Full chapter
Quantum Physics
All 7 topics and the revision summary on one page.
01
Evidence for photons
Make a guess. You are not marked.
Ultraviolet light makes a clean zinc plate emit electrons. The intensity of the light is doubled at the same frequency. What happens to the maximum kinetic energy of the emitted electrons?
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It stays the same, because each photon still carries the same energy.
Each electron gains its energy from one photon, of energy hf. The frequency is unchanged, so hf and the maximum kinetic energy hf - work function are unchanged. Twice the intensity gives about twice as many photons, so about twice as many electrons.
Light produces interference and diffraction patterns, but it also transfers energy in individual quanta. A successful explanation must account for both kinds of evidence.
In the photoelectric effect, electrons can leave a material when it absorbs suitable electromagnetic radiation. For a given surface there is a threshold frequency: below it, increasing ordinary illumination does not cause emission. Above it, emission is possible even with weak illumination.
A photon is one quantum of electromagnetic radiation. Its energy depends on frequency:
Here h is the Planck constant, 6.63 × 10-34 J s. At fixed frequency, higher intensity means more incident photons per second for a fixed illuminated area. It does not increase each photon's energy.
Compare frequency and intensity separately
Consider a surface with supplied threshold f0 = 5.00 × 1014 Hz. Keep the surface, geometry and collection conditions unchanged. The comparisons below use the ordinary single-photon model and, above threshold, conditions without saturation.
| Frequency | Illumination | Photoelectric result |
|---|---|---|
| 4.00 × 1014 Hz | Lower intensity | No emission: insufficient energy per photon. |
| 4.00 × 1014 Hz | Higher intensity | Still no emission in this model. |
| 6.00 × 1014 Hz | Lower intensity | Emission is possible. |
| 6.00 × 1014 Hz | Higher intensity | The emission rate can increase. |
Below the threshold: greater intensity is insufficient
Compare frequencies while holding the material and collection conditions fixed. The arrows show incoming photons and, where possible, outgoing electrons. Icon counts do not specify an efficiency or a measured rate.
Above the threshold: emission is possible
Compare frequencies while holding the material and collection conditions fixed. The arrows show incoming photons and, where possible, outgoing electrons. Icon counts do not specify an efficiency or a measured rate.
The threshold supports quantised energy exchange: adding more photons whose individual energies are too small does not meet the single-photon condition. This does not remove the wave evidence from interference and diffraction. Nor does it claim that one ordinary single-photon model covers arbitrarily intense fields.
The work function is an energy
The work function, written here as Φwork, is the minimum energy needed to remove an electron from the specified surface. At threshold:
= (6.63 × 10-34)(5.00 × 1014)
= 3.315 × 10-19 J
= 2.07 eV
One electronvolt is the energy gained by a particle of charge magnitude e accelerated through a potential difference of 1 V. Thus 1 eV = e × 1 V = 1.60 × 10-19 J using the supplied elementary charge. An eV is an energy unit, not a voltage. Φwork is measured in joules; it is distinct from magnetic flux Φ measured in webers.
Worked photon calculation
Energy, wavelength and momentum
An optical frequency of order 5 × 1014 Hz with h of order 7 × 10-34 J s suggests a few times 10-19 J per photon: a few eV. Expect a wavelength of a few hundred nanometres and momentum of order 10-27 kg m/s.
For f = 6.00 × 1014 Hz, use c = 3.00 × 108 m/s in vacuum:
E = hf = 3.978 × 10-19 J
E = (3.978 × 10-19)/(1.60 × 10-19)
= 2.48625 eV ≈ 2.49 eV
A photon has zero rest mass but nonzero momentum. Its momentum points along propagation, with magnitude:
= 1.326 × 10-27 kg m/s
≈ 1.33 × 10-27 N s
The non-relativistic formula p = mv for a massive particle is not a way to set photon momentum to zero. Check that energy divided by speed has momentum units.
A monochromatic beam with supplied power 1.989 µW at this frequency carries P/E = 5.00 × 1012 photons/s. That incident photon rate is not automatically a recorded electron rate: absorption, emission and collection efficiencies matter.
Try it step by step
Work it out in steps: Photon energy E = hc / lambda in J, then Photon energy in eV, then Maximum kinetic energy in eV.
