AP Physics 2 · Topic 15.1

Topic 15.1: Quantum Theory and Wave-Particle Duality

Unit 15: Modern Physics12-15% of the multiple-choice section

Wave-particle duality is the claim that one object can behave as a particle in some situations and as a wave in others. Light does it: photons carry energy hf and momentum, yet also interfere. Matter does it too, with a de Broglie wavelength equal to Planck's constant divided by momentum.

AP Physics: Unit 15 (topics 15.1 Quantum Theory and Wave-Particle Duality). AP Physics 2 Unit 15, Topic 15.1. One learning objective, 15.1.A, describe the properties and behavior of an object that exhibits both particle-like and wave-like behavior. Five essential knowledge statements: 15.1.A.1 (quantum theory was developed to explain observations of matter and energy that could not be explained using classical mechanics, including but not limited to atomic spectra, blackbody radiation, and the photoelectric effect), with 15.1.A.1.i and 15.1.A.1.ii; 15.1.A.2 (light can be modeled both as a wave and as discrete particles, called photons), with 15.1.A.2.i defining a photon as massless and electrically neutral with energy proportional to frequency and citing E = hf and lambda = c/f, and 15.1.A.2.ii on photons travelling in straight lines unless they interact with matter; 15.1.A.3 (the speed of a photon depends on the medium), with 15.1.A.3.i giving c = 3.00e8 m/s in free space and 15.1.A.3.ii making photon speed inversely proportional to the index of refraction; 15.1.A.4 (particles can demonstrate wave properties, as shown by variations of Young's double-slit experiment), with 15.1.A.4.i citing the de Broglie wavelength lambda = h/p and noting it increases as momentum decreases, and 15.1.A.4.ii stating quantum theory is necessary where the de Broglie wavelength is comparable to the size of the system; and 15.1.A.5 (energy and momentum have discrete, or quantized, values for bound systems described by quantum theory). The topic has NO boundary statement. Suggested skills: 1.A, 2.A, 2.B, 3.B. Three of the ten equations in the Modern Physics group of the AP Physics 2 equation sheet are cited here: E = hf, lambda = c/f, lambda = h/p. The combined form E = hc/lambda is not printed. Relevant printed constants: h = 6.63e-34 J s = 4.14e-15 eV s, hc = 1.99e-25 J m = 1240 eV nm, c = 3.00e8 m/s, 1 eV = 1.60e-19 J, m_e = 9.11e-31 kg. The CED's sample free-response question 4, worth 8 points, is aligned to 15.1.A together with 15.5.A, with skills 2.A, 2.D, 3.B and 3.C.

What Topic 15.1 requires

Topic 15.1 carries a single learning objective, 15.1.A: describe the properties and behavior of an object that exhibits both particle-like and wave-like behavior. Five essential knowledge statements sit under it, and the topic prints no boundary statement, so nothing here is explicitly ruled out.

15.1.A.1 Quantum theory was developed to explain observations of matter and energy that could not be explained using classical mechanics. These phenomena include, but are not limited to, atomic spectra, blackbody radiation, and the photoelectric effect.
15.1.A.1.i Quantum theory is necessary to describe the properties of matter at atomic and subatomic scales.
15.1.A.1.ii In quantum theory, fundamental particles can exhibit both particle-like and wave-like behavior.
15.1.A.2 Light can be modeled both as a wave and as discrete particles, called photons.
15.1.A.2.i A photon is a massless, electrically neutral particle with energy proportional to the photon's frequency. Relevant equations: $E = hf$ and $\lambda = \frac{c}{f}$ .
15.1.A.2.ii Photons travel in straight lines unless they interact with matter.
15.1.A.3 The speed of a photon depends on the medium through which the photon travels.
15.1.A.3.i The speed of all photons in free space is equal to the speed of light, $c = 3.00 \times 10^8$ m/s.
15.1.A.3.ii In general, the speed of photons through a given medium is inversely proportional to the index of refraction of that medium.
15.1.A.4 Particles can demonstrate wave properties, as shown by variations of Young's double-slit experiment.
15.1.A.4.i A wave model of matter is quantified by the de Broglie wavelength, which increases as the momentum of a particle decreases. Relevant equation: $\lambda = \frac{h}{p}$ .
15.1.A.4.ii Quantum theory is necessary to describe systems where the de Broglie wavelength is comparable to the size of the system.
15.1.A.5 Values of energy and momentum have discrete, or quantized, values for bound systems described by quantum theory.

