

. Multiplying

6.626\times10^{-34}\ \mathrm{J\,s}

by

5.00\times10^{14}\ \mathrm{s^{-1}}

gives

3.31\times10^{-19}\ \mathrm{J}

.

Question 2

For a photon scattered through

0^\circ

, what is the Compton wavelength shift?

Zero
$h/(m_ec)$
Twice $h/(m_ec)$
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Show answer and explanation

Zero

At zero degrees,

1-\cos0^\circ=0

, so the wavelength shift is zero.

Question 3

If a particle’s momentum magnitude doubles, what happens to its de Broglie wavelength?

It is halved.
It doubles.
It remains unchanged.
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Show answer and explanation

It is halved.

Since

\lambda=h/p

, doubling momentum halves the wavelength.

Key terms

Work function: The minimum energy needed to remove an electron from a material.
Maximum kinetic energy: The greatest kinetic energy of electrons emitted in a photoelectric interaction.
Compton wavelength shift: The change in a photon’s wavelength after it scatters from an electron.
de Broglie wavelength: The wavelength assigned to a moving particle by the matter-wave model.
Momentum: A vector quantity equal to mass times velocity for the non-relativistic particle model used here.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Physics (SPH4U), expectation F2.2. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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