Show the answer
Problem 1.
Photon energy E = hc / lambda in J: E = 6.63 x 10-34 x 3.00 x 108 / (250 x 10-9) = 7.96 x 10-19 J
Photon energy in eV: E = 7.96 x 10-19 / 1.60 x 10-19 = 4.97 eV
Maximum kinetic energy in eV: Maximum kinetic energy = 4.97 - 4.3 = 0.67 eV
Problem 2.
Photon energy E = hc / lambda in J: E = 6.63 x 10-34 x 3.00 x 108 / (200 x 10-9) = 9.95 x 10-19 J
Photon energy in eV: E = 9.95 x 10-19 / 1.60 x 10-19 = 6.22 eV
Maximum kinetic energy in eV: Maximum kinetic energy = 6.22 - 4.3 = 1.92 eV
Problem 3.
Photon energy E = hc / lambda in J: E = 6.63 x 10-34 x 3.00 x 108 / (300 x 10-9) = 6.63 x 10-19 J
Photon energy in eV: E = 6.63 x 10-19 / 1.60 x 10-19 = 4.14 eV
Maximum kinetic energy in eV: Maximum kinetic energy = 4.14 - 2.3 = 1.84 eV
Check your understanding
A surface has threshold frequency 5.00 x 1014 Hz. In the ordinary single-photon model, what happens if the intensity at 4.00 x 1014 Hz increases while frequency and other conditions stay fixed?
Show the answer
The photon rate increases, but each photon still has insufficient energy for photoelectric emission.
Intensity at fixed frequency changes incident energy per second and photon rate, not energy per photon. The frequency remains below the supplied threshold, so the ordinary single-photon condition is not met.
02
Matter waves and detections
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An electron and a proton travel at the same speed. Which has the longer de Broglie wavelength?
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The electron, because it has less momentum at that speed.
The de Broglie wavelength is h / p, and p = mv. At the same speed the electron, with about 1/1840 of the proton's mass, has much less momentum and so a much longer wavelength.
An electron can produce a localised detection while an accumulated pattern displays wave behaviour. The detection position and the state used to predict its distribution are different parts of the explanation.
Two distinct observations
In an electron-diffraction demonstration, a beam encounters a thin crystalline specimen and produces a diffraction pattern at a detector. Its ordered structure supplies a suitable spatial scale. Thin polycrystalline graphite can produce rings; a ring pattern is not universal to every crystal and orientation.
In a single-particle double-slit experiment, individual detections accumulate into an interference pattern when the alternatives remain coherent. A sufficiently weak beam still builds the pattern one detection at a time. Collisions between neighbouring electrons are not required to explain it.
Electron diffraction from polycrystalline graphite
The detector pattern is evidence of electron diffraction. This is a schematic of the arrangement and pattern type; the ring spacing is not supplied data. The enlarged face view uses its own drawing scale.
Two openings and a position-sensitive detector
A detection records a position. Repeating the coherent two-opening experiment builds a spatial distribution. Sparse particles can still produce an interference pattern; neighbouring particles do not need to collide.
A detection dot tells us where an interaction was recorded. It does not reveal a definite classical path between the source and detector. An electron is not a tiny ball following a sine curve or two half-charge balls passing separately through the slits. Obtaining which-path information changes the experimental conditions; simply looking at an already recorded screen does not cause that change.
Worked de Broglie wavelength
Choose the momentum model first
For a particle with momentum magnitude p, the de Broglie wavelength is:
An electron of mass about 10-30 kg moving at about 106 m/s has momentum of order 10-24 kg m/s and wavelength of order 1 nm. Its speed is well below c.
Use me = 9.11 × 10-31 kg, h = 6.63 × 10-34 J s and v = 1.50 × 106 m/s. Here v/c = 0.00500, so the non-relativistic p = mv approximation is suitable:
λ = (6.63 × 10-34)/(1.3665 × 10-24)
= 4.85181 × 10-10 m ≈ 0.485 nm
Doubling v doubles p and halves λ to 0.243 nm. Kinetic energy instead scales as v2 and quadruples. Initially it is (1/2)mev2 = 1.024875 × 10-18 J, or about 6.41 eV. Do not use the photon's E = pc as this electron's kinetic-energy equation.