The suggested skills for the topic are 1.A (create diagrams, tables, charts, or schematics to represent physical situations), 2.A (derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway), 2.B (calculate or estimate an unknown quantity with units from known quantities), and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Read 15.1.A.1 again and notice what it is doing. It lists atomic spectra, blackbody radiation and the photoelectric effect, which are Topic 15.3, Topic 15.4 and Topic 15.5. The unit announces its own table of contents in its first sentence. The hedge "but are not limited to" is in the CED text and matters: the list is not a closed set, and Compton scattering is a fourth example the unit gets to later.

Duality is a statement about behavior, not about identity

The unhelpful version of this topic is "light is both a particle and a wave", which sounds like a contradiction and gives you nothing to write in an answer.

The CED's version is narrower and more useful. 15.1.A.2 says light can be modeled both as a wave and as discrete particles. 15.1.A.1.ii says fundamental particles can exhibit both particle-like and wave-like behavior. Both sentences are about which model reproduces which observation, not about what light secretly is underneath.

So the exam-usable form of duality is a rule about model selection:

When the observation is about where energy arrives, in what amounts, one lump at a time, the particle model is the one that works. Electrons ejected above a threshold frequency and not below it. A photon scattering off an electron and conserving momentum.
When the observation is about spreading, overlapping and cancelling, the wave model works. Two slits producing a fringe pattern. Light bending round an edge.

Nothing in Topic 15.1 asks you to say which model is true. It asks you to say which one the situation calls for, and why. That is skill 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim, and it is the skill the topic lists.

The symmetric half is the surprising one. 15.1.A.4 states that particles can demonstrate wave properties, as shown by variations of Young's double-slit experiment. The CED names the experiment by name, which is a pointer back at Topic 14.8: the same two-slit geometry that produces fringes with light produces fringes with electrons. The fringes are the evidence, and the de Broglie wavelength is the number that predicts their spacing.

The photon, defined by three properties and one behavior

15.1.A.2.i is a definition worth memorising in the CED's own words, because every word in it carries weight. A photon is:

massless. Not "very light". This is why a photon's energy cannot be written as $\frac{1}{2}mv^2$ , and why the photon's momentum has to come from $\lambda = h/p$ rather than from $p = mv$ .
electrically neutral. A photon is not deflected by the electric and magnetic fields you spent Units 10 to 12 on.
carrying energy proportional to the photon's frequency, with the relevant equations $E = hf$ and $\lambda = c/f$ .

The behavior is 15.1.A.2.ii: photons travel in straight lines unless they interact with matter. That single sentence is the bridge back to Unit 13, where the whole ray model rests on light going straight until a surface changes its direction.

A photon does carry momentum, even though it has no mass. The CED says so in 15.6.A.2.ii, where the transfer of a photon's energy to an electron results in the energy, momentum, frequency, and wavelength of the photon changing. The number comes from the printed relation $\lambda = h/p$ read backwards as $p = h/\lambda$ , and for a photon that equals $E/c$ . The worked examples below check that equality on real numbers.

One word to be careful with: intensity. Turning a light source up makes more photons per second, but it does not change what one photon carries. $E = hf$ has no intensity in it. That distinction is the entire content of Topic 15.5, and it is set up here.

Two printed equations, one photon, and the unit trap

Topic 15.1 cites three of the ten equations in the Modern Physics group of the AP Physics 2 equation sheet:

E = hf \qquad \lambda = \frac{c}{f} \qquad \lambda = \frac{h}{p}

They are printed exactly like that, with no combined form. The combination that does most of the practical work, $E = hc/\lambda$ , is not printed anywhere, so make it yourself by eliminating $f$ between the first two. That is a one-line derivation and it is skill 2.A.

