DP Physics HL E.2 Quantum Physics

DP Physics HL E.2 Quantum Physics - Full notes

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DP Physics HL | E.2 Quantum Physics

DP PHYSICS · THEME E: NUCLEAR AND QUANTUM PHYSICS

E.2

Quantum Physics

Full notes for HL · IB-style

Contents

1	The photoelectric effect
2	Photons as particles
3	Matter waves
4	Wave–particle duality
5	Applications
6	Practical work and data analysis
7	Exam technique and further practice
8	Common misconceptions and links
9	Exam-style question with markscheme
10	Check your understanding
11	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe the photoelectric effect and why the wave model fails
use Einstein's equation hf = Φ + Ek,max and stopping potential
interpret Ek,max–frequency graphs to find h, f₀ and Φ
use photon momentum and describe Compton scattering
use the de Broglie wavelength and describe electron diffraction
discuss wave–particle duality and probability

E.2 is an Additional Higher Level (AHL) subtopic. It is studied by HL students only.

Guiding questions

How can light behave as both a wave and a particle? How did quantum ideas resolve the failures of classical physics? What evidence is there for the wave nature of matter?

How this topic is examined

E.2 is HL only, tested in Papers 1 and 2. Expect photoelectric effect explanations and calculations, graph interpretation, de Broglie wavelength and Compton scattering.

Keep energies in consistent units (J or eV) and use h = 6.63 × 10⁻³⁴ J s.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 The photoelectric effect

Light ejecting electrons.

Think about it

Why can dim ultraviolet light eject electrons from zinc when very bright red light cannot?

At the end of the 19th century, light was firmly understood as a wave. The photoelectric effect showed that this picture was incomplete: light also behaves as a stream of particles.

1.1 Observations

When light of high enough frequency shines on a clean metal surface in a vacuum, electrons (photoelectrons) are emitted. Four observations are key:

•No electrons are emitted below a threshold frequency f₀, however intense the light.
•Above f₀, emission is instantaneous, even for very dim light.
•The maximum kinetic energy of the electrons depends on frequency, not intensity.
•The number of electrons emitted per second is proportional to the intensity.

Figure 1. Apparatus for investigating the photoelectric effect.

Quick check

1. What happens to the photocurrent if the intensity doubles (above threshold)?

Answers: 1. it doubles

1.2 Why the wave model fails

The wave model predicts that energy arrives continuously and depends on intensity. So any frequency should eventually eject electrons if bright enough, dim light should involve a time delay while energy builds up, and brighter light should give faster electrons. None of these predictions is observed.

Figure 2. Intensity changes the photocurrent but not the maximum kinetic energy.

WORKED EXAMPLE 1 Wave model (exam level)

Explain why the existence of a threshold frequency cannot be explained by the wave model.

Solution

Wave model: energy delivered continuously, depends on intensityR1

So low-frequency, high-intensity light should eventually eject electrons, which is not observedR1

1.3 How long would the wave model take?

Classically, an electron could only absorb the energy falling on an area about the size of an atom. For dim light the predicted delay before emission is very long, yet emission is observed within nanoseconds.

WORKED EXAMPLE 2 Classical delay (exam level)

Light of intensity 1.0 × 10⁻³ W m⁻² falls on sodium (Φ = 2.3 eV). Estimate how long an electron absorbing over an area of 1.0 × 10⁻²⁰ m² would take to gain enough energy.

Solution

Power absorbed = 1.0 × 10⁻³ × 1.0 × 10⁻²⁰ = 1.0 × 10⁻²³ WM1

t = 3.7 × 10⁻¹⁹/1.0 × 10⁻²³ = 3.7 × 10⁴ s (about 10 hours), yet emission is instantA1

1.4 Einstein's photon explanation

Einstein (1905) proposed that light consists of photons of energy hf, and each photon interacts with one electron. The electron needs at least the work function Φ to escape from the surface; any excess becomes kinetic energy: Ek,max = hf − Φ. The threshold frequency is f₀ = Φ/h.

Figure 3. Energy of a photoelectron: the photon energy minus the work function.

Key point

One photon, one electron: intensity changes the number of photons, frequency changes the energy of each.

WORKED EXAMPLE 3 Kinetic energy (standard)

Light of wavelength 250 nm falls on a metal with work function 3.2 eV. Find the maximum kinetic energy of the electrons.