Try it step by step
Work it out in steps: Momentum p = mv in kg m/s, then Wavelength lambda = h / p in m.
Show the answer
Problem 1.
Momentum p = mv in kg m/s: p = 9.11 x 10-31 x 2.0 x 106 = 1.82 x 10-24 kg m/s
Wavelength lambda = h / p in m: lambda = 6.63 x 10-34 / 1.82 x 10-24 = 3.64 x 10-10 m
Problem 2.
Momentum p = mv in kg m/s: p = 9.11 x 10-31 x 4.0 x 106 = 3.64 x 10-24 kg m/s
Wavelength lambda = h / p in m: lambda = 6.63 x 10-34 / 3.64 x 10-24 = 1.82 x 10-10 m
Problem 3.
Momentum p = mv in kg m/s: p = 1.67 x 10-27 x 2.0 x 103 = 3.34 x 10-24 kg m/s
Wavelength lambda = h / p in m: lambda = 6.63 x 10-34 / 3.34 x 10-24 = 1.99 x 10-10 m
Check your understanding
An electron at 1.50 x 106 m/s has de Broglie wavelength 0.485 nm. Its speed doubles and the non-relativistic approximation remains valid. What is the new wavelength?
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About 0.243 nm.
With p = mv in the stated approximation, doubling speed doubles momentum and halves wavelength: 0.485/2 is about 0.243 nm.
Interpret a finite collection of detections
The following constructed illustrative counts use nine detector-position bins of width 1.0 mm. Centres run from -4 to +4 mm, with edges from -4.5 to +4.5 mm. They illustrate accumulation; they are not a reported experiment or an exact diffraction calculation.
First 20 detections
The small sample gives limited information about the distribution. An empty early bin does not establish zero underlying probability.
The same constructed run after 200 detections
The count distribution becomes clearer after more detections. These are constructed illustrative counts with fixed positions, not a fitted diffraction model or authentic measurement record.
Count per equal-width bin after 200 detections
The middle bin contains 88 of 200 detections: relative frequency 0.44 over -0.5 to +0.5 mm. Histogram height here is a count, not probability density. A density estimate would also divide the relative frequency by the bin width.
| Centre / mm | First 20 | After 200 |
|---|---|---|
| -4 | 1 | 10 |
| -3 | 1 | 4 |
| -2 | 3 | 34 |
| -1 | 0 | 8 |
| 0 | 10 | 88 |
| 1 | 1 | 8 |
| 2 | 3 | 34 |
| 3 | 0 | 4 |
| 4 | 1 | 10 |
The middle bin contains 88/200 = 0.44 of the later detections. This estimates probability over the finite interval -0.5 to +0.5 mm, not probability at exactly x = 0. An empty bin in the smaller sample does not establish zero underlying probability there.
Paper 4 practice: spreadsheet
Keep bin probability separate from density
Download the constructed illustrative count CSV. Import it as comma-delimited numeric columns, preserving raw A2:E10. The headers are centre, left edge and right edge in mm, then counts after 20 and 200 detections. The spreadsheet workflow gives the entry and import steps.
Add F1 = width / mm, G1 = early relative frequency, H1 = later relative frequency and I1 = later density estimate / mm^-1. Enter:
| Cell | Formula |
|---|---|
| F2 | =C2-B2 |
| G2 | =D2/SUM($D$2:$D$10) |
| H2 | =E2/SUM($E$2:$E$10) |
| I2 | =H2/F2 |
Fill F2:I2 through row 10. Check that the count totals are 20 and 200, both relative-frequency columns sum to 1, and every later count is at least its earlier count. Use bars over the actual bin edges. These supplied values do not determine a unique fitted sine curve.
Interpret the middle bin
Its later relative frequency is 0.44. Dividing by its 1.0 mm width gives a bin-average density estimate of 0.44 mm-1. Multiplying that density estimate by the same width returns the dimensionless bin probability estimate. The coordinate unit here is mm, separate from the nm model on the probability page.
03
Wavefunctions and probability
Make a guess. You are not marked.
