What is the difference between transverse and longitudinal waves in the context of Edexcel International A Levels Physics?

In Edexcel International A Levels Physics, transverse waves are characterized by oscillations that are perpendicular to the direction of wave propagation, such as light waves. Longitudinal waves, like sound waves, have oscillations parallel to the direction of wave travel. Understanding these differences is crucial for analyzing wave behaviors in various mediums.

How does the photoelectric effect demonstrate the particle nature of light?

The photoelectric effect, a key concept in Edexcel International A Levels Physics, shows that light can be thought of as particles called photons. When light shines on a metal surface, it can eject electrons if the photons have enough energy. This phenomenon supports the particle theory of light, as it cannot be explained by wave theory alone.

What role do electrons play in the conduction of electricity?

Electrons are fundamental to the conduction of electricity, as they are the charge carriers in conductive materials. In metals, electrons move freely between atoms, allowing electric current to flow. This concept is essential in understanding circuits and electrical components in the Edexcel International A Levels curriculum.

How do wave-particle duality and photons relate to each other?

Wave-particle duality is a fundamental principle in Physics, particularly in the study of photons. It posits that particles like photons exhibit both wave-like and particle-like properties. This duality is crucial for understanding phenomena such as diffraction and interference, which are part of the Edexcel International A Levels Physics syllabus.

What is the significance of wave speed, frequency, and wavelength in wave equations?

In Edexcel International A Levels Physics, wave speed, frequency, and wavelength are interconnected through the wave equation: wave speed = frequency × wavelength. This relationship helps in calculating one of these parameters if the other two are known, and is vital for solving problems related to sound, light, and other wave phenomena.

Why is "Calculating the slit spacing $d$ directly from 'lines per mm' without converting to lines per metre" a common mistake in Waves, Electrons and Photons?

Why it happens: Students take the reciprocal of the labelled figure (e.g. 1/300) and forget that the rating is per millimetre, not per metre. How to avoid it: **Convert to lines per metre first** (×1000), then take the reciprocal: for 300 lines/mm, N=3.00×10^5\ m^-1 and d=1/N=3.33×10^-6\ m. Check that d is of order 10^-6\ m. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Leaving the calculator in radian mode (or forgetting the inverse sine) when finding the grating angle" a common mistake in Waves, Electrons and Photons?

Why it happens: After computing = n/d, students either quote that number as the angle or evaluate ^-1 in radians. How to avoid it: n/d gives , **not ** — take the inverse sine, and make sure the calculator is in **degree mode**. A first-order angle for visible light is typically only a few tens of degrees, so a 'huge' answer signals a mode error. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Mixing up the photon equation $E = hf$ with the de Broglie equation $\lambda = h/p$" a common mistake in Waves, Electrons and Photons?

Why it happens: Both contain the Planck constant h, so students grab whichever they remember and substitute the wrong quantities. How to avoid it: Keep them separate: **E = hf = hc/ gives a photon's ENERGY** (photons of light); ** = h/p gives the WAVELENGTH of a moving particle** (electrons). Ask 'am I finding an energy or a particle's wavelength?' before choosing. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Multiplying instead of dividing (or vice versa) when converting between joules and electronvolts" a common mistake in Waves, Electrons and Photons?

Why it happens: Students memorise '1.60×10^-19' but not the direction of the conversion. How to avoid it: Remember the eV is the **smaller** unit, so there are *many* eV in a joule: to go **J → eV divide** by 1.60×10^-19; to go **eV → J multiply** by 1.60×10^-19. Sanity-check: a few-eV visible photon should be a few ×10^-19\ J. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Rounding the maximum-order calculation $n_{\max}=d/\lambda$ UP to the next integer" a common mistake in Waves, Electrons and Photons?

Why it happens: Students round 4.17 to 4.2 → 'about 4 or 5' or treat it like an ordinary number and round up. How to avoid it: The order must satisfy \le1, so you must round d/ **DOWN** to the nearest whole number. A non-integer like 4.17 means the highest visible order is **4** (the 5th would need >1). Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Substituting a velocity straight into $\lambda = h/p$ without first calculating the momentum $p = mv$" a common mistake in Waves, Electrons and Photons?

Why it happens: Students see = h/p and put the speed where the momentum belongs, omitting the mass. How to avoid it: Always compute the **momentum p = mv first** (include the mass), then divide h by it. When the particle is accelerated through a p.d., find v from eV=\tfrac12mv^2 before forming p. Source: Edexcel IAL Physics Unit 2 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Waves, Electrons and Photons

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Short Notes - Waves, Electrons and Photons

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Detailed Study Notes

Detailed notes on Waves and Electricity for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Waves, Electrons and Photons — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 2)

Diffraction and the diffraction grating equation $n\lambda = d\sin\theta$ (including Core Practical 6 — measuring the wavelength of laser light); the photon model of light, $E = hf = hc/\lambda$ , and the electronvolt; wave–particle duality and the wave nature of electrons through electron diffraction and the de Broglie relation $\lambda = h/p$ ; and estimating the Planck constant.

At a glance

Diffraction is the spreading of waves through a gap or round an edge — most noticeable when the gap is comparable to the wavelength.
Grating equation: $n\lambda = d\sin\theta$ . The slit spacing $d = \dfrac{1}{N}$ where $N$ is the number of lines per metre (always convert from per-mm first).
A photon is a discrete packet of electromagnetic energy: $E = hf = \dfrac{hc}{\lambda}$ , with $h = 6.63\times10^{-34}\ \text{J s}$ .
The electronvolt is a unit of energy: $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ . J → eV: divide by $1.60\times10^{-19}$ ; eV → J: multiply.
Wave–particle duality: light and matter each show both wave behaviour (diffraction/interference) and particle behaviour (photons / mass, charge, momentum).
Electron diffraction (a wave effect) is direct evidence that electrons have a wave nature.
de Broglie wavelength: $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$ — find the momentum first. Electrons show diffraction because $\lambda\sim10^{-10}\ \text{m}$ , comparable to atomic spacing.
$E = hf$ gives an energy; $\lambda = h/p$ gives a wavelength — both contain $h$ , so do not confuse them.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Recall that diffraction occurs when waves pass through a gap or round an obstacle, and that it is greatest when the gap is comparable to the wavelength (Huygens' construction qualitatively).
Use the diffraction grating equation $n\lambda = d\sin\theta$ , obtaining the slit spacing from the number of lines per unit length, $d = 1/N$ .
CP6 — Determine the wavelength of light from a laser using a diffraction grating, including a suitable safety precaution.
Describe the photon model of electromagnetic radiation and use $E = hf = hc/\lambda$ ; use the electronvolt as a unit of energy ( $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ ).
Explain wave–particle duality, and describe electron diffraction as evidence for the wave nature of electrons.
Use the de Broglie equation $\lambda = h/p$ , and explain why electrons (but not everyday objects) show observable diffraction.

Diffraction and the diffraction grating

Many slits, sharp maxima: $n\lambda = d\sin\theta$ , with $d = 1/N$ .

Diffraction is the spreading out of a wave when it passes through a gap or travels around the edge of an obstacle. Using Huygens' construction — treating every point on a wavefront as a source of secondary wavelets — you can picture how the wavefronts bend at an edge. Diffraction is most pronounced when the gap is about the same size as the wavelength: a wide gap gives little spreading, a gap close to $\lambda$ gives a lot.