Solution

hf = hc/λ = 4.97 eVM1

Ek,max = 4.97 − 3.2 = 1.8 eVA1

WORKED EXAMPLE 4 Threshold (standard)

Find the threshold wavelength for this metal.

Solution

λ₀ = hc/ΦM1

= 388 nmA1

1.5 Work functions

The work function is the minimum energy needed to remove an electron from the surface of a metal. It differs between metals.

Metal	Work function / eV	Threshold wavelength / nm
caesium	2.1	590
sodium	2.3	540
calcium	2.9	430
zinc	4.3	290
copper	4.7	260
platinum	5.6	220

Quick check

1. Which metals in the table emit electrons with green (530 nm) light?

Answers: 1. caesium and sodium

1.6 Stopping potential

A reverse potential difference on the collector repels photoelectrons. The stopping potential Vs is just large enough to stop the fastest: eVs = Ek,max. Increasing the intensity increases the saturation current but leaves Vs unchanged; increasing the frequency increases Vs.

Figure 4. Photocurrent against collector potential: intensity changes the current; frequency changes the stopping potential.

WORKED EXAMPLE 5 Stopping potential (standard)

For light giving Ek,max = 1.8 eV, find the stopping potential.

Solution

Vs = Ek,max/eM1

= 1.8 VA1

1.7 Graphs of maximum kinetic energy against frequency

A graph of Ek,max (or eVs) against frequency is a straight line with gradient h, x-intercept f₀ and y-intercept −Φ. Every metal gives a parallel line, because h is a universal constant.

Figure 5. Maximum kinetic energy against frequency for sodium and zinc.

WORKED EXAMPLE 6 From the graph (exam level)

A line crosses the frequency axis at 5.6 × 10¹⁴ Hz. Find the work function.

Solution

Φ = hf₀ = 6.63 × 10⁻³⁴ × 5.6 × 10¹⁴M1

= 3.7 × 10⁻¹⁹ J = 2.3 eVA1

1.8 Different metals on one graph

Lines for different metals are parallel (same gradient h) but cross the frequency axis at different threshold frequencies f₀ = Φ/h.

Figure 6. Maximum kinetic energy against frequency for three metals.

WORKED EXAMPLE 7 Threshold (standard)

Find the threshold frequency for calcium (Φ = 2.9 eV).

Solution

f₀ = Φ/h = 2.9/4.14 × 10⁻¹⁵M1

= 7.0 × 10¹⁴ HzA1

1.9 Millikan's data

Millikan measured stopping potentials for sodium at several frequencies. His straight-line graph gave h within 0.5% of the modern value, confirming Einstein's equation.

f / 10¹⁴ Hz	5.49	6.91	7.41	8.22	9.58	11.8
Vs / V	0.00	0.58	0.79	1.13	1.69	2.60

Figure 7. Stopping potential against frequency (Millikan-style data for sodium).

WORKED EXAMPLE 8 Planck's constant (exam level)

The gradient of the graph is 4.14 × 10⁻¹⁵ V s. Find h and the work function.

Solution

h = e × gradient = 6.6 × 10⁻³⁴ J sM1 A1

Φ = hf₀ = 4.14 × 10⁻¹⁵ × 5.49 × 10¹⁴ = 2.3 eVA1

2 Photons as particles

Momentum and scattering.

2.1 Photon momentum

Photons have no rest mass but carry momentum p = E/c = h/λ. This explains radiation pressure and the recoil seen in Compton scattering.

WORKED EXAMPLE 9 Momentum (standard)

Find the momentum of a 600 nm photon.

Solution

p = 6.63 × 10⁻³⁴/6.0 × 10⁻⁷M1

= 1.1 × 10⁻²⁷ kg m s⁻¹A1

WORKED EXAMPLE 10 Radiation force (exam level)

A 1.0 W laser beam is completely absorbed by a surface. Find the force on it.

Solution

F = rate of momentum transfer = P/cM1

= 3.3 × 10⁻⁹ NA1

In the real world

Solar sails use the tiny momentum of sunlight photons to propel spacecraft without fuel; Japan's IKAROS probe demonstrated this in 2010.

