22.1Quantum Theory and the Atom
A Particle Model of Waves — American System Physics Revision Notes & Topical Questions
10 min read 7 parts Lesson 3 of 5 1 quiz
Lesson summary
Describe the spectrum of a hot (blackbody) object and how it changes with temperature.Photon energy = Planck constant × frequencyE = hf (E in J, h = 6.63 × 10−34 J s, f in Hz).Using c = fλ: photon energy E = hc/λ (λ in m, c = 3.00 × 108 m/s).Useful shortcutE (in eV) ≈ 1240 / λ (i…
- Einstein's photoelectric equation
- Converting
- Photon momentum = Planck constant ÷ wavelength
- Planck's quantisation
- Einstein extended the idea
- Light therefore shows wave-particle duality
Section 1 of 7A Particle Model of Waves
Essential knowledge
Every object above absolute zero emits electromagnetic radiation. A heated metal first glows dull red, then orange, yellow and finally bluish-white: as temperature rises, total emitted power increases steeply and the peak of the spectrum moves to shorter (higher-frequency) wavelengths.
Classical wave theory predicted that a hot body should emit enormous (unlimited) amounts of ultraviolet radiation, which does not happen.
- Max Planck solved this in 1900 by assuming the energy of vibrating atoms is quantised: it can only take values nhf,
- and energy is emitted or absorbed only in jumps of hf.
Einstein extended the idea
Einstein extended the idea light itself travels as photons. A photon has no rest mass, travels at c in a vacuum and carries energy hf. Higher frequency (shorter wavelength) means more energetic photons, which is why ultraviolet and X-rays can damage cells but radio waves cannot.
In the photoelectric effect, each photon interacts with a single electron and gives up all of its energy.
- Part of this energy (the work function W) is used to free the electron from the metal surface;
- the rest becomes kinetic energy.
- Electrons from deeper inside lose extra energy, so hf − W is the MAXIMUM kinetic energy.
If hf < W, no electrons are emitted at all, however intense the light, because electrons cannot collect energy from several photons. This is why red light may not eject electrons from zinc while weak ultraviolet does.
Increasing intensity (at a fixed frequency above threshold) increases the number of photons per second, so the photocurrent increases, but KEmax and the stopping potential stay the same.
A graph of KEmax against f is a straight line of gradient h for every metal; different metals give parallel lines with different x-intercepts (f0) because they have different work functions.
The electron-volt is a convenient energy unit at atomic scale: 1 eV = 1.60 × 10−19 J. Visible photons have energies of roughly 1.6 eV (red) to 3.1 eV (violet).
Arthur Compton fired X-rays at graphite and found the scattered X-rays had longer wavelengths.
- Treating the X-ray as a particle with momentum h/λ colliding elastically with an electron explains this: the electron recoils, gaining energy and momentum,
- so the photon loses energy and its wavelength increases.
Light therefore shows wave-particle duality
Light therefore shows wave-particle duality interference and diffraction need the wave model, while the photoelectric and Compton effects need the photon model. Neither model alone describes all behaviour.
Section 2 of 7Process & sequence
1
Shine monochromatic light of known frequency onto a clean metal plate (the emitter) inside an evacuated photocell.
2
Connect the collector electrode to a variable power supply so that it is made negative relative to the emitter (reverse potential difference), with a sensitive ammeter in series.
3
Slowly increase the reverse potential difference until the photocurrent just falls to zero; record this stopping potential V0.
4
Calculate KEmax = qV0 (in eV this is numerically equal to V0).
5
Repeat with several different frequencies (use filters or LEDs) and plot KEmax (y-axis) against frequency f (x-axis).
6
The gradient of the straight line equals the Planck constant h; the x-intercept is the threshold frequency f0; the y-intercept is −W.
How a photon releases a photoelectronwrite these steps in this order; each step is a marking point.
1
A photon of energy hf reaches the metal surface.
2
The photon is absorbed by one electron and gives it all its energy.
3
Energy equal to the work function W is used to escape the surface.
4
If hf > W, the electron leaves with kinetic energy up to hf − W.
5
A reverse voltage V0 can just stop the fastest electrons: qV0 = KEmax.
Section 3 of 7Compare
| Observation | Wave model predicts | Photon model explains |
|---|---|---|
| No emission below f0, however bright | Bright enough light of any frequency should eventually eject electrons | One photon gives all its energy to one electron; if hf < W no electron escapes |
| Emission is instant | A time delay while energy builds up | A single photon delivers its energy at once |
| Brighter light (above f0) gives more electrons, not faster ones | Brighter light should give faster electrons | Brighter = more photons per second, so more electrons per second; each photon still has energy hf |
| KEmax increases linearly with frequency | No link with frequency | KEmax = hf − W |
Intensity controls the NUMBER of photoelectrons; frequency controls their maximum ENERGY.
| Radiation | Typical wavelength | Photon energy |
|---|---|---|
| Radio | 1 m | about 1 × 10−6 eV |
| Infrared | 10 µm | about 0.1 eV |
| Visible (green) | 500 nm | about 2.5 eV |
| Ultraviolet | 100 nm | about 12 eV |
| X-ray | 0.1 nm | about 12 000 eV |
Section 4 of 7Worked examples
What the symbols mean:
Calculate the energy of a photon of wavelength 450 nm in (a) joules and (b) electron-volts.