A diffraction grating is a slide ruled with a very large number of equally spaced parallel slits (often hundreds per millimetre). When monochromatic light (a single wavelength, e.g. from a laser) passes through, the light from all the slits interferes. Bright maxima appear only at the special angles where the path difference between adjacent slits is a whole number of wavelengths, so that every slit is exactly in phase. Those angles obey the grating equation:

$n\lambda = d\sin\theta$

$n$ = the order (an integer: $0$ for the central maximum, $1$ for the first order, …),
$\lambda$ = wavelength of the light,
$d$ = the slit spacing (distance between adjacent slit centres),
$\theta$ = the angle of that order from the straight-through (zero-order) direction.

Getting $d$ right. Gratings are labelled in lines per millimetre. To find $d$ you must first convert to lines per metre ( $N$ ), then take the reciprocal:

$d = \frac{1}{N}$

For example, 300 lines/mm $= 3.00\times10^{5}$ lines/m, so $d = 1/(3.00\times10^{5}) = 3.33\times10^{-6}\ \text{m}$ .

A diffraction grating splits monochromatic light into a central zero-order beam and symmetric higher orders. Each order n appears at the angle θ given by n λ = d sin θ; higher orders sit at larger angles. The slit spacing is d = 1/N, where N is the number of lines per metre.

Because $\sin\theta$ can never exceed 1, there is a maximum order: $n_{\max}$ is the largest whole number satisfying $n \le d/\lambda$ . (Always round this down.)

Diffraction is greatest when the gap ≈ the wavelength.
Grating equation: $n\lambda = d\sin\theta$ ; $n$ is the integer order.
Slit spacing $d = 1/N$ — convert lines per mm to lines per metre first.
Maximum order: largest integer $n \le d/\lambda$ (since $\sin\theta\le1$ ).

See the full worked example for waves, electrons and photons →

Core Practical 6 — wavelength of laser light with a grating

Shine, measure the angle, rearrange $n\lambda = d\sin\theta$ for $\lambda$ — and protect your eyes.

Core Practical 6 uses a diffraction grating to measure the wavelength of monochromatic laser light.

Method (outline)

Mount the grating so the laser beam strikes it normally (at 90°).
Let the diffracted orders fall on a screen a measured distance away.
For an order $n$ , measure the angle $\theta$ from the straight-through spot (e.g. measure the distance along the screen and the grating-to-screen distance, then use trigonometry, or use a spectrometer for direct angle reading).
Rearrange the grating equation for the wavelength:

$\lambda = \frac{d\sin\theta}{n}$

Improving accuracy

Use a higher order $n$ where possible: the maxima are spread further apart, so the angle is measured more precisely.
Measure across symmetric orders (e.g. from the $+n$ spot to the $-n$ spot) and halve, to remove any error from the grating not being exactly normal.
Repeat and average; keep the grating-to-screen distance large.

Safety (a B mark you must not skip). A laser can permanently damage the retina. Never look directly into the beam or at a specular (mirror-like) reflection, keep the beam below eye level, and avoid shiny surfaces in its path. Examiners frequently award one mark purely for a valid laser-safety precaution.

The same arrangement run in reverse — known wavelength, measured angle — lets you find an unknown grating spacing, so the experiment links directly to the algebra of $n\lambda = d\sin\theta$ .

Beam must hit the grating normally; measure the order angle $\theta$ .
Rearrange: $\lambda = d\sin\theta / n$ .
Use a higher order and symmetric measurements for precision.
Quote a laser-safety precaution — it is a genuine mark.

See the full worked example for waves, electrons and photons →

The photon model and the electronvolt

Light comes in packets: $E = hf = hc/\lambda$ ; energies often quoted in eV.

Diffraction shows light behaving as a wave, but other experiments (notably the photoelectric effect) show light delivering energy in discrete packets. A single packet is a photon — a quantum of electromagnetic energy. The energy of one photon is:

$E = hf = \frac{hc}{\lambda}$

where $h = 6.63\times10^{-34}\ \text{J s}$ is the Planck constant, $f$ the frequency, $c = 3.00\times10^{8}\ \text{m s}^{-1}$ the speed of light, and $\lambda$ the wavelength. The two forms are linked by the wave equation $c = f\lambda$ . Higher frequency (shorter wavelength) means a more energetic photon — which is why ultraviolet photons carry more energy than visible ones.

The electronvolt. Photon energies in joules are tiny (around $10^{-19}\ \text{J}$ for visible light), so physicists use a more convenient unit, the electronvolt (eV). One electronvolt is the energy gained by an electron accelerated through a potential difference of 1 V:

$1\ \text{eV} = e \times 1\ \text{V} = 1.60\times10^{-19}\ \text{J}$

Converting between the two units trips many candidates, so fix the direction firmly:

joules → electronvolts: DIVIDE by $1.60\times10^{-19}$ ,
electronvolts → joules: MULTIPLY by $1.60\times10^{-19}$ .

A quick sanity check: a visible-light photon is a few eV (a few $\times10^{-19}\ \text{J}$ ). If your eV answer is enormous or microscopic, you have converted the wrong way.

The same light can be described as a wave of wavelength λ or as a stream of photons, each carrying energy E = hf = hc/λ. Photon energies are conveniently expressed in electronvolts, where 1 eV = 1.60 × 10⁻¹⁹ J.

The power of a light source equals the energy per second, so the number of photons emitted per second is $\dfrac{P}{E}$ — a favourite exam calculation.

Photon energy: $E = hf = hc/\lambda$ , with $h = 6.63\times10^{-34}\ \text{J s}$ .
Use $c = f\lambda$ to switch between the frequency and wavelength forms.
$1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ ; J → eV divide, eV → J multiply.
Photons per second from a source $= P/E$ .

See the full worked example for waves, electrons and photons →

Wave–particle duality and the de Broglie wavelength

Electrons diffract; $\lambda = h/p$ tells you why — find $p$ first.

If light (a 'wave') can behave as particles (photons), the converse question is natural: can particles behave as waves? Louis de Broglie proposed that any moving particle has an associated wavelength:

$\lambda = \frac{h}{p} = \frac{h}{mv}$

where $p = mv$ is the particle's momentum. This is the heart of wave–particle duality: both light and matter show wave-like and particle-like behaviour depending on the experiment.

Evidence — electron diffraction. Fire a beam of electrons at a thin layer of polycrystalline graphite and a pattern of concentric bright rings appears on a fluorescent screen. Diffraction is a wave property, so the rings prove electrons have a wave nature. The regular atomic planes of the crystal act like a natural grating; because the electron's de Broglie wavelength ( $\sim10^{-10}\ \text{m}$ ) is comparable to the atomic spacing, strong diffraction occurs.

Electrons passing through a thin crystal produce concentric diffraction rings — a wave effect. Speeding the electrons up increases their momentum, so by λ = h/p their wavelength shortens and the rings move closer to the centre.

Why everyday objects don't diffract. Diffraction is only noticeable when $\lambda$ is comparable to the size of the gap. A tennis ball has an enormous momentum compared with an electron, so its de Broglie wavelength is around $10^{-34}\ \text{m}$ — unimaginably smaller than any gap it could pass through, so its wave nature is completely undetectable.

Accelerated electrons. A charged particle accelerated from rest through a potential difference $V$ gains kinetic energy $eV = \tfrac12 mv^2$ . Find $v$ , then $p = mv$ , then $\lambda = h/p$ . This three-step chain is a classic Unit 2 question.