2.2 Compton scattering

When X-ray photons scatter from free electrons, the scattered photons have longer wavelengths. Compton explained this by treating the photon as a particle with momentum colliding with an electron: Δλ = (h/mec)(1 − cos θ). Energy and momentum are both conserved.

Figure 8. Compton scattering of a photon by an electron.

Figure 9. Compton wavelength shift against scattering angle.

WORKED EXAMPLE 11 Shift (exam level)

Calculate the wavelength shift for X-rays scattered at 90°.

Solution

Δλ = 6.63 × 10⁻³⁴/(9.11 × 10⁻³¹ × 3.0 × 10⁸) × 1M1

= 2.43 × 10⁻¹² mA1

WORKED EXAMPLE 12 Energy lost (exam level)

A 50.0 pm X-ray photon is scattered at 180°. Find its new wavelength and the energy transferred to the electron.

Solution

λ′ = 50.0 + 4.86 = 54.9 pmM1

ΔE = hc(1/λ − 1/λ′) = 3.5 × 10⁻¹⁶ J (2.2 keV)A1

2.3 Photon number and intensity

The number of photons per second N = P/hf. A source's intensity depends on photon number; its colour (frequency) determines the energy per photon.

WORKED EXAMPLE 13 Laser photons (standard)

How many photons per second does a 5.0 mW, 650 nm laser emit?

Solution

E = hc/λ = 3.06 × 10⁻¹⁹ JM1

N = 5.0 × 10⁻³/3.06 × 10⁻¹⁹ = 1.6 × 10¹⁶ s⁻¹A1

2.4 Measuring Planck's constant with LEDs

An LED begins to emit light at a switch-on voltage V where eV ≈ hc/λ. Measuring V for LEDs of several colours and plotting V against 1/λ gives a gradient of hc/e.

Figure 10. Switch-on voltage against 1/λ for four LEDs.

WORKED EXAMPLE 14 Gradient (exam level)

The gradient is 1.25 × 10⁻⁶ V m. Find h.

Solution

h = gradient × e/c = 1.25 × 10⁻⁶ × 1.6 × 10⁻¹⁹/3.0 × 10⁸M1

= 6.7 × 10⁻³⁴ J sA1

Try this

Find the switch-on voltage of red, green and blue LEDs using a potentiometer and a dark tube to view each LED. Plot V against 1/λ and calculate h.

3 Matter waves

Particles behaving as waves.

Think about it

If light waves can behave as particles, can particles such as electrons behave as waves?

3.1 The de Broglie hypothesis

De Broglie (1924) proposed that all particles have a wavelength λ = h/p = h/mv. For everyday objects it is immeasurably small; for electrons accelerated through modest voltages it is similar to the spacing of atoms in crystals.

WORKED EXAMPLE 15 Electron (exam level)

Find the de Broglie wavelength of an electron accelerated through 150 V.

Solution

p = √(2meV) = 6.6 × 10⁻²⁴ kg m s⁻¹M1

λ = h/p = 1.0 × 10⁻¹⁰ mA1

WORKED EXAMPLE 16 Ball (standard)

Find the de Broglie wavelength of a 0.15 kg ball at 30 m s⁻¹.

Solution

λ = 6.63 × 10⁻³⁴/4.5M1

= 1.5 × 10⁻³⁴ m (unobservable)A1

3.2 Wavelength and accelerating voltage

For an electron accelerated from rest through V, p = √(2meV), so λ = h/√(2meV): the wavelength falls as the voltage rises (λ ∝ 1/√V).

Figure 11. de Broglie wavelength of electrons against accelerating voltage.

Quick check

1. By what factor does λ change if the accelerating voltage is quadrupled?

Answers: 1. halves

3.3 Electron diffraction

Electrons fired through a thin film of graphite produce bright rings on a fluorescent screen: a diffraction pattern, because their wavelength is similar to the spacing of carbon atoms. Increasing the voltage gives shorter wavelengths and smaller rings.

Figure 12. An electron diffraction tube.

Figure 13. Diffraction rings at a lower and a higher accelerating voltage.

WORKED EXAMPLE 17 Ring size (exam level)

Explain why the diffraction rings shrink when the accelerating voltage is increased.

Solution

Electrons gain more momentumR1

Shorter de Broglie wavelength, so less diffraction (smaller angles)R1

In the real world

Davisson and Germer confirmed electron diffraction in 1927, earning de Broglie the Nobel Prize just five years after his hypothesis.