The constants come from the Table of Information at the front of the sheet:

Constant	Printed value
Planck's constant	$h = 6.63 \times 10^{-34}$ J s $= 4.14 \times 10^{-15}$ eV s
$hc$	$1.99 \times 10^{-25}$ J m $= 1240$ eV nm
Speed of light	$c = 3.00 \times 10^8$ m/s
Electron volt	$1$ eV $= 1.60 \times 10^{-19}$ J
Electron mass	$m_e = 9.11 \times 10^{-31}$ kg

The sheet prints $h$ twice, in joule seconds and in electron volt seconds, and it prints $hc$ twice, in joule metres and in electron volt nanometres. That is a deliberate convenience, and using it properly is a large part of the arithmetic skill this unit asks for. Pick the units first, then pick the constant.

Answer wanted in electron volts, wavelength given in nanometres: use $E = \frac{1240 \text{ eV nm}}{\lambda}$ and do not convert anything.
Answer wanted in joules, wavelength given in metres: use $E = \frac{1.99 \times 10^{-25} \text{ J m}}{\lambda}$ , or go through $f = c/\lambda$ and $E = hf$ .

The factor of $10^9$ between metres and nanometres is where this goes wrong, and it goes wrong in a way that looks plausible: an answer nine orders of magnitude out still has a familiar-looking mantissa.

One honest wrinkle. The sheet's own constants are not perfectly self-consistent, because they are each rounded to three significant figures. Multiply the printed $h$ by the printed $c$ and divide by the printed elementary charge and you get about $1243$ eV nm, not $1240$ . The gap is roughly a quarter of a percent, so two significant figures are safe and the third digit can move depending on which route you take. Do not chase it, and do not assume you have made an error when a two-route check disagrees in the last place.

The speed of a photon is not always c

15.1.A.3 is short and easy to skim past: the speed of a photon depends on the medium through which the photon travels. It then splits into two claims that are worth keeping separate.

15.1.A.3.i pins the vacuum case: the speed of all photons in free space is equal to the speed of light, $c = 3.00 \times 10^8$ m/s. The word doing the work there is all. A gamma photon and a radio photon travel at the same speed in free space; they differ in energy, not in speed.

15.1.A.3.ii handles media: in general, the speed of photons through a given medium is inversely proportional to the index of refraction of that medium. That is the same statement as $n = c/v$ from Topic 13.3, reached from the photon side rather than the wave side.

What changes and what does not, when light crosses into glass:

Quantity	In the medium	Why
Speed	Decreases to $v = c/n$	15.1.A.3.ii
Frequency	Unchanged	14.3.A.1.iv: the frequency of a wave does not change when it travels from one medium to another
Photon energy	Unchanged, since $E = hf$ and $f$ is unchanged	15.1.A.2.i
Wavelength	Decreases, since $\lambda = v/f$ with a smaller $v$	Waves group of the sheet

Notice which printed equation you are allowed to use where. $\lambda = c/f$ in the Modern Physics group is the vacuum relation. Inside a medium the relation is $\lambda = v/f$ , which the Waves, Sound, and Optics group of the same sheet prints. Substituting $c$ for a photon travelling in glass is a quiet, common error, and it produces a wavelength that is too long by a factor of $n$ .

The cross-link to remember for Topic 14.3 is that this is exactly why refraction bends light: the wave slows, the frequency holds, so the wavelength has to shrink, and a wavefront meeting the surface at an angle has to change direction to stay continuous.

Matter waves and the de Broglie wavelength

Run the argument backwards. Light was a wave; the unit's experiments make it also behave like a particle. So ask whether an electron, which was a particle, also behaves like a wave. 15.1.A.4 answers yes, and cites the evidence: particles can demonstrate wave properties, as shown by variations of Young's double-slit experiment.

The quantitative statement is 15.1.A.4.i, and it is worth quoting whole because the second half is the part that gets dropped: a wave model of matter is quantified by the de Broglie wavelength, which increases as the momentum of a particle decreases.

\lambda = \frac{h}{p}

That inverse relationship is what the exam tests, more often than the number itself. Slow the particle down, or make it lighter, and the wavelength grows. Speed it up and the wavelength shrinks toward nothing, which is the classical limit.