Don't confuse the two $h$ equations. $E = hf$ gives a photon's energy; $\lambda = h/p$ gives a particle's wavelength. Decide which you need before substituting.

de Broglie: $\lambda = h/p = h/mv$ — compute momentum $p = mv$ first.
Electron diffraction (rings) is evidence of the wave nature of electrons.
Electrons diffract because $\lambda\sim10^{-10}\ \text{m}\approx$ atomic spacing.
Accelerated electron: $eV = \tfrac12 mv^2 \to v \to p \to \lambda$ .

See the full worked example for waves, electrons and photons →

Quick recap

Diffraction spreads waves through gaps/round edges; it is greatest when the gap ≈ the wavelength.
Grating equation $n\lambda = d\sin\theta$ with $d = 1/N$ (convert lines per mm → per metre); maximum order is the largest integer $n \le d/\lambda$ .
Core Practical 6: rearrange to $\lambda = d\sin\theta/n$ ; use higher/symmetric orders for precision and always state a laser-safety precaution.
A photon is a packet of EM energy, $E = hf = hc/\lambda$ ; energies are often quoted in eV, where $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ (J → eV divide, eV → J multiply).
Wave–particle duality: light and matter both show wave and particle behaviour; electron diffraction is evidence of the wave nature of electrons.
de Broglie wavelength $\lambda = h/p = h/mv$ — find the momentum first; electrons diffract because their wavelength is comparable to atomic spacing.
$E = hf$ gives an energy and $\lambda = h/p$ gives a wavelength — keep the two equations apart, and quote answers to 3 s.f. with units.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Diffraction grating: $n\lambda = d\sin\theta$ (booklet); $d = 1/N$ with $N$ in lines per metre.
Maximum order: largest integer $n \le d/\lambda$ (round DOWN, since $\sin\theta \le 1$ ).
Photon energy: $E = hf = \dfrac{hc}{\lambda}$ ; $h = 6.63\times10^{-34}\ \text{J s}$ , $c = 3.00\times10^{8}\ \text{m s}^{-1}$ .
Electronvolt: $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ . J → eV divide; eV → J multiply.
de Broglie wavelength: $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$ — momentum $p = mv$ first.
Electron diffraction (rings) = evidence of the wave nature of electrons; $\lambda \approx$ atomic spacing.
Accelerated electron: $eV = \tfrac12 mv^2 \Rightarrow v \Rightarrow p \Rightarrow \lambda$ .
Laser safety: never look into the beam or its reflections (a marked B point in CP6).

How it’s examined

Waves, electrons and photons is Unit 2 content (WPH12, Waves and Electricity). It appears as: short calculations using the grating equation (find an angle, a wavelength, the spacing, or the maximum order — 2–4 marks, M and A marks); a Core Practical 6 data/method question (determine $\lambda$ from $n\lambda = d\sin\theta$ , evaluate precision, and state a laser-safety precaution — a B mark); photon-energy calculations with $E = hf = hc/\lambda$ and J ↔ eV conversion (2–4 marks); a 'photons per second $=P/E$ ' problem; an explain question on electron diffraction as evidence for the wave nature of electrons and how the pattern changes with speed (4 marks, AO1/AO2); de Broglie calculations including the accelerate-then-find- $\lambda$ chain (4–6 marks); and an extended/levels-style discussion of wave–particle duality requiring specific evidence for both light and electrons. ALWAYS show the equation, the substitution with units, and a 3 s.f. answer; keep full precision until the final step; convert 'lines per mm' to 'per metre' before finding $d$ ; take the inverse sine in degree mode; and never confuse $E = hf$ with $\lambda = h/p$ .

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 2 examiner reports; Edexcel IAL Physics Unit 2 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

Take this whole topic with you

Step-by-step worked examples — Waves, Electrons and Photons

Step-by-step solutions to past-paper-style questions on waves, electrons and photons, written exactly the way a tutor would explain them at the board.

Question type:

1Energy of a single photon (2 marks)
Getting startedDirect calculation• photon, AO2, Unit 2
Question
Green light has a frequency of $5.6 \times 10^{14}\ \text{Hz}$ . Take the Planck constant $h = 6.63 \times 10^{-34}\ \text{J s}$ . Calculate the energy of one photon of this light. (2 marks)
Step-by-step solution
1. Step 1
  Photon energy is $E = hf$ . The frequency is given, so substitute directly.
  $E = hf = (6.63 \times 10^{-34})(5.6 \times 10^{14})$
2. Step 2
  Evaluate, keeping standard form tidy.
  $E = 3.7128 \times 10^{-19}\ \text{J}$
Answer
$E = 3.71 \times 10^{-19}\ \text{J}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) correct equation/substitution $E=hf$ (M); (2) $3.71 \times 10^{-19}\ \text{J}$ (A). The most common slip is a power-of-ten error in the standard-form multiplication; write the powers separately ( $10^{-34}\times10^{14}=10^{-20}$ ) as a check.
2Slit spacing from lines per millimetre (2 marks)
Getting startedDirect calculation• diffraction grating, AO2, Unit 2
Question
A diffraction grating is labelled '300 lines per mm'. Calculate the spacing $d$ between adjacent slits. (2 marks)
Step-by-step solution
1. Step 1
  Convert lines per millimetre to lines per metre: $300\ \text{mm}^{-1} = 300 \times 1000 = 3.00 \times 10^{5}\ \text{m}^{-1}$ .
  $N = 300 \times 10^{3} = 3.00 \times 10^{5}\ \text{lines m}^{-1}$
2. Step 2
  The slit spacing is the reciprocal of the number of lines per metre, $d = 1/N$ .
  $d = \frac{1}{N} = \frac{1}{3.00 \times 10^{5}} = 3.33 \times 10^{-6}\ \text{m}$
Answer
$d = 3.33 \times 10^{-6}\ \text{m}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) convert to lines per metre ( $3.00\times10^{5}$ ); (2) $d=1/N=3.33\times10^{-6}\ \text{m}$ . The dominant error is forgetting the mm → m conversion and computing $d=1/300=3.33\times10^{-3}\ \text{m}$ — a thousand times too big.
3de Broglie wavelength of a moving electron (3 marks)
Getting startedDirect calculation• de Broglie, AO2, Unit 2
Question
An electron travels at $7.0 \times 10^{6}\ \text{m s}^{-1}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and the electron mass $m_e = 9.11 \times 10^{-31}\ \text{kg}$ . Calculate its de Broglie wavelength. (3 marks)
Step-by-step solution
1. Step 1
  First find the momentum $p = mv$ of the electron.
  $p = m_e v = (9.11 \times 10^{-31})(7.0 \times 10^{6}) = 6.377 \times 10^{-24}\ \text{kg m s}^{-1}$
2. Step 2
  Apply the de Broglie equation $\lambda = h/p$ .
  $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.377 \times 10^{-24}}$
3. Step 3
  Evaluate the wavelength.
  $\lambda = 1.04 \times 10^{-10}\ \text{m}$
Answer
$\lambda = 1.04 \times 10^{-10}\ \text{m}$ (about 0.1 nm — comparable to atomic spacing).
Examiner tip
Mark scheme: (1) $p=mv$ (M); (2) $\lambda=h/p$ with substitution (M); (3) $1.04\times10^{-10}\ \text{m}$ (A). Note this wavelength is of the same order as atomic spacings in a crystal (~ $10^{-10}\ \text{m}$ ), which is why electrons diffract through crystals.
4Diffraction angle of a grating order (3 marks)
Building confidenceDirect calculation• Adapted from Edexcel IAL Physics Unit 2 (Core Practical 6)• diffraction grating, AO2, Unit 2
Question
Monochromatic laser light of wavelength 633 nm is shone normally onto a grating with 300 lines per mm. Calculate the angle of the first-order ( $n = 1$ ) diffracted beam. (3 marks)
Step-by-step solution
1. Step 1
  Find the slit spacing $d$ from the number of lines per metre.
  $d = \frac{1}{300 \times 10^{3}} = 3.333 \times 10^{-6}\ \text{m}$
2. Step 2
  Rearrange the grating equation $n\lambda = d\sin\theta$ for $\sin\theta$ with $n=1$ .
  $\sin\theta = \frac{n\lambda}{d} = \frac{1 \times 633 \times 10^{-9}}{3.333 \times 10^{-6}} = 0.1899$
3. Step 3
  Take the inverse sine (work in degrees) to find the angle.
  $\theta = \sin^{-1}(0.1899) = 10.9°$
Answer
$\theta = 10.9°$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) $d=3.33\times10^{-6}\ \text{m}$ ; (2) $\sin\theta=n\lambda/d=0.190$ (M); (3) $\theta=10.9°$ (A). Keep the calculator in degree mode — answering $0.190\ \text{rad}$ or forgetting the inverse sine are the usual errors.
5Wavelength of a photon from its energy (3 marks)
Building confidenceDirect calculation• photon, AO2, Unit 2
Question
A photon has energy $3.06 \times 10^{-19}\ \text{J}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ . Calculate its wavelength and state the part of the visible spectrum it lies in. (3 marks)
Step-by-step solution
1. Step 1
  Start from $E = \dfrac{hc}{\lambda}$ and rearrange for the wavelength.
  $\lambda = \frac{hc}{E}$
2. Step 2
  Substitute the values with units.
  $\lambda = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{3.06 \times 10^{-19}}$
3. Step 3
  Evaluate. The numerator is $1.989 \times 10^{-25}\ \text{J m}$ .
  $\lambda = \frac{1.989 \times 10^{-25}}{3.06 \times 10^{-19}} = 6.50 \times 10^{-7}\ \text{m}$
Answer
$\lambda = 6.50 \times 10^{-7}\ \text{m} = 650\ \text{nm}$ — in the red part of the visible spectrum.
Examiner tip
Mark scheme: (1) rearrange $\lambda=hc/E$ (M); (2) substitution (M); (3) $6.50\times10^{-7}\ \text{m}$ (A) and 'red'. Recognising that 650 nm is red (400 nm violet → 700 nm red) earns the final mark; an answer in metres without identifying the colour loses it if the question asks.
6Converting a photon energy to electronvolts (4 marks)
Building confidenceMulti-step problem• electronvolt, photon, AO2, Unit 2
Question
Blue light has a wavelength of 450 nm. Take $h = 6.63 \times 10^{-34}\ \text{J s}$ , $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ and $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ . Calculate the energy of one photon (a) in joules and (b) in electronvolts. (4 marks)
Step-by-step solution
1. Step 1
  Use $E = \dfrac{hc}{\lambda}$ with $\lambda = 450 \times 10^{-9}\ \text{m}$ .
  $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{450 \times 10^{-9}}$
2. Step 2
  Evaluate the energy in joules.
  $E = \frac{1.989 \times 10^{-25}}{4.50 \times 10^{-7}} = 4.42 \times 10^{-19}\ \text{J}$
3. Step 3
  To convert joules to electronvolts, divide by $1.60 \times 10^{-19}\ \text{J eV}^{-1}$ .
  $E_{\text{eV}} = \frac{4.42 \times 10^{-19}}{1.60 \times 10^{-19}}$
4. Step 4
  Evaluate the energy in electronvolts.
  $E_{\text{eV}} = 2.76\ \text{eV}$
Answer
(a) $E = 4.42 \times 10^{-19}\ \text{J}$ . (b) $E = 2.76\ \text{eV}$ .
Examiner tip
Mark scheme: (1) $E=hc/\lambda$ (M); (2) $4.42\times10^{-19}\ \text{J}$ (A); (3) divide by $1.60\times10^{-19}$ (M); (4) $2.76\ \text{eV}$ (A). To go from J to eV you DIVIDE by $1.60\times10^{-19}$ ; multiplying (the reverse conversion) is a frequent error.
7Finding laser wavelength from a higher order (5 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Physics Unit 2 (Core Practical 6)• Core Practical 6, diffraction grating, AO3, Unit 2
Question
In Core Practical 6 a student shines a laser normally onto a grating of 500 lines per mm. The second-order ( $n = 2$ ) maximum is measured at an angle of 38.0° from the straight-through direction. Determine the wavelength of the laser light, and state one safety precaution for the experiment. (5 marks)
Step-by-step solution
1. Step 1
  Find the slit spacing from 500 lines per mm.
  $d = \frac{1}{500 \times 10^{3}} = 2.00 \times 10^{-6}\ \text{m}$
2. Step 2
  Rearrange $n\lambda = d\sin\theta$ for $\lambda$ , with $n=2$ and $\theta=38.0°$ .
  $\lambda = \frac{d\sin\theta}{n} = \frac{(2.00 \times 10^{-6})\sin 38.0°}{2}$
3. Step 3
  Evaluate $\sin 38.0° = 0.6157$ .
  $\lambda = \frac{(2.00 \times 10^{-6})(0.6157)}{2} = \frac{1.2314 \times 10^{-6}}{2}$
4. Step 4
  Complete the division.
  $\lambda = 6.16 \times 10^{-7}\ \text{m} = 616\ \text{nm}$
5. Step 5
  State a safety precaution. Lasers can damage the retina, so never look directly into the beam or its reflections.
Answer
$\lambda = 6.16 \times 10^{-7}\ \text{m}$ ( $616\ \text{nm}$ ). Safety: do not look directly into the laser beam (or its specular reflections).
Examiner tip
Mark scheme: (1) $d=2.00\times10^{-6}\ \text{m}$ ; (2) rearrange $\lambda=d\sin\theta/n$ (M); (3)/(4) $6.16\times10^{-7}\ \text{m}$ (A); (5) a valid laser-safety precaution (B). Using a higher order $n$ spreads the maxima further apart, so the angle is measured more precisely — a stretch point examiners reward.
8de Broglie wavelength after acceleration (6 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Physics Unit 2• de Broglie, energy, AO2, Unit 2
Question
An electron, initially at rest, is accelerated through a potential difference of 200 V. Take $e = 1.60 \times 10^{-19}\ \text{C}$ , $m_e = 9.11 \times 10^{-31}\ \text{kg}$ and $h = 6.63 \times 10^{-34}\ \text{J s}$ . Calculate (a) the kinetic energy gained (in joules), (b) the speed of the electron, and (c) its de Broglie wavelength. (6 marks)
Step-by-step solution
1. Step 1
  The work done on the electron equals its gain in kinetic energy: $E_k = eV$ .
  $E_k = eV = (1.60 \times 10^{-19})(200) = 3.20 \times 10^{-17}\ \text{J}$
2. Step 2
  Use $E_k = \tfrac{1}{2}m_e v^2$ and rearrange for $v$ .
  $v = \sqrt{\frac{2E_k}{m_e}} = \sqrt{\frac{2 \times 3.20 \times 10^{-17}}{9.11 \times 10^{-31}}}$
3. Step 3
  Evaluate the speed.
  $v = \sqrt{7.025 \times 10^{13}} = 8.38 \times 10^{6}\ \text{m s}^{-1}$
4. Step 4
  Find the momentum $p = m_e v$ .
  $p = (9.11 \times 10^{-31})(8.38 \times 10^{6}) = 7.636 \times 10^{-24}\ \text{kg m s}^{-1}$
5. Step 5
  Apply the de Broglie equation $\lambda = h/p$ .
  $\lambda = \frac{6.63 \times 10^{-34}}{7.636 \times 10^{-24}} = 8.68 \times 10^{-11}\ \text{m}$
Answer
(a) $E_k = 3.20 \times 10^{-17}\ \text{J}$ . (b) $v = 8.38 \times 10^{6}\ \text{m s}^{-1}$ . (c) $\lambda = 8.68 \times 10^{-11}\ \text{m}$ .
Examiner tip
Mark scheme: (1) $E_k=eV$ (M+A); (2)/(3) $v=8.38\times10^{6}\ \text{m s}^{-1}$ (M+A); (4)/(5) $\lambda=8.68\times10^{-11}\ \text{m}$ (M+A). Carry full precision through $v$ before computing $\lambda$ ; rounding $v$ early shifts the final wavelength and risks the accuracy mark.
9Maximum number of orders from a grating (6 marks)
StretchMulti-step problem• diffraction grating, AO3, Unit 2
Question
Monochromatic light of wavelength 600 nm is shone normally onto a grating with 400 lines per mm. (a) Show that the slit spacing is $2.5 \times 10^{-6}\ \text{m}$ . (b) Determine the highest order of maximum that can be observed. (6 marks)
Step-by-step solution
1. Step 1
  (a) Convert 400 lines per mm to lines per metre and take the reciprocal.
  $d = \frac{1}{400 \times 10^{3}} = 2.5 \times 10^{-6}\ \text{m}$
2. Step 2
  (b) The largest possible angle is $\theta = 90°$ , where $\sin\theta = 1$ . Set $\sin\theta = 1$ in the grating equation to find the maximum order.
  $n\lambda = d\sin\theta \;\Rightarrow\; n_{\max} = \frac{d}{\lambda}$
3. Step 3
  Substitute the values.
  $n_{\max} = \frac{2.5 \times 10^{-6}}{600 \times 10^{-9}} = 4.17$
4. Step 4
  The order $n$ must be a whole number, and the beam cannot diffract beyond 90°, so round down to the nearest integer.
  $n = 4$
Answer
(a) $d = 2.5 \times 10^{-6}\ \text{m}$ as shown. (b) The highest order observable is the 4th order ( $n = 4$ ).
Examiner tip
Mark scheme: (a) 2 marks: conversion to m⁻¹ and $d=2.5\times10^{-6}\ \text{m}$ . (b) 4 marks: set $\sin\theta=1$ (M); $n_{\max}=d/\lambda$ (M); $=4.17$ (A); round DOWN to $n=4$ (A). Rounding 4.17 up to 5 is the classic error — the 5th order would need $\sin\theta>1$ , which is impossible.