3.4 Wavelengths of different particles

Diffraction needs wavelengths comparable to the spacing of the obstacles. Slow (thermal) neutrons and low-energy electrons have wavelengths similar to atomic spacings, which is why both are used to study crystal structures. Even molecules as large as C₆₀ have been diffracted.

Figure 14. de Broglie wavelengths of various particles (log scale).

WORKED EXAMPLE 18 Thermal neutron (exam level)

A neutron has kinetic energy 0.025 eV. Find its de Broglie wavelength.

Solution

p = √(2mE) = √(2 × 1.67 × 10⁻²⁷ × 4.0 × 10⁻²¹) = 3.66 × 10⁻²⁴M1

λ = h/p = 1.8 × 10⁻¹⁰ mA1

3.5 Electron microscopes

Fast electrons have wavelengths far shorter than visible light, so electron microscopes can resolve details down to fractions of a nanometre, revealing viruses, proteins and even individual atoms.

WORKED EXAMPLE 19 Microscope wavelength (exam level)

Find the de Broglie wavelength of electrons accelerated through 50 kV (ignore relativistic effects) and compare with visible light.

Solution

λ = h/√(2meV) = 5.5 × 10⁻¹² mM1 A1

About 100 000 times shorter than green lightA1

3.6 Comparing photons and electrons

Property	Photon	Electron
rest mass	zero	9.11 × 10⁻³¹ kg
charge	zero	−e
energy	E = hf = hc/λ	Ek = p²/2m (non-relativistic)
momentum	p = h/λ = E/c	p = mv = h/λ
speed	c	less than c

4 Wave–particle duality

Two descriptions of one reality.

4.1 Duality

Light shows wave behaviour (diffraction, interference) and particle behaviour (photoelectric effect, Compton scattering). Matter does too: electrons, neutrons and even large molecules diffract. Which behaviour we observe depends on the experiment.

Phenomenon	Wave or particle?	Evidence for
diffraction, interference	wave	light and electrons
photoelectric effect	particle	light
Compton scattering	particle	light (X-rays)
electron diffraction	wave	matter

Quick check

1. Give one experiment showing light behaving as (a) a wave; (b) a particle.

Answers: 1. (a) double slit; (b) photoelectric effect

4.2 Probability and the wave function

In quantum physics a particle is described by a wave function; the square of its amplitude gives the probability of finding the particle at a location. Firing electrons one at a time through a double slit, each lands at a single point, but an interference pattern gradually builds up.

Figure 15. Single electrons arriving one at a time build up an interference pattern.

Quick check

1. What builds up when single electrons pass through a double slit one at a time?

Answers: 1. an interference pattern

4.3 Interpreting duality

Wave and particle are models, each successful in its own domain. Light travels and interferes like a wave but is emitted and absorbed in discrete photons. Quantum theory combines both descriptions: the wave describes where a particle is likely to be detected; the detection itself is particle-like.

WORKED EXAMPLE 20 Model choice (standard)

State which model best explains (a) the colours in a soap film; (b) the release of electrons by UV light.

Solution

(a) wave model (interference)A1

(b) photon modelA1

4.4 Quantisation from waves

An electron confined in an atom can only have wavelengths that fit, like standing waves on a string. This restricts its momentum and energy to particular values, explaining the discrete energy levels of E.1.

Quick check

1. How does confining an electron lead to discrete energies?

Answers: 1. only certain standing-wave wavelengths fit, giving certain momenta and energies

4.5 Where it began: black-body radiation

Classical physics predicted that a hot object should emit ever-increasing intensity at short wavelengths (the 'ultraviolet catastrophe'). Planck (1900) matched the observed spectrum only by assuming energy is emitted in quanta of size hf: the birth of quantum physics.

Figure 16. Observed black-body spectrum at 3000 K compared with the classical prediction.

Quick check

1. What assumption did Planck make to explain black-body radiation?

Answers: 1. energy is emitted in discrete quanta E = hf

4.6 Historical development

Planck (1900) introduced energy quanta to explain black-body radiation; Einstein (1905) explained the photoelectric effect with photons; Millikan (1916) confirmed Einstein's equation experimentally despite initially doubting it; Compton (1923) showed photon momentum; de Broglie (1924) proposed matter waves, confirmed by Davisson and Germer (1927).