A particle's wavefunction has amplitude 2A near point X and amplitude A near point Y, over equal small intervals. How does the chance of detecting the particle near X compare with near Y?
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Four times as likely, because probability goes with amplitude squared.
The probability of detection in a small interval is proportional to the square of the wavefunction's amplitude there. (2A)2 = 4A2, which is four times A2.
A wavefunction represents a particle's state. Its amplitude is not a displacement of matter. Squared amplitude gives a probability density; area under that density gives an interval probability.
For a real wavefunction ψ, density is ψ2. More generally it is the modulus squared, |ψ|2. A negative real amplitude is allowed, but its squared density is nonnegative. Reversing the sign of the whole real wavefunction leaves its density unchanged; relative signs matter when combining contributions.
In one dimension, the probability of a position interval is the area under |ψ|2 over that interval. Normalisation sets the total area over all possible positions to 1. A density value alone is not a probability: it has an inverse-length unit and can be numerically greater than 1 without contradiction.
| Coordinate x | Amplitude ψ | Density |ψ|2 |
|---|---|---|
| m | m-1/2 | m-1 |
| nm | nm-1/2 | nm-1 |
Density multiplied by a width in the matching unit is dimensionless. An atomic electron cloud instead represents a three-dimensional spatial probability distribution, not a classical orbit. Its full volume density has unit m-3, and probability comes from a volume, rather than a one-dimensional area under a curve.
Square-profile normalisation
Determine the height before finding an interval area
Let ψ = C for 0 < x < 4.00 nm, with zero amplitude outside. For this constant profile:
C = 1/√(4.00 nm) = 0.500 nm-1/2
|ψ|2 = 0.250 nm-1
Between 1.00 and 2.50 nm the width is 1.50 nm, giving:
Square the amplitude first, then find the density area. Squaring the area under ψ gives a different result with the wrong units. Choosing C = -0.500 nm-1/2 would give the same probabilities.
In metre units the same amplitude is 1.58114 × 104 m-1/2, and density is 2.50 × 108 m-1. The interval width is then 1.50 × 10-9 m, so its probability remains 0.375.
This abrupt square profile is an ideal normalisation exercise. The values at its isolated endpoints do not change its area. It is not an infinite-well stationary state.
Check your understanding
A normalised real wavefunction is 0.500 nm^(-1/2) for 0 < x < 4.00 nm and zero outside. What is the probability of finding the particle between 1.00 and 2.50 nm?
Show the answer
0.375.
Square the amplitude first: density is 0.250 nm-1. Its area over the 1.50 nm interval is 0.250 x 1.50 = 0.375, a dimensionless probability.
Sinusoidal-profile normalisation
Use the mean square over the whole interval
Now let ψ = C sin(πx/L) within 0 < x < L, where L = 4.00 nm, and zero outside. The mean of sin2 over this half-wave is 1/2:
C = √(2/L) = 0.707107 nm-1/2
|ψ|2 = 0.500 sin2(πx/4.00) nm-1
The last expression uses x in nm, making πx/4.00 a dimensionless angle in radians. Symmetry gives probability 0.500 in each half of the interval. The central half, 1.00 to 3.00 nm, contains more probability than the combined two outer quarters because its density is higher on average.
The different coefficient from the square profile is essential. Using 1/√L here would leave total probability 1/2, not 1.
Check your understanding
For 0 < x < L, psi = C sin(pi x/L), with L = 4.00 nm and zero amplitude outside. The mean of sin2 over this interval is 1/2. Which positive C normalises the state?
Show the answer
0.707 nm^(-1/2).
C2 L/2 = 1 gives C = sqrt(2/L). With L entered in nm, C = sqrt(2/4.00) = 0.707 nm^(-1/2), and density times a width in nm is dimensionless.
Square profile: signed amplitude
This discontinuous square profile is an ideal normalization exercise, not an infinite-well stationary state. Values at single endpoints do not change an interval area.
Square profile: density and interval probability
Shaded width: 1.50 nm. Density: 0.250 per nm. Their product is probability 0.375, with no unit. The total width of 4.00 nm gives total probability 1.
Sinusoidal profile: signed amplitude
The coefficient is sqrt(2/L), with L = 4.00 nm. Amplitude is not a material displacement, and area under this amplitude graph is not probability.