Model Answers — Waves, Electrons and Photons

High-scoring sample answers for waves, electrons and photons on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). State what is meant by a photon, and write down the equation for the energy of a photon.
Model answer
A photon is a quantum (discrete packet) of electromagnetic energy — light and other electromagnetic radiation is delivered in indivisible bundles rather than continuously.

The energy of a photon is given by:

$E = hf = \frac{hc}{\lambda}$

where $h$ is the Planck constant, $f$ the frequency, $c$ the speed of light and $\lambda$ the wavelength.
Why this scores
Why this scores 2/2. (1) photon = a quantum/discrete packet of electromagnetic energy; (2) correct equation $E=hf$ (or $hc/\lambda$ ). Saying 'a particle of light' alone is weak — examiners want the idea of a discrete amount of energy.
Question 2
2 marks
Q (2 marks). Define the electronvolt and state its value in joules.
Model answer
One electronvolt (eV) is the energy transferred to (or by) an electron when it moves through a potential difference of one volt.

Numerically, $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ (equal to the magnitude of the electron's charge multiplied by 1 V).
Why this scores
Why this scores 2/2. (1) energy transferred when an electron passes through a p.d. of 1 V; (2) $1.60\times10^{-19}\ \text{J}$ . The eV is a unit of energy, not charge or voltage — a common confusion that costs the first mark.
Question 3
3 marks
Q (3 marks). A diffraction grating has 600 lines per mm. (a) Calculate the slit spacing $d$ . (b) Light of wavelength 500 nm strikes the grating normally. Show that no fifth-order maximum can be seen.
Model answer
(a) Converting 600 lines per mm to lines per metre gives $N = 600 \times 10^{3} = 6.00 \times 10^{5}\ \text{m}^{-1}$ , so:

$d = \frac{1}{N} = \frac{1}{6.00 \times 10^{5}} = 1.67 \times 10^{-6}\ \text{m}$

(b) For the fifth order ( $n = 5$ ) the grating equation needs:

$\sin\theta = \frac{n\lambda}{d} = \frac{5 \times 500 \times 10^{-9}}{1.67 \times 10^{-6}} = 1.50$

Since $\sin\theta$ cannot exceed 1, this is impossible: the fifth-order beam would have to diffract beyond 90°, so it does not exist.
Why this scores
Why this scores 3/3. (a) 1 mark: $d=1.67\times10^{-6}\ \text{m}$ (with mm → m conversion). (b) 2 marks: substitute $n=5$ to get $\sin\theta=1.5$ , then state that $\sin\theta\le1$ so the order cannot exist. Just writing 'impossible' without the value of $\sin\theta$ would not earn full credit.
Question 4
4 marks
Q (4 marks). A laser pointer emits red light of wavelength 650 nm at a power of 1.0 mW. Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ . Calculate the number of photons emitted each second.
Model answer
First find the energy of one photon using $E = \dfrac{hc}{\lambda}$ :

$E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{650 \times 10^{-9}} = 3.06 \times 10^{-19}\ \text{J}$

Power is energy per second, so the number of photons per second is the total energy emitted each second divided by the energy of one photon:

$\text{rate} = \frac{P}{E} = \frac{1.0 \times 10^{-3}}{3.06 \times 10^{-19}} = 3.27 \times 10^{15}\ \text{photons s}^{-1}$

So roughly $3.3 \times 10^{15}$ photons leave the laser every second.
Why this scores
Why this scores 4/4. (1) $E=hc/\lambda$ set up; (2) $E=3.06\times10^{-19}\ \text{J}$ ; (3) rate $=P/E$ ; (4) $3.27\times10^{15}\ \text{s}^{-1}$ . Watch the mW → W conversion ( $1.0\ \text{mW}=1.0\times10^{-3}\ \text{W}$ ); omitting it gives an answer 1000× too large.
Question 5
4 marks
Q (4 marks). When a beam of electrons is fired at a thin polycrystalline graphite target, a pattern of concentric bright rings is seen on a fluorescent screen. (a) Explain what this observation tells us about the nature of electrons. (b) State and explain what happens to the rings if the speed of the electrons is increased.
Model answer
(a) The rings are a diffraction pattern. Diffraction is a property of waves, so the fact that electrons produce a diffraction pattern is evidence that electrons (normally thought of as particles) also show wave-like behaviour — they have a de Broglie wavelength. The rings arise because the regularly spaced atomic planes in the graphite act like a diffraction grating, and the electron wavelength is comparable to the atomic spacing.