Quick check

1. Why was Millikan's confirmation significant?

Answers: 1. he set out to disprove it, but his precise data supported it

5 Applications

Quantum physics at work.

5.1 Photocells and image sensors

Photocells, light meters, photomultipliers and the image sensors in cameras use light to release electrons. Solar cells use the related photovoltaic effect, in which photons free electrons inside a semiconductor.

Quick check

1. Why do some photocells respond only to ultraviolet?

Answers: 1. their work function is too large for visible photons

5.2 X-ray production

In an X-ray tube, electrons accelerated through a large pd strike a metal target and decelerate, emitting X-ray photons. The maximum photon energy equals the electron's kinetic energy, so the minimum wavelength is λmin = hc/eV: the photoelectric effect in reverse.

Figure 17. Minimum X-ray wavelength against tube voltage.

WORKED EXAMPLE 21 Minimum wavelength (exam level)

Find the minimum X-ray wavelength from a tube operating at 30 kV.

Solution

λ = hc/eV = 6.63 × 10⁻³⁴ × 3.0 × 10⁸/(1.6 × 10⁻¹⁹ × 3.0 × 10⁴)M1

= 4.1 × 10⁻¹¹ mA1

5.3 Solar cells and photon energy

A silicon solar cell needs photons of at least 1.12 eV (wavelength below about 1110 nm) to free an electron. Longer-wavelength sunlight passes unused, and any excess photon energy above 1.12 eV is lost as heat, which limits the efficiency of a single-junction cell to about 33%.

Figure 18. Solar spectrum (black-body approximation) with the silicon cut-off wavelength.

WORKED EXAMPLE 22 Cut-off (standard)

Show that photons longer than about 1110 nm cannot be used by silicon (1.12 eV).

Solution

λ = 1240/1.12M1

= 1107 nmA1

5.4 Photon rate and wavelength

For a fixed power, longer-wavelength light contains more (lower-energy) photons per second: N = Pλ/hc.

Figure 19. Number of photons per second in a 1 mW beam against wavelength.

Quick check

1. Does a 1 mW red laser or a 1 mW blue laser emit more photons per second?

Answers: 1. the red laser

5.5 Changing variables: summary

Change (above threshold)	Photocurrent	Maximum kinetic energy
increase intensity	increases	no change
increase frequency	no change (same photon rate)	increases
metal with larger work function	may stop if f < f₀	decreases
reverse pd increased	decreases	no change

Quick check

1. What happens to the stopping potential if frequency increases?

Answers: 1. it increases

6 Practical work and data analysis

Quantum experiments in the laboratory.

6.1 Photocell investigation

A vacuum photocell illuminated through coloured filters (or with LEDs of known wavelength) is connected to a variable reverse pd and a sensitive ammeter. For each colour, the reverse pd is increased until the current just falls to zero: this is the stopping potential. Plotting Vs against f gives h and Φ.

WORKED EXAMPLE 23 Uncertainty (exam level)

Explain why it is difficult to judge the stopping potential precisely.

Solution

The current falls gradually towards zero (electrons have a range of energies)R1

Small leakage currents make the exact zero hard to identifyR1

Try this

Use a photocell simulation to vary intensity and frequency independently. Record the stopping potential and current each time and check them against Einstein's equation.

6.2 Electron diffraction measurements

With a diffraction tube, measure the diameter of a ring on the screen for several accelerating voltages. Since λ ∝ 1/√V and the ring radius is roughly proportional to λ, a graph of ring radius against 1/√V should be a straight line through the origin.

Figure 20. Ring radius against 1/√V for an electron diffraction tube.

Quick check

1. What does a straight line through the origin confirm?

Answers: 1. the ring size (wavelength) is proportional to 1/√V, as de Broglie predicts

7 Exam technique and further practice

Applying quantum ideas.

7.1 Method checklist

•Explain the photoelectric effect using 'one photon gives all its energy to one electron'.
•Keep units consistent: J with h in J s, or eV with h = 4.14 × 10⁻¹⁵ eV s.
•Read h from the gradient and f₀ from the intercept of Ek,max–f graphs.
•For electrons from rest through V, use p = √(2meV).
•State clearly which experiment shows wave and which particle behaviour.