For a non-relativistic particle of mass $m$ and kinetic energy $K$ , momentum comes from $K = \frac{p^2}{2m}$ , so

p = \sqrt{2mK} \qquad \text{and} \qquad \lambda = \frac{h}{\sqrt{2mK}}

Neither of those is printed. Both follow in one step from $K = \frac{1}{2}mv^2$ and $p = mv$ , which are printed in the Mechanics and Fluids group, so deriving them is fair game and is exactly what skill 2.A asks for.

The scaling is worth internalising before any numbers. For a photon, energy is inversely proportional to wavelength: halve $\lambda$ and you double $E$ . For a slow massive particle, kinetic energy is inversely proportional to the square of the wavelength: halve $\lambda$ and you quadruple $K$ . Same wavelength, different energy, because the two objects get their momentum from different places.

When quantum theory is actually necessary (15.1.A.4.ii)

This is the sentence to have exactly: quantum theory is necessary to describe systems where the de Broglie wavelength is comparable to the size of the system. 15.1.A.1.i says the same thing in different words: quantum theory is necessary to describe the properties of matter at atomic and subatomic scales.

That gives you a test you can actually run:

Compute the de Broglie wavelength of the object, $\lambda = h/p$ .
Compare it to the size of the thing the object has to fit through or inside.
If the two are comparable, wave behaviour shows up and the classical model fails. If the wavelength is vastly smaller, classical physics is fine.

The answer to "why do we never see a person diffract through a doorway" is not that the rule stops applying to large objects. It is that $h = 6.63 \times 10^{-34}$ J s is small and a person's momentum is not, so the wavelength comes out around $10^{-35}$ m, which is not comparable to a doorway by any stretch. The worked examples put numbers on both ends of that comparison.

This is also the sentence that justifies Topic 15.2. An electron bound in an atom has a de Broglie wavelength of the same order as the atom, which is precisely the condition 15.1.A.4.ii names, and it is why the standing-wave condition in 15.2.A.3.ii is a sensible thing to impose.

Bound systems are quantized (15.1.A.5)

The topic's last statement is a single sentence and it is the hinge for the next two topics: values of energy and momentum have discrete, or quantized, values for bound systems described by quantum theory.

Three words in it are load-bearing.

Bound. The quantization applies to systems that are held together, such as an electron trapped by the attraction of a nucleus. A free electron crossing a vacuum tube is not bound and its energy is not restricted to a list.

Discrete. Not "small", not "approximate". A bound system's energy takes values from a list with gaps between them, and values in the gaps do not occur.

Energy and momentum. The CED names both quantities, not just energy.

Everything the atom does in Topic 15.2 and Topic 15.3 is a consequence of this one line. If the energies are a list, then the differences between them are also a list, and 15.3.A.2 says an atom can only absorb or emit an amount of energy that corresponds to the energy difference between two of its states. A spectrum with a fixed set of lines in it is what a list with gaps looks like from the outside.

The mechanism the CED offers for why the list exists is the standing-wave condition of 15.2.A.3.ii, and that condition is built on the de Broglie wavelength from this topic. So the chain runs: matter has a wavelength (15.1.A.4.i), a bound electron's wave has to fit around its orbit (15.2.A.3.ii), only certain orbits let it fit, therefore only certain energies exist (15.1.A.5), therefore only certain photon energies come out (15.3.A.2).

How Topic 15.1 is tested

The clearest evidence in the CED is its own sample free-response question 4, the Qualitative/Quantitative Translation question. It is worth 8 points and is aligned to two learning objectives at once, 15.1.A and 15.5.A, with skills 2.A, 2.D, 3.B and 3.C. Its three parts run: predict whether a quantity increases, decreases or stays the same and justify it with reasoning beyond algebra; derive a symbolic expression for that quantity, beginning from a fundamental principle or an equation from the reference information; then say whether the derived expression agrees with the earlier justification.

That shape tells you what to practise. Not "compute the photon energy", but:

2.A, derive. Get from the printed $E = hf$ and $\lambda = c/f$ to a symbolic answer in terms of the quantities the question names, before substituting anything.
3.B, apply a model to make a claim. Say which model, particle or wave, the situation requires, and say why.
3.C, justify with evidence. Point at the observation. The threshold frequency, the interference pattern, the momentum transfer.
2.B, calculate. With units chosen deliberately, per the table above.
1.A, represent. Sketch the situation. For duality questions that usually means a beam, a barrier and what arrives on the far side.