Sinusoidal profile: density and interval probability
Only area under density gives an interval probability. The central 1-3 nm area is 0.818309886...; the total density area is 1. The shaded area is not the square of the area under amplitude.
Paper 4 practice: spreadsheet
Generate the sine model and refine its area estimate
Download the probability starter workbook. Its Coarse, Fine and Finer sheets contain coordinates for this generated 4.00 nm model, not experimental observations. The calculation cells are blank for you to complete.
All three sheets use A for x / nm, B for ψ / nm-1/2, C for density / nm-1 and D for interval probability. G2 contains L = 4; I2 contains the coordinate step. H2 is the blank normalisation coefficient. Enter these first formulas on each sheet:
| Cell | Formula |
|---|---|
| H2 | =SQRT(2/$G$2) |
| B2 | =$H$2*SIN(PI()*A2/$G$2) |
| C2 | =B2^2 |
| D2 | =(A3-A2)*(C2+C3)/2 |
Keep the fixed references when copying. B and C have a value at every coordinate. D is the trapezium area between two consecutive coordinates, so its last formula belongs one row earlier:
| Sheet and step | Coordinates; B/C fill | D interval fill |
|---|---|---|
| Coarse: 0.5 nm | A2:A10; B2:C10 | D2:D9 |
| Fine: 0.25 nm | A2:A18; B2:C18 | D2:D17 |
| Finer: 0.125 nm | A2:A34; B2:C34 | D2:D33 |
Plot B against numeric A and C against numeric A in separate XY graphs, with both vertical quantities and units labelled. The sine argument is in radians. For area, density in nm-1 must multiply widths in nm; mixing an unchanged nm-1 density with metre widths would change the answer incorrectly.
Use a blank cell for each of the following sums. Preserve the stored values while formatting display precision.
| Sheet | Probability sums |
|---|---|
| Coarse | Full: =SUM(D2:D9)1-3 nm: =SUM(D4:D7) |
| Fine | Full: =SUM(D2:D17)1-3 nm: =SUM(D6:D13) |
| Finer | Full: =SUM(D2:D33)1-3 nm: =SUM(D10:D25) |
Compare the three area estimates
The central-interval estimates are 0.8017766953, 0.8142087183 and 0.8172865746, respectively. Compare them with the supplied exact model area 1/2 + 1/π = 0.8183098862. The shortfalls are about 0.0165332, 0.00410117 and 0.00102331.
The full-range sums happen to equal 1, apart from numerical round-off, on all three unrounded regular grids. That coincidence does not make each partial area exact. The exact smooth density and the straight-sided trapezia are different representations.
Refining the step reduces a numerical approximation error here. It is distinct from experimental uncertainty and from the quantum position-momentum width relation. Every interval probability must stay between 0 and 1; a dimensional density has no such numerical upper limit.
04
Adding probability amplitudes
Make a guess. You are not marked.
Single photons pass one at a time through a double slit to a sensitive screen. Where do the first few photons land?
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At scattered single points; a fringe pattern builds up after many detections.
Each photon is detected at one point. Where it lands is random, but the probability is greatest where the two amplitudes add. Many detections together show the bright and dark fringes.
Coherent contributions combine as probability amplitudes. Add them first, then square the result to find density. Adding the separate densities omits interference.
A wavefunction describes one particle's state. In a double-slit arrangement, the coherent alternatives contribute to the amplitude at a detector position. They are not two independently detected fractions of the electron.
For locally real amplitude contributions a and b:
= a2 + b2 + 2ab
The 2ab term depends on the relative signs. The same separate squares can therefore produce different combined densities. For general amplitudes the rule is the modulus squared of their sum.
Worked local cancellation
A zero amplitude at one point
At one detector position, suppose the already weighted contributions are +0.300 and -0.300 mm-1/2:
|a + b|2 = 0 mm-1
Adding their separate squares would give 0.180 mm-1 and miss the cancellation. These are local contributions, not separately normalised complete states. Zero density at an ideal point does not establish zero probability throughout a finite detector bin.
Check your understanding
At one detector position, two coherent, already weighted real amplitude contributions are +0.300 and -0.300 mm^(-1/2). What is the resulting probability density there?