7.2 Multi-step problem: photoelectrons

WORKED EXAMPLE 24 Stopping potential data (exam level)

Photoelectrons from a metal under 400 nm light have stopping potential 0.80 V. Find (a) the work function; (b) the threshold frequency; (c) the de Broglie wavelength of the fastest electrons.

Solution

(a) hf = 3.11 eV; Φ = 2.3 eVM1 A1

(b) f₀ = Φ/h = 5.6 × 10¹⁴ HzA1

7.3 Multi-step problem: Compton and photon energy

WORKED EXAMPLE 25 X-ray scattering (exam level)

X-rays of wavelength 0.071 nm scatter from carbon at 90°. Find (a) the incident photon energy in keV; (b) the scattered wavelength; (c) the energy given to the electron.

Solution

(a) E = 1240/0.071 eV = 17.5 keVA1

(b) λ′ = 71.0 + 2.43 = 73.4 pmA1

7.4 Multi-step problem: LED and Planck's constant

WORKED EXAMPLE 26 LED analysis (exam level)

A blue LED (470 nm) switches on at 2.62 V and a red LED (630 nm) at 1.96 V. Use the two points to estimate h, assuming eV = hc/λ.

Solution

eΔV = hcΔ(1/λ): 1.6 × 10⁻¹⁹ × 0.66 = h × 3.0 × 10⁸ × (2.128 − 1.587) × 10⁶M1

h = 1.056 × 10⁻¹⁹/1.62 × 10¹⁴ = 6.5 × 10⁻³⁴ J sA1

Close to the accepted 6.63 × 10⁻³⁴ J s; switch-on judged by eye adds uncertaintyR1

7.5 Graded practice

WORKED EXAMPLE 27 Threshold wavelength (standard)

Calculate the threshold wavelength for a metal of work function 2.6 eV.

Solution

λ₀ = 1240/2.6 nmM1

= 477 nmA1

WORKED EXAMPLE 28 Neutron (standard)

Find the de Broglie wavelength of a neutron moving at 2.0 km s⁻¹ (m = 1.67 × 10⁻²⁷ kg).

Solution

λ = h/mvM1

= 2.0 × 10⁻¹⁰ mA1

WORKED EXAMPLE 29 Same wavelength (exam level)

An electron and a proton have the same de Broglie wavelength. Compare their momenta and kinetic energies.

Solution

Same momentum (λ = h/p)R1

Ek = p²/2m: electron has about 1840 times more kinetic energyR1

7.6 Graded practice: photoelectric effect

WORKED EXAMPLE 30 Work function from data (standard)

Light of 300 nm gives photoelectrons with maximum kinetic energy 1.2 eV. Find the work function.

Solution

hf = 1240/300 = 4.13 eVM1

Φ = 2.9 eVA1

WORKED EXAMPLE 31 Speed of photoelectrons (exam level)

Find the maximum speed of photoelectrons with Ek,max = 1.2 eV.

Solution

v = √(2E/m) = √(2 × 1.92 × 10⁻¹⁹/9.11 × 10⁻³¹)M1

= 6.5 × 10⁵ m s⁻¹A1

WORKED EXAMPLE 32 Photocurrent (exam level)

A photocell receives 2.0 × 10¹⁵ photons per second; 5% eject electrons. Find the photocurrent.

Solution

Electrons per second = 1.0 × 10¹⁴M1

I = 16 μAA1

7.7 Graded practice: matter waves

WORKED EXAMPLE 33 Same wavelength as light (exam level)

Find the accelerating voltage that gives electrons a de Broglie wavelength of 500 nm.

Solution

p = h/λ = 1.33 × 10⁻²⁷; E = p²/2m = 9.7 × 10⁻²⁶ JM1

V = E/e = 6.0 × 10⁻⁶ VA1

WORKED EXAMPLE 34 Doubling speed (standard)

An electron's speed doubles (non-relativistic). What happens to its wavelength and kinetic energy?

Solution

λ ∝ 1/v: halvesA1

Ek ∝ v²: quadruplesA1

7.8 Graded practice: graphs

WORKED EXAMPLE 35 Reading intercepts (exam level)

An Ek,max–f graph has y-intercept −3.0 eV and x-intercept 7.2 × 10¹⁴ Hz. Find Φ and h.