(b) Increasing the electron speed increases the momentum $p = mv$ . From the de Broglie equation $\lambda = h/p$ , a larger momentum means a shorter wavelength. A shorter wavelength diffracts through a smaller angle, so the rings move closer together (the pattern shrinks towards the centre).
Why this scores
Why this scores 4/4. (a) 2 marks: diffraction is a wave property → electrons have wave nature / a de Broglie wavelength comparable to atomic spacing. (b) 2 marks: faster → larger $p$ → smaller $\lambda$ (via $\lambda=h/p$ ) → rings closer together. The link 'shorter wavelength → smaller diffraction angle' is the discriminating step.
Question 6
5 marks
Q (5 marks). (a) Calculate the de Broglie wavelength of an electron ( $m = 9.11 \times 10^{-31}\ \text{kg}$ ) moving at $2.0 \times 10^{6}\ \text{m s}^{-1}$ . (b) Calculate the de Broglie wavelength of a 0.058 kg tennis ball moving at $50\ \text{m s}^{-1}$ . (c) Use your answers to explain why we observe diffraction for electrons but not for tennis balls. Take $h = 6.63 \times 10^{-34}\ \text{J s}$ .
Model answer
(a) For the electron, $p = mv = (9.11 \times 10^{-31})(2.0 \times 10^{6}) = 1.822 \times 10^{-24}\ \text{kg m s}^{-1}$ , so:

$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.822 \times 10^{-24}} = 3.64 \times 10^{-10}\ \text{m}$

(b) For the tennis ball, $p = (0.058)(50) = 2.9\ \text{kg m s}^{-1}$ , so:

$\lambda = \frac{6.63 \times 10^{-34}}{2.9} = 2.29 \times 10^{-34}\ \text{m}$

(c) Diffraction is only noticeable when the wavelength is comparable to the size of the gap or obstacle. The electron's wavelength ( $\sim 10^{-10}\ \text{m}$ ) is about the same as the atomic spacing in a crystal, so electrons diffract appreciably. The tennis ball's wavelength ( $\sim 10^{-34}\ \text{m}$ ) is vastly smaller than any aperture it could pass through, so its diffraction is completely undetectable — which is why everyday objects appear purely particle-like.
Why this scores
Why this scores 5/5. (a) 2 marks: $p=mv$ and $\lambda=3.64\times10^{-10}\ \text{m}$ . (b) 1 mark: $\lambda=2.3\times10^{-34}\ \text{m}$ . (c) 2 marks: diffraction needs $\lambda \approx$ gap size; electron $\lambda \approx$ atomic spacing (diffracts) but tennis-ball $\lambda$ is far too small (no diffraction). The comparison to atomic spacing is the key marking point.
Question 7
4 marks
Q (4 marks). A proton moves with a speed of $3.0 \times 10^{5}\ \text{m s}^{-1}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and the proton mass $m_p = 1.67 \times 10^{-27}\ \text{kg}$ . (a) Calculate the de Broglie wavelength of the proton. (b) An electron ( $m_e = 9.11 \times 10^{-31}\ \text{kg}$ ) moves at the same speed. State, without further calculation, whether its de Broglie wavelength is longer or shorter than the proton's, and explain why.
Model answer
(a) First find the momentum of the proton:

$p = m_p v = (1.67 \times 10^{-27})(3.0 \times 10^{5}) = 5.01 \times 10^{-22}\ \text{kg m s}^{-1}$

Then apply the de Broglie equation:

$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.01 \times 10^{-22}} = 1.32 \times 10^{-12}\ \text{m}$

(b) The electron's de Broglie wavelength is longer. At the same speed the electron has a much smaller momentum ( $p = mv$ ) because its mass is far smaller (about 1800 times lighter). Since $\lambda = h/p$ , a smaller momentum gives a larger wavelength, so the electron's wavelength is longer than the proton's.
Why this scores
Why this scores 4/4. (a) 3 marks: $p=m_p v=5.01\times10^{-22}\ \text{kg m s}^{-1}$ ; $\lambda=h/p$ ; $1.32\times10^{-12}\ \text{m}$ . (b) 1 mark: electron has smaller $p$ (smaller mass at same $v$ ) → larger $\lambda$ via $\lambda=h/p$ . The reasoning must link mass → momentum → wavelength, not just assert 'lighter means longer'.
Question 8
5 marks
Q (5 marks). In an experiment to estimate the Planck constant, several LEDs of different colours are each connected in turn and the minimum 'switch-on' voltage $V$ at which the LED just lights is recorded. For each LED the photon energy is taken as $eV \approx hf$ . A graph of $V$ against frequency $f$ is plotted and found to be a straight line through the origin. (a) Show how the gradient of this graph is related to the Planck constant. (b) The gradient is measured as $4.10 \times 10^{-15}\ \text{V s}$ . Estimate the Planck constant ( $e = 1.60 \times 10^{-19}\ \text{C}$ ).
Model answer
(a) When the LED just lights, the energy delivered to each electron, $eV$ , equals the energy of the photon it emits, $hf$ :

$eV = hf$

Rearranging to make $V$ the subject (the plotted $y$ -quantity):

$V = \left(\frac{h}{e}\right) f$

Comparing with the straight-line form $y = mx$ (with $y = V$ , $x = f$ ), the gradient $m = \dfrac{h}{e}$ . So a graph of $V$ against $f$ has gradient $h/e$ , and $h = e \times \text{gradient}$ .

(b) Substituting the measured gradient:

$h = e \times \text{gradient} = (1.60 \times 10^{-19})(4.10 \times 10^{-15}) = 6.56 \times 10^{-34}\ \text{J s}$

This is close to the accepted value $6.63 \times 10^{-34}\ \text{J s}$ , so the experiment gives a reasonable estimate of the Planck constant.
Why this scores
Why this scores 5/5. (a) 3 marks: $eV=hf$ ; rearrange to $V=(h/e)f$ ; compare to $y=mx$ → gradient $=h/e$ . (b) 2 marks: $h=e\times$ gradient and $6.56\times10^{-34}\ \text{J s}$ . Comparing the estimate with the accepted value (and commenting it is close) shows the evaluative skill examiners reward.
Question 9
6 marks
Q (6 marks). Monochromatic light of wavelength 589 nm is shone normally onto a diffraction grating. The first-order maximum is observed at 20.5°. (a) Calculate the number of lines per millimetre on the grating. (b) Calculate the angle of the second-order maximum. (c) Explain why a diffraction grating gives sharper, brighter maxima than a double slit.
Model answer
(a) Use $n\lambda = d\sin\theta$ with $n = 1$ , $\lambda = 589 \times 10^{-9}\ \text{m}$ and $\theta = 20.5°$ :

$d = \frac{n\lambda}{\sin\theta} = \frac{1 \times 589 \times 10^{-9}}{\sin 20.5°} = \frac{589 \times 10^{-9}}{0.3502} = 1.682 \times 10^{-6}\ \text{m}$

The number of lines per metre is $N = 1/d = 5.945 \times 10^{5}\ \text{m}^{-1}$ , which is $5.945 \times 10^{2} = 595\ \text{lines mm}^{-1}$ (about 600 lines/mm).