Show the answer
Zero at that point.
The total amplitude is +0.300 - 0.300 = 0, so its square is zero. This local statement does not establish zero probability throughout a finite detector bin.
Normalise a complete combined state
For a separate 4.00 nm interval, let L = 4.00 nm and define two normalised profiles, zero outside that interval:
v = √(2/L) sin(2πx/L)
Each squared profile has area 1. The product uv has equal positive and negative areas, which cancel over the whole interval. Consequently the total squared area of u + v, or of u - v, is 2. Dividing the amplitude by √2 normalises either combined state:
ψ- = (u - v)/√2
The factor works because of these particular normalised components and their cancelling product area. It is not an automatic coefficient for every pair of overlapping wavefunctions.
Start with two signed component amplitudes
The component profiles u and v each have squared area 1 and are orthogonal over this interval. Their relative signs matter when amplitudes are combined.
Combine amplitudes and normalize the state
The factor 1/sqrt(2) normalizes these particular orthogonal equal-weight combinations. Changing the relative sign changes the spatial density. A mixed-energy superposition generally evolves; this is one instant.
Square the combined amplitude to obtain density
Both full densities have area 1. Their relative sign information came from amplitude addition before squaring. A negative amplitude is not a negative probability.
At x = 1.00 nm, the plus-state density is about 0.728553 nm-1, while the minus-state density is about 0.0214466 nm-1. At 3.00 nm they swap. In particular, ψ+ there is -0.146447 nm-1/2, whose square is positive. At the midpoint both densities are 0.250 nm-1.
These are state profiles at a specified instant. A superposition of different energy states generally has a density that changes with time; it is not a stationary particle tracing the plotted curve.
Connect interference and standing shapes
Coherent alternative paths produce interference because their amplitudes combine. In a confined region, combining appropriate oppositely travelling amplitude components can produce standing spatial shapes that meet the boundaries. Both use amplitude superposition, but neither is obtained by adding probability densities.
The infinite-well model makes the boundary restriction explicit. Its standing wavefunction does not require a particle to have a definite classical leftward or rightward trajectory.
05
Localisation and momentum spread
Make a guess. You are not marked.
A beam of electrons passes through a slit, which is then made narrower. What happens to the angular spread of the beam beyond the slit?
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It widens, because a narrower position range means a larger momentum spread.
The slit limits where each electron can be across the beam. A narrower slit means a smaller spread in position, so a larger spread in sideways momentum, and the beam fans out more.
A sharply localised state needs a spread of momentum components. This is a property of the state, not merely poor measuring technique or an uncertainty about one definite classical speed.
A single perfectly defined travelling-wave momentum does not form a localised packet. Combining momentum components can localise the state, with a narrower position distribution requiring a broader momentum distribution along the same axis.
Here Δx and Δp are characteristic width scales, and the relation means greater than or of the order of h. Use it for an order-of-magnitude estimate. It is not an exact equality and does not require a state to attain a minimum product. It is also not an instrument specification.
Compare position widths on one common scale
State B has half the position width of A and twice its peak density. Both mean positions are zero. These relative Gaussian shapes illustrate corresponding spreads; they do not replace the separate syllabus order-of-magnitude estimate.
The narrower-position state has a broader momentum distribution
Follow the same state labels: B is twice as broad in momentum, although both mean momenta remain zero. Its peak is half as high so the full density area remains 1. The tails continue beyond this crop.
The momentum spread describes the distribution of possible results for similarly prepared states. Its mean can remain zero while its width is large. Reducing ruler or detector errors does not remove this state-level relationship.
Worked width-scale estimate
Estimate momentum spread before a velocity scale
Take an electron localised on the scale Δx = 2.0 × 10-10 m. With h = 6.63 × 10-34 J s:
= (6.63 × 10-34)/(2.0 × 10-10)
= 3.315 × 10-24 kg m/s
≈ 3.3 × 10-24 kg m/s
Using the non-relativistic electron mass me = 9.11 × 10-31 kg gives a corresponding velocity-spread scale:
= 3.64 × 106 m/s
This is a spread scale, not a measured speed or a required mean velocity. Halving Δx to 1.0 × 10-10 m doubles the characteristic required momentum and velocity spreads under the same convention. It does not require the mean momentum to double.