Solution

Φ = 3.0 eVA1

h = Φ/f₀ = 4.8 × 10⁻¹⁹/7.2 × 10¹⁴ = 6.7 × 10⁻³⁴ J sA1

WORKED EXAMPLE 36 Stopping potential change (standard)

Light of 400 nm gives Vs = 1.0 V. Find Vs for 300 nm light on the same metal.

Solution

Extra photon energy = 1240/300 − 1240/400 = 1.03 eVM1

Vs = 2.0 VA1

WORKED EXAMPLE 37 Intensity doubled (standard)

The intensity of the 400 nm light is doubled. State the new stopping potential and how the saturation current changes.

Solution

Vs unchanged at 1.0 VA1

Saturation current doublesA1

7.9 Graded practice: explanations

WORKED EXAMPLE 38 Instantaneous emission (exam level)

Explain why photoemission begins instantly even in very dim light.

Solution

Energy arrives in photons; each photon gives all its energy to one electron at onceR1

If hf > Φ the electron escapes immediately; dim light simply means fewer photonsR1

WORKED EXAMPLE 39 Saturation current (exam level)

Explain why the photocurrent reaches a maximum (saturation) as the collector potential is made more positive.

Solution

All emitted photoelectrons are already reaching the collectorR1

The rate of emission is fixed by the photon rate, so the current cannot rise furtherR1

WORKED EXAMPLE 40 Rings and spacing (exam level)

Explain how electron diffraction rings could be used to measure the spacing of atoms in graphite.

Solution

Measure ring angle; know λ from the accelerating voltageR1

Use the grating-like relation nλ = d sin θ to find dR1

7.10 Graded practice: mixed

WORKED EXAMPLE 41 Photon and electron (exam level)

A photon and an electron both have wavelength 0.20 nm. Find the energy of each in eV.

Solution

Photon: E = 1240/0.20 = 6200 eVA1

Electron: p = 3.3 × 10⁻²⁴; E = p²/2m = 6.0 × 10⁻¹⁸ J = 38 eVM1 A1

WORKED EXAMPLE 42 UV and sunburn (standard)

Explain why UV light can damage skin cells while far more intense visible light does not.

Solution

UV photons have more energy each (higher f)R1

Enough to break chemical bonds in DNA; visible photons individually are notR1

7.11 Self-test 2

More quick questions on quantum physics.

Quick check

1. Why is emission instantaneous even in dim light?

2. What does the x-intercept of an Ek,max–f graph give?

3. Find the photon momentum for 0.10 nm X-rays.

4. What is the maximum Compton shift?

5. Name two particles that have been diffracted.

6. Why do everyday objects not show wave behaviour?

Answers: 1. each photon transfers its energy to one electron immediately 2. the threshold frequency f₀ 3. 6.6 × 10⁻²⁴ kg m s⁻¹ 4. 2h/mec = 4.86 pm 5. electrons and neutrons (also atoms, molecules) 6. their wavelengths are far too small

7.12 Key terms

Term	Meaning
photon	quantum of electromagnetic energy, E = hf
work function Φ	minimum energy to remove an electron from a surface
threshold frequency	minimum frequency for photoemission, f₀ = Φ/h
stopping potential	reverse pd that just stops the fastest photoelectrons
de Broglie wavelength	λ = h/p for any particle
Compton scattering	wavelength increase of photons scattered by electrons
wave–particle duality	light and matter show both wave and particle behaviour

7.13 Self-test

Try these without notes, then check the answers.

Quick check

1. What does increasing intensity change in the photoelectric effect?

2. Find Ek,max for 5.0 eV photons on a 3.0 eV metal.

3. State the gradient of an Ek,max–f graph.

4. What is the momentum of a 1.0 nm photon?

5. Why do electron microscopes resolve smaller details than light microscopes?

6. Which experiment confirmed matter waves?

Answers: 1. the number of photoelectrons (current) 2. 2.0 eV 3. Planck's constant h 4. 6.6 × 10⁻²⁶ kg m s⁻¹ 5. electrons have much shorter wavelengths 6. electron diffraction (Davisson–Germer)

7.14 Common exam errors

•Saying brighter light gives faster photoelectrons.
•Treating the work function as a property of the light rather than the metal.
•Using eV and J inconsistently.
•Forgetting to square-root when finding p from kinetic energy.
•Claiming electron diffraction shows electrons are only waves.