Skill 2.D, predict new values or factors of change using functional dependence, is not listed for Topic 15.1 itself but is flagged for the unit as a whole and appears on the sample free-response question, so expect "the wavelength doubles, what happens to the photon energy" phrasing. The answers follow straight from the printed forms: photon energy goes as $1/\lambda$ , de Broglie wavelength goes as $1/p$ .

What this topic does not ask for: no Heisenberg uncertainty principle, no wavefunction, no probability density, no relativistic momentum. None of those words appear in the topic. The atom's structure is a different topic and comes with its own boundary statement limiting it to energy levels. Topic 15.1 is about one object, one wavelength, one energy, and the choice of which model to reach for.

Two de Broglie wavelengths, and the 15.1.A.4.ii test

(a) An electron starts at rest and is accelerated through a potential difference of $100 \text{ V}$ , giving it a kinetic energy of $100 \text{ eV}$ . Find its de Broglie wavelength. (b) Find the de Broglie wavelength of a $0.145 \text{ kg}$ baseball moving at $40 \text{ m/s}$ . (c) Apply the test in 15.1.A.4.ii to each.

(a) Convert the kinetic energy to joules first, using the printed conversion $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$ : $K = (100)(1.60 \times 10^{-19}) = 1.60 \times 10^{-17} \text{ J}$ .
Get the momentum. Nothing on the sheet gives $p$ from $K$ directly, so build it: $K = \frac{1}{2}mv^2$ and $p = mv$ combine to $p = \sqrt{2mK}$ . Then $p = \sqrt{2(9.11 \times 10^{-31} \text{ kg})(1.60 \times 10^{-17} \text{ J})} = \sqrt{2.915 \times 10^{-47}} = 5.40 \times 10^{-24} \text{ kg} \cdot \text{m/s}$ .
Check the non-relativistic assumption before trusting that: $v = p/m = (5.40 \times 10^{-24})/(9.11 \times 10^{-31}) = 5.93 \times 10^6 \text{ m/s}$ , which is about 2 percent of $c$ . Safe.
Now the printed de Broglie relation: $\lambda = \dfrac{h}{p} = \dfrac{6.63 \times 10^{-34} \text{ J} \cdot \text{s}}{5.40 \times 10^{-24} \text{ kg} \cdot \text{m/s}} = 1.23 \times 10^{-10} \text{ m}$ , which is $0.123 \text{ nm}$ .
(b) The baseball needs no conversion. $p = mv = (0.145 \text{ kg})(40 \text{ m/s}) = 5.8 \text{ kg} \cdot \text{m/s}$ , so $\lambda = \dfrac{6.63 \times 10^{-34}}{5.8} = 1.1 \times 10^{-34} \text{ m}$ .
(c) Apply 15.1.A.4.ii: quantum theory is necessary where the de Broglie wavelength is comparable to the size of the system. The electron's $0.123 \text{ nm}$ is the same order as the size of an atom: the smallest allowed orbit radius that Topic 15.2 derives from printed constants is $5.30 \times 10^{-11} \text{ m}$ , so an atom is around $10^{-10} \text{ m}$ across. An electron at this wavelength therefore meets matter on its own scale, which is the condition 15.1.A.4.ii names and the setting for the double-slit variations 15.1.A.4 points at. The baseball's $10^{-34} \text{ m}$ is smaller than a nucleus by about twenty orders of magnitude and is not comparable to anything the ball could pass through, so its wave behaviour is undetectable and classical mechanics is complete.
The moral is in the algebra, not the arithmetic. Both objects obey $\lambda = h/p$ . The baseball's momentum is about $10^{24}$ times the electron's, so its wavelength is about $10^{24}$ times smaller. Nothing switched off; the number just left the measurable range.

(a) $\lambda = 1.23 \times 10^{-10} \text{ m} = 0.123 \text{ nm}$ , comparable to atomic spacing, so the wave model is required. (b) $\lambda = 1.1 \times 10^{-34} \text{ m}$ , comparable to nothing, so classical mechanics suffices.