- (a) E = hc/λ = (6.63 × 10−34 × 3.00 × 108) / (450 × 10−9) = 4.42 × 10−19 J.What the symbols mean: E = photon energy (J) · h = 6.63 × 10−34 J s · c = 3.00 × 108 m s−1 · λ = wavelength (m)
- (b) E = 4.42 × 10−19 / 1.60 × 10−19 = 2.76 eV (check: 1240/450 = 2.76 eV).
0.83 eV
Sodium has a work function of 2.28 eV. Light of frequency 7.50 × 1014 Hz falls on it. Find (a) the threshold frequency, (b) the maximum kinetic energy of the photoelectrons in eV and (c) the stopping potential.
- (a) W = 2.28 × 1.60 × 10−19 = 3.65 × 10−19 J; f0 = W/h = 3.65 × 10−19 / 6.63 × 10−34 = 5.50 × 1014 Hz.
- (b) hf = 6.63 × 10−34 × 7.50 × 1014 = 4.97 × 10−19 J = 3.11 eV;
- KEmax = 3.11 − 2.28 = 0.83 eV (1.3 × 10−19 J).
- (c) V0 = KEmax/q = 0.83 V.
- (b) hf = 6.63 × 10−34 × 7.50 × 1014 = 4.97 × 10−19 J = 3.11 eV;
A lamp emits 5.0 W of light of wavelength 600 nm. How many photons does it emit each second?
- Energy per photon E = hc/λ = (6.63 × 10−34 × 3.00 × 108)/(600 × 10−9) = 3.32 × 10−19 J.
- Number per second = power / energy per photon = 5.0 / 3.32 × 10−19 = 1.5 × 1019 photons per second.
Section 5 of 7Key terms
Section 6 of 7Command terms
Command terms
Identify / Describe (FRQ)
A short, specific statement; describing requires relevant characteristics, not a single word.
Explain / Justify (FRQ)
Give the cause-and-effect reasoning or evidence; a restated fact does not score.
Calculate
Show setup, substitution and the answer with units.
Predict
State the expected outcome; add a reason only if asked to justify.
MCQ technique
Read every option; eliminate distractors that are true statements but do not answer the question; watch for NOT/EXCEPT.
Section 7 of 7Exam technique
Keywords the mark scheme looks for in this lesson
Keywords the mark scheme looks for in this lesson
Blackbody · Thermal (blackbody) radiation · Quantum · Photon · Planck constant (h) · Photoelectric effect · Threshold frequency (f0) · Work function (W or φ) · Electron-volt (eV) · Maximum kinetic energy of photoelectrons (KEmax) · Stopping potential (V0) · Compton effect · Wave-particle duality
Learn each definition in the Key words tab exactly; a missing keyword in a definition usually loses the mark.
How to get full marks (examiner guidance)
1
Answer exactly what is asked; extra correct material does not earn credit, and contradictory statements can cancel credit (AP FRQ).
2
Use precise vocabulary, never “it/stuff/things”.
3
Show units and working in calculations.
4
In MCQs, predict the answer before reading the options, then eliminate.
Words to avoid
1
brighter light makes the electrons fasterbrighter light gives more photons per second, so more electrons are emitted per second, but their maximum kinetic energy is unchanged.
2
the work function is the energy of the electronthe work function is the minimum energy needed to remove an electron from the surface of the metal.
3
KE = hf − W for all electronshf − W is the MAXIMUM kinetic energy of the photoelectrons.
4
electrons store up energy from the light until they escapeeach electron absorbs the energy of a single photon, all at once.
5
Don't mix units: never subtract a work function in eV from a photon energy in J — convert both to the same unit first.
Instant feedbackCheck yourself
Q1. What is the energy of a photon of frequency 5.0 × 1014 Hz?
1
A. 3.3 × 10−19 J
2
B. 1.3 × 10−48 J
3
C. 7.5 × 1047 J
4
D. 3.3 × 10−20 J
Show answer
A. E = hf = 6.63 × 10−34 × 5.0 × 1014 = 3.3 × 10−19 J.
What the symbols mean: E = energy of one photon (J) · h = Planck constant (6.63 × 10−34 J s) · f = frequency (Hz)
Q2. Light of frequency above the threshold frequency is made twice as intense. What happens?
1
A. KEmax doubles
2
B. The threshold frequency halves
3
C. The number of electrons emitted per second doubles
4
D. No electrons are emitted
Show answer
C. Doubling intensity doubles the photons per second, so twice as many electrons per second; each photon energy (and KEmax) is unchanged.
Q3. A metal has work function 3.0 eV. Photons of 4.5 eV fall on it. What is the stopping potential?
- A. 7.5 V
- B. 3.0 V
- C. 4.5 V
- D. 1.5 V
Show answer
D. KEmax = 4.5 − 3.0 = 1.5 eV, so V0 = 1.5 V.
Q4. On a graph of KEmax against frequency, what is the gradient?
1
A. The work function
2
B. The Planck constant
3
C. The threshold frequency
4
D. The speed of light
Show answer
B. KEmax = hf − W is of the form y = mx + c with m = h.
Q5. As a blackbody gets hotter, the peak of its emission spectrum
1
A. moves to longer wavelength
2
B. stays at the same wavelength
3
C. moves to shorter wavelength
4
D. disappears
Show answer
C. Hotter bodies emit more at all wavelengths and their peak shifts to shorter wavelength (red → white/blue).
Q6. In the Compton effect, the scattered X-ray photon has
1
A. a longer wavelength than before
2
B. a shorter wavelength than before
3
C. a higher frequency than before
4
D. more energy than before
Show answer
A. The photon gives some energy and momentum to the electron, so its energy falls and its wavelength increases.
Marked practiceQuizzes for this lesson
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