(b) For the second order, $n = 2$ :

$\sin\theta = \frac{n\lambda}{d} = \frac{2 \times 589 \times 10^{-9}}{1.682 \times 10^{-6}} = 0.7004$

$\theta = \sin^{-1}(0.7004) = 44.5°$

(c) A diffraction grating has many thousands of slits rather than two. The light from all these slits interferes, and a bright maximum forms only where every slit is exactly in phase. Away from these precise angles the contributions from the many slits cancel almost completely, so each maximum is much narrower (sharper); and because the light from far more slits adds together, each maximum is much brighter.
Why this scores
Why this scores 6/6. (a) 2 marks: $d=1.68\times10^{-6}\ \text{m}$ and ~595 lines/mm. (b) 2 marks: $\sin\theta=0.700$ , $\theta=44.5°$ . (c) 2 marks: many slits → maxima only where all are in phase → sharper; more slits add → brighter. Keep the un-rounded $d$ for part (b); using a rounded 600 lines/mm shifts the angle slightly and can lose the accuracy mark.
Question 10
6 marks
Q (6 marks). 'Light and electrons both show wave–particle duality.' Discuss this statement, referring to specific evidence for both the wave nature and the particle nature in each case.
Model answer
Wave–particle duality is the idea that both light and matter can behave as waves in some experiments and as particles in others; neither model alone describes everything.

Light. The wave nature of light is shown by diffraction and interference — for example, the sharp maxima produced when light passes through a diffraction grating ( $n\lambda = d\sin\theta$ ). These effects can only be explained by waves superposing. The particle nature of light is shown by the photoelectric effect and by the photon model: light delivers energy in discrete packets of energy $E = hf$ , and a single photon can eject a single electron. A purely wave picture cannot explain why there is a threshold frequency, but the photon (particle) picture can.

Electrons. Electrons were originally regarded as particles: they have a definite mass and charge, are deflected by electric and magnetic fields, and carry momentum. However, when a beam of electrons passes through a thin crystal it produces a diffraction pattern of rings — a wave effect. This is explained by the de Broglie wavelength $\lambda = h/p$ : each electron has an associated wavelength comparable to the atomic spacing, so it diffracts.

Conclusion. Both light and electrons therefore exhibit both behaviours. Which behaviour we observe depends on the experiment: diffraction/interference reveals the wave aspect, while the photoelectric effect (light) and deflection by fields or carrying momentum (electrons) reveal the particle aspect.
Why this scores
Why this scores 6/6. This is a levels-style/extended response. Full marks need, for light: a wave evidence (diffraction/interference) AND a particle evidence (photoelectric effect / photons $E=hf$ ); and for electrons: a particle property (mass/charge/momentum) AND a wave evidence (electron diffraction via $\lambda=h/p$ ). Continuous, correctly linked prose with the right terminology — not a bare list — secures the top level.

Key Formulae — Waves, Electrons and Photons

The formulae you need to memorise for waves, electrons and photons on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Diffraction grating equation
$n\lambda = d\sin\theta$
$n$
order of the maximum (integer 0, 1, 2, …)
$\lambda$
wavelength of the light / m
$d$
slit (grating) spacing / m
$\theta$
angle of the order from the straight-through direction
When to use
Finding the angle of a diffracted order, the wavelength of light, the grating spacing, or the maximum order ( $\sin\theta \le 1$ ). In the Edexcel formulae booklet (Appendix 7).
Example
300 lines/mm, $n=1$ , $\lambda=633\ \text{nm}$ : $\sin\theta=633\times10^{-9}/3.33\times10^{-6}=0.190$ , $\theta=10.9°$ .
Grating slit spacing
$d = \frac{1}{N}$
$d$
spacing between adjacent slits / m
$N$
number of lines (slits) per metre / m⁻¹
When to use
Converting a grating's 'lines per mm' rating into the slit spacing $d$ in metres before using $n\lambda=d\sin\theta$ . Must be recalled.
Example
500 lines/mm $=5.00\times10^{5}\ \text{m}^{-1}$ : $d=1/(5.00\times10^{5})=2.00\times10^{-6}\ \text{m}$ .
Photon energy
$E = hf = \frac{hc}{\lambda}$
$E$
photon energy / J
$h$
Planck constant $=6.63\times10^{-34}\ \text{J s}$
$f$
frequency / Hz
$c$
speed of light $=3.00\times10^{8}\ \text{m s}^{-1}$
$\lambda$
wavelength / m
When to use
Finding the energy carried by one photon from its frequency or wavelength (use $c=f\lambda$ to switch between the two forms). In the formulae booklet.
Example
$\lambda=500\ \text{nm}$ : $E=hc/\lambda=3.98\times10^{-19}\ \text{J}=2.49\ \text{eV}$ .
Electronvolt conversion
$E_{\text{(J)}} = E_{\text{(eV)}} \times 1.60 \times 10^{-19}$
$1\ \text{eV}$
energy gained by an electron accelerated through 1 V $=1.60\times10^{-19}\ \text{J}$
When to use
Converting between joules and electronvolts. To go J → eV, DIVIDE by $1.60\times10^{-19}$ ; to go eV → J, MULTIPLY by $1.60\times10^{-19}$ . Must be recalled.
Example
$4.42\times10^{-19}\ \text{J} \div 1.60\times10^{-19} = 2.76\ \text{eV}$ .
de Broglie wavelength
$\lambda = \frac{h}{p} = \frac{h}{mv}$
$\lambda$
de Broglie wavelength / m
$h$
Planck constant / J s
$p$
momentum $=mv$ / kg m s⁻¹
$m$
mass of the particle / kg
$v$
speed of the particle / m s⁻¹
When to use
Finding the wavelength associated with a moving particle (e.g. an electron). The wavelength is significant only when comparable to the size of the obstacle/gap (atomic spacing for electron diffraction). In the formulae booklet.
Example
Electron at $7.0\times10^{6}\ \text{m s}^{-1}$ : $\lambda=h/(m_e v)=1.04\times10^{-10}\ \text{m}$ .
Kinetic energy from an accelerating p.d.
$eV = \tfrac{1}{2}mv^2$
$e$
charge of the electron $=1.60\times10^{-19}\ \text{C}$
$V$
accelerating potential difference / V
$m$
mass of the particle / kg
$v$
final speed / m s⁻¹
When to use
Finding the speed (and hence the de Broglie wavelength) of a charged particle accelerated from rest through a known potential difference. Must be recalled (from work–energy).
Example
Electron through 200 V: $E_k=eV=3.20\times10^{-17}\ \text{J}$ , so $v=8.38\times10^{6}\ \text{m s}^{-1}$ .