Same wavelength, photon versus electron

The electron from the previous example has a de Broglie wavelength of $0.123 \text{ nm}$ and a kinetic energy of $100 \text{ eV}$ . Find the energy of a photon whose wavelength is also $0.123 \text{ nm}$ , and explain why the two energies are so different.

Use the printed constant in the units the question is already in. With $hc = 1240 \text{ eV} \cdot \text{nm}$ and $\lambda$ in nanometres: $E = \dfrac{1240 \text{ eV} \cdot \text{nm}}{0.123 \text{ nm}} = 1.01 \times 10^4 \text{ eV}$ , that is about $10.1 \text{ keV}$ .
Cross-check by the other route. $f = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^8}{1.23 \times 10^{-10}} = 2.44 \times 10^{18} \text{ Hz}$ , then $E = hf = (6.63 \times 10^{-34})(2.44 \times 10^{18}) = 1.62 \times 10^{-15} \text{ J}$ , which is $1.62 \times 10^{-15}/1.60 \times 10^{-19} = 1.01 \times 10^4 \text{ eV}$ . The routes agree.
So the photon carries about $10100 \text{ eV}$ while the electron with the same wavelength carries $100 \text{ eV}$ , a factor of about 100.
The reason is that the two objects convert momentum into energy differently. Both have the same momentum, since both have the same wavelength and $p = h/\lambda$ applies to both. But for a massless photon, $E = pc$ , linear in $p$ . For a slow massive electron, $K = \dfrac{p^2}{2m}$ , quadratic in $p$ and divided by a mass.
Check that against the numbers: the photon's $E = pc = (5.40 \times 10^{-24})(3.00 \times 10^8) = 1.62 \times 10^{-15} \text{ J}$ , matching the value above exactly, as it must.
The practical consequence, and a good 3.B claim: to probe structure at a given length scale you need a wavelength of that size, and an electron gets you there far more cheaply in energy than a photon does. That is the reason electron microscopes exist.

The photon carries $1.01 \times 10^4 \text{ eV}$ , about 100 times the electron's $100 \text{ eV}$ , because a photon's energy is $pc$ while a slow electron's is $p^2/2m$ .

A photon crossing into glass: what changes and what does not

A photon of vacuum wavelength $600 \text{ nm}$ enters a block of glass with index of refraction $n = 1.50$ . Find (a) its frequency in the glass, (b) its speed in the glass, (c) its wavelength in the glass, and (d) its energy in the glass.

Start in vacuum, where $\lambda = c/f$ from the Modern Physics group applies: $f = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^8 \text{ m/s}}{600 \times 10^{-9} \text{ m}} = 5.00 \times 10^{14} \text{ Hz}$ .
(a) The frequency in the glass is the same $5.00 \times 10^{14} \text{ Hz}$ . That is 14.3.A.1.iv: the frequency of a wave does not change when it travels from one medium to another. Frequency is set by the source, not by the material.
(b) From 15.1.A.3.ii, the speed is inversely proportional to the index of refraction, the same relation as $n = c/v$ : $v = \dfrac{c}{n} = \dfrac{3.00 \times 10^8 \text{ m/s}}{1.50} = 2.00 \times 10^8 \text{ m/s}$ .
(c) Inside the medium the relation to use is $\lambda = v/f$ from the Waves, Sound, and Optics group, not $\lambda = c/f$ : $\lambda_{glass} = \dfrac{2.00 \times 10^8 \text{ m/s}}{5.00 \times 10^{14} \text{ Hz}} = 4.00 \times 10^{-7} \text{ m} = 400 \text{ nm}$ . Using $c$ here instead of $v$ would have returned $600 \text{ nm}$ , too long by exactly the factor $n$ .
(d) Energy comes from $E = hf$ , and the frequency did not change, so the energy did not change: $E = (6.63 \times 10^{-34})(5.00 \times 10^{14}) = 3.32 \times 10^{-19} \text{ J}$ , or $2.07 \text{ eV}$ .
Sanity check on (d) using the shortcut with the vacuum wavelength: $1240/600 = 2.07 \text{ eV}$ . Note that the shortcut needs the vacuum wavelength, because $hc$ was built from $c$ . Feeding it the $400 \text{ nm}$ in-glass wavelength would have given $3.10 \text{ eV}$ , which is wrong, and wrong in a way that would also break conservation of energy at the surface.