8 Common misconceptions and links

Students often think…	Actually…
brighter light gives faster photoelectrons	intensity changes the number, not the maximum energy
the work function depends on the light	it is a property of the metal
photons have mass	photons have zero rest mass but carry momentum
only light shows wave–particle duality	matter does too
electron diffraction shows electrons are only waves	duality: behaviour depends on the experiment

How E.2 connects to the rest of the course

Topic	Connection
E.1 Structure of the atom	photons and energy levels
C.3 Wave phenomena	diffraction and interference
B.1 Thermal energy	black-body radiation and Planck
D.2 Electric fields	accelerating electrons through pd

9 Exam-style question with markscheme

Try this under exam conditions (about 20 minutes) before reading the markscheme.

Ultraviolet radiation of wavelength 220 nm falls on a clean zinc surface (work function 4.3 eV).

(a) Calculate the photon energy in eV. [2]

(b) Calculate the maximum kinetic energy of the photoelectrons. [1]

(d) Explain why no electrons are emitted with red light of any intensity. [2]

(e) Calculate the de Broglie wavelength of the fastest photoelectrons. [3]

(f) Outline how electron diffraction provides evidence for matter waves. [2]

Markscheme

Part	Answer	Marks
(a)	E = hc/λ = 9.04 × 10⁻¹⁹ J	M1
	5.65 eV	A1
(b)	1.35 eV	A1
(c)	1.35 V	A1
(d)	red photon energy less than work function	R1
	one photon per electron; intensity only increases photon number	R1
(e)	Ek = 2.16 × 10⁻¹⁹ J; p = √(2mEk)	M1
	p = 6.27 × 10⁻²⁵ kg m s⁻¹	A1
	λ = 1.06 × 10⁻⁹ m	A1
(f)	electrons passing through graphite produce rings	R1
	diffraction is a wave property; ring size matches de Broglie wavelength	R1

Examiner's comments. In (d), answers mentioning only 'not enough energy' without reference to individual photons did not gain full credit.

10 Check your understanding

Section A: multiple choice

1. The maximum kinetic energy of photoelectrons depends on

A. frequency of light B. intensity of light C. area of the surface D. time of exposure

2. The gradient of an Ek,max–f graph is

A. Φ B. e C. h D. f₀

3. The de Broglie wavelength of a particle is

A. p/h B. h/p C. hp D. h/E

4. Photon momentum equals

A. hλ B. hf C. λ/h D. h/λ

5. Compton scattering shows that photons

A. have rest mass B. carry momentum C. are waves only D. slow down in vacuum

6. Increasing an electron's accelerating voltage

A. decreases its de Broglie wavelength B. increases it C. has no effect D. makes it zero

7. The work function is

A. the photon energy B. the kinetic energy of electrons C. the stopping potential D. the minimum energy to remove an electron from the surface

8. Electron diffraction rings provide evidence for

A. the particle nature of light B. the nuclear model C. the wave nature of electrons D. Compton scattering

Section B: structured

9. Calculate the threshold wavelength for a metal of work function 2.6 eV. [2]

10. Explain the photoelectric effect using the photon model. [4]

11. Calculate the de Broglie wavelength of a neutron moving at 2.0 km s⁻¹ (m = 1.67 × 10⁻²⁷ kg). [2]

12. Describe one piece of evidence for the particle nature of light and one for its wave nature. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. λ₀ = hc/Φ = 478 nm.

10. One photon gives all its energy to one electron; needs ≥ Φ; excess as kinetic energy; threshold frequency; instantaneous emission.

11. λ = h/mv = 2.0 × 10⁻¹⁰ m.

12. Particle: photoelectric effect/Compton; wave: diffraction/interference.

11 Summary and topic map

Idea	Remember
Photoelectric	threshold f; instant; Ek depends on f
Einstein	hf = Φ + Ek,max; eVs = Ek,max
Graph	gradient h; intercept f₀
Photon momentum	p = h/λ; Compton
de Broglie	λ = h/p
Duality	waves and particles

Before the exam, make sure you…

•explain using one photon per electron
•convert eV and J consistently
•read h and f₀ from graphs
•use p = √(2mEk) for electrons

Quantum Physics on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 21. Topic map for Quantum Physics.

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