Key Definitions and Keywords — Waves, Electrons and Photons

Definitions to memorise and the exact keywords mark schemes credit for waves, electrons and photons answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Diffraction
Examiner keyword
The spreading of a wave as it passes through a gap or around the edge of an obstacle. It is most pronounced when the gap is comparable in size to the wavelength.
Huygens' construction
A model in which every point on a wavefront acts as a source of secondary spherical wavelets; the new wavefront is the surface that touches all these wavelets, which explains how waves diffract and bend round edges.
Diffraction grating
Examiner keyword
An optical component with a very large number of equally spaced parallel slits (or lines). Light passing through produces sharp, bright maxima at angles given by $n\lambda = d\sin\theta$ .
Grating equation
Examiner keyword
$n\lambda = d\sin\theta$ , relating the wavelength $\lambda$ , the slit spacing $d$ , the order $n$ (an integer) and the angle $\theta$ of a maximum from the straight-through direction.
Order (of a maximum)
The integer $n$ in $n\lambda = d\sin\theta$ . $n = 0$ is the central (zero-order) maximum, $n = 1$ the first order, and so on. The maximum possible order satisfies $\sin\theta \le 1$ .
Slit (grating) spacing, d
Examiner keyword
The distance between the centres of adjacent slits on a grating, equal to the reciprocal of the number of lines per metre: $d = 1/N$ .
Monochromatic light
Light of a single wavelength (and hence a single frequency/colour), such as the light from a laser. Required for a clean grating pattern with sharp maxima.
Photon
Examiner keyword
A quantum (discrete packet) of electromagnetic energy. Each photon carries energy $E = hf = hc/\lambda$ .
Planck constant, h
Examiner keyword
The fundamental constant relating a photon's energy to its frequency, $E = hf$ . Its value is $h = 6.63 \times 10^{-34}\ \text{J s}$ .
Electronvolt (eV)
Examiner keyword
A unit of energy equal to the energy transferred when an electron moves through a potential difference of 1 V: $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ .
Wave–particle duality
Examiner keyword
The principle that both electromagnetic radiation and matter exhibit both wave-like and particle-like properties, depending on the experiment performed.
de Broglie wavelength
Examiner keyword
The wavelength associated with any moving particle, $\lambda = h/p = h/mv$ , where $p$ is the particle's momentum. It is significant only when comparable to the size of the gap or obstacle.
Electron diffraction
Examiner keyword
The production of a diffraction pattern (e.g. concentric rings) when a beam of electrons passes through a thin crystal. It is direct evidence that electrons have a wave nature.
Momentum (in de Broglie)
The product of mass and velocity, $p = mv$ , of a particle. A larger momentum gives a shorter de Broglie wavelength ( $\lambda = h/p$ ).
Quantum (of energy)
The smallest discrete amount by which a quantity (such as the energy carried by electromagnetic radiation) can change. Light energy comes in whole numbers of photon-sized quanta, not continuously.

Common Mistakes and Misconceptions — Waves, Electrons and Photons

The traps other students keep falling into on waves, electrons and photons questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Calculating the slit spacing $d$ directly from 'lines per mm' without converting to lines per metre
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students take the reciprocal of the labelled figure (e.g. $1/300$ ) and forget that the rating is per millimetre, not per metre.
How to avoid it
Convert to lines per metre first (×1000), then take the reciprocal: for 300 lines/mm, $N=3.00\times10^{5}\ \text{m}^{-1}$ and $d=1/N=3.33\times10^{-6}\ \text{m}$ . Check that $d$ is of order $10^{-6}\ \text{m}$ .
✕Leaving the calculator in radian mode (or forgetting the inverse sine) when finding the grating angle
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
After computing $\sin\theta = n\lambda/d$ , students either quote that number as the angle or evaluate $\sin^{-1}$ in radians.
How to avoid it
$n\lambda/d$ gives $\sin\theta$ , not $\theta$ — take the inverse sine, and make sure the calculator is in degree mode. A first-order angle for visible light is typically only a few tens of degrees, so a 'huge' answer signals a mode error.
✕Mixing up the photon equation $E = hf$ with the de Broglie equation $\lambda = h/p$
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Both contain the Planck constant $h$ , so students grab whichever they remember and substitute the wrong quantities.
How to avoid it
Keep them separate: $E = hf = hc/\lambda$ gives a photon's ENERGY (photons of light); $\lambda = h/p$ gives the WAVELENGTH of a moving particle (electrons). Ask 'am I finding an energy or a particle's wavelength?' before choosing.
✕Multiplying instead of dividing (or vice versa) when converting between joules and electronvolts
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students memorise ' $1.60\times10^{-19}$ ' but not the direction of the conversion.
How to avoid it
Remember the eV is the smaller unit, so there are many eV in a joule: to go J → eV divide by $1.60\times10^{-19}$ ; to go eV → J multiply by $1.60\times10^{-19}$ . Sanity-check: a few-eV visible photon should be a few $\times10^{-19}\ \text{J}$ .
✕Rounding the maximum-order calculation $n_{\max}=d/\lambda$ UP to the next integer
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students round 4.17 to 4.2 → 'about 4 or 5' or treat it like an ordinary number and round up.
How to avoid it
The order must satisfy $\sin\theta\le1$ , so you must round $d/\lambda$ DOWN to the nearest whole number. A non-integer like 4.17 means the highest visible order is 4 (the 5th would need $\sin\theta>1$ ).
✕Substituting a velocity straight into $\lambda = h/p$ without first calculating the momentum $p = mv$
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students see $\lambda = h/p$ and put the speed where the momentum belongs, omitting the mass.
How to avoid it
Always compute the momentum $p = mv$ first (include the mass), then divide $h$ by it. When the particle is accelerated through a p.d., find $v$ from $eV=\tfrac12mv^2$ before forming $p$ .

Waves, Electrons and Photons — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Waves, Electrons and Photons

Short Notes - Waves, Electrons and Photons

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Energy of a single photon (2 marks)

2Slit spacing from lines per millimetre (2 marks)

3de Broglie wavelength of a moving electron (3 marks)

4Diffraction angle of a grating order (3 marks)

5Wavelength of a photon from its energy (3 marks)

6Converting a photon energy to electronvolts (4 marks)

7Finding laser wavelength from a higher order (5 marks)

8de Broglie wavelength after acceleration (6 marks)

9Maximum number of orders from a grating (6 marks)

Diffraction grating equation

Grating slit spacing

Photon energy

Electronvolt conversion

de Broglie wavelength

Kinetic energy from an accelerating p.d.

Diffraction

Huygens' construction

Diffraction grating

Grating equation

Order (of a maximum)

Slit (grating) spacing, d

Monochromatic light

Photon

Planck constant, h

Electronvolt (eV)

Wave–particle duality

de Broglie wavelength

Electron diffraction

Momentum (in de Broglie)

Quantum (of energy)

✕Calculating the slit spacing ddd directly from 'lines per mm' without converting to lines per metre

✕Leaving the calculator in radian mode (or forgetting the inverse sine) when finding the grating angle

✕Mixing up the photon equation E=hfE = hfE=hf with the de Broglie equation λ=h/p\lambda = h/pλ=h/p

✕Multiplying instead of dividing (or vice versa) when converting between joules and electronvolts

✕Rounding the maximum-order calculation nmax⁡=d/λn_{\max}=d/\lambdanmax​=d/λ UP to the next integer

✕Substituting a velocity straight into λ=h/p\lambda = h/pλ=h/p without first calculating the momentum p=mvp = mvp=mv

Waves, Electrons and Photons — frequently asked questions

✕Calculating the slit spacing $d$ directly from 'lines per mm' without converting to lines per metre

✕Mixing up the photon equation $E = hf$ with the de Broglie equation $\lambda = h/p$

✕Rounding the maximum-order calculation $n_{\max}=d/\lambda$ UP to the next integer

✕Substituting a velocity straight into $\lambda = h/p$ without first calculating the momentum $p = mv$