(a) $5.00 \times 10^{14} \text{ Hz}$ , unchanged. (b) $2.00 \times 10^8 \text{ m/s}$ . (c) $400 \text{ nm}$ . (d) $3.32 \times 10^{-19} \text{ J}$ , or $2.07 \text{ eV}$ , unchanged.

Frequently asked questions

What is wave-particle duality in AP Physics 2?

Wave-particle duality is the idea that a single object can be modeled as a wave in some situations and as a particle in others. The AP Physics 2 CED states it twice in Topic 15.1: essential knowledge 15.1.A.1.ii says fundamental particles can exhibit both particle-like and wave-like behavior, and 15.1.A.2 says light can be modeled both as a wave and as discrete particles called photons. The exam asks which model a given observation requires, not which one is really true. Energy arriving in fixed lumps calls for the particle model; interference and diffraction call for the wave model.

What is the de Broglie wavelength formula?

The de Broglie wavelength is lambda = h / p, Planck's constant divided by momentum, and it is printed in the Modern Physics group of the AP Physics 2 equation sheet. Essential knowledge 15.1.A.4.i adds the part that is easiest to drop: the de Broglie wavelength increases as the momentum of a particle decreases. For a non-relativistic particle you usually need p = sqrt(2mK), which is not printed but follows in one step from K = mv-squared over 2 and p = mv, both of which are on the sheet.

Why don't everyday objects show wave behavior?

Because their de Broglie wavelengths are far too small to be comparable to anything they interact with. Essential knowledge 15.1.A.4.ii sets the test: quantum theory is necessary to describe systems where the de Broglie wavelength is comparable to the size of the system. A baseball of mass 0.145 kg moving at 40 m/s has momentum 5.8 kg m/s, so its de Broglie wavelength is about 1.1 times 10 to the minus 34 metres, roughly twenty orders of magnitude smaller than a nucleus. An electron accelerated through 100 volts has a wavelength of about 0.12 nanometres, comparable to the size of an atom, so an electron meets matter on its own scale and a baseball does not.

Does a photon have mass or momentum?

A photon has no mass and does have momentum. Essential knowledge 15.1.A.2.i defines a photon as a massless, electrically neutral particle with energy proportional to its frequency. Its momentum follows from the printed relation lambda = h / p read as p = h / lambda, and for a photon that value also equals E divided by c. The AP Physics 2 CED relies on photon momentum in Topic 15.6, where 15.6.A.2.i says Compton scattering is explained by treating the photon as a particle and applying conservation of energy and conservation of momentum to its collision with an electron.

Is the speed of light always 3.00 times 10 to the 8 metres per second?

Only in free space. Essential knowledge 15.1.A.3.i says the speed of all photons in free space equals the speed of light, c = 3.00 times 10 to the 8 metres per second, and 15.1.A.3.ii says that in general the speed of photons through a given medium is inversely proportional to the index of refraction of that medium. So in glass with n = 1.50 a photon travels at 2.00 times 10 to the 8 metres per second. Its frequency and its energy are unchanged, per 14.3.A.1.iv and E = hf, so the wavelength shrinks by the factor n instead.

How do you convert a photon wavelength to energy in electron volts?

Divide 1240 by the wavelength in nanometres. The AP Physics 2 Table of Information prints hc = 1240 eV nm, and combining the printed equations E = hf and lambda = c/f gives E = hc / lambda, so a 500 nm photon carries 1240 / 500 = 2.48 eV. Keep the wavelength in nanometres for this route and do not convert to metres. If you need joules instead, either multiply the answer by 1.60 times 10 to the minus 19, or work in SI throughout with hc = 1.99 times 10 to the minus 25 joule metres.

Which Topic 15.1 equations are on the AP Physics 2 equation sheet?

Three of the ten equations in the Modern Physics group are cited by Topic 15.1: E = hf and lambda = c/f, both listed under essential knowledge 15.1.A.2.i, and lambda = h/p, listed under 15.1.A.4.i. The combined form E = hc/lambda is not printed and has to be assembled from the first two. There is no boundary statement for Topic 15.1, and the topic's suggested skills are 1.A, 2.A, 2.B and 3.B.