Why the width convention matters
The estimate above uses characteristic widths and suppresses numerical factors. If position and momentum widths are defined specifically as standard deviations σx and σp, their precise lower bound is σxσp ≥ h/(4π). Do not substitute standard-deviation numbers into the rough-width form while treating it as an exact theorem.
Check your understanding
Using the same order-of-magnitude position-momentum width convention, a particle is localised to half its previous position-width scale. How does the characteristic required momentum-spread scale change?
Show the answer
It doubles; the mean momentum need not double.
The required spread scale is of order h/Delta x. Halving the position scale doubles that momentum-spread scale. A distribution can retain zero mean while becoming wider.
06
A particle in a box
Make a guess. You are not marked.
An electron in a one-dimensional box moves from its n = 1 energy level to its n = 3 level. By what factor does its energy increase?
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Nine, because the energy is proportional to n squared.
In a box, E = n2 h2 / (8 m L2). Going from n = 1 to n = 3 multiplies n2 from 1 to 9, so the energy becomes nine times as large.
Confinement restricts the wavefunctions that can satisfy the boundaries. In a one-dimensional infinite well, only certain standing shapes and corresponding energies are allowed.
Define the well and its boundaries
The potential energy V is zero inside 0 < x < L and is an ideal infinite barrier outside. Here V means potential energy, not voltage. The particle cannot penetrate the barrier, so its wavefunction vanishes outside and at x = 0 and x = L.
A nonzero standing shape must fit n half-wavelengths between the walls:
λn = 2L/n
ψn = √(2/L) sin(nπx/L)
n = 1, 2, 3, ...
The coefficient normalises each state. An n = 0 sine profile would be zero everywhere and could not have total probability 1. Each state has n - 1 interior nodes, in addition to its two boundary nodes.
n = 1: no interior node
The upper curve is a signed state amplitude, not a moving particle or a material displacement. Its square gives the lower density. Each stationary-state density has total area 1 and remains fixed in time; it is not multiplied by an oscillating probability factor.
n = 2: 1 interior node
The upper curve is a signed state amplitude, not a moving particle or a material displacement. Its square gives the lower density. Each stationary-state density has total area 1 and remains fixed in time; it is not multiplied by an oscillating probability factor.
n = 3: 2 interior nodes
The upper curve is a signed state amplitude, not a moving particle or a material displacement. Its square gives the lower density. Each stationary-state density has total area 1 and remains fixed in time; it is not multiplied by an oscillating probability factor.
The second state has an interior node at x/L = 1/2; the third has nodes at 1/3 and 2/3. Negative amplitude lobes still give positive density. These stationary states have time-independent position densities. Their full wavefunctions can change phase, but a time factor must not make total probability oscillate between zero and one.
Connect the allowed wavelength to energy
The spatial wavelength sets a momentum scale h/λn = hn/(2L). Combining this scale with the non-relativistic energy relationship gives the allowed well energies:
= h2n2/(8mL2)
This is an energy calculation for a confined standing state. It does not give one definite measured classical momentum, or restrict all momentum measurements to just two values. Confinement requires a momentum spread.
An infinite-well potential is an ideal energy model
Inside: V = 0. Outside: the ideal potential energy is infinite. The wavefunction vanishes at both boundaries and outside. The electron example uses L = 1.00 nm, independently of the earlier 4.00 nm normalization profiles.
Allowed energies have the ratio 1 : 4 : 9
For the supplied 1.00 nm electron box, E1 is about 0.377 eV. The first three energies are about 0.377, 1.51 and 3.39 eV. The increasing gaps are genuine; the levels have not been evenly spaced for convenience.
Worked electron well
Keep the width distinct from the earlier profiles
An electron confined across about 1 nm has a ground-energy scale of order 10-19 J, below a few eV. Doubling the width should reduce that energy scale by four.
Use L = 1.00 nm = 1.00 × 10-9 m, me = 9.11 × 10-31 kg and h = 6.63 × 10-34 J s. This is a new width, distinct from the 4.00 nm normalisation examples.
/ [8(9.11 × 10-31)(1.00 × 10-9)2]
= 6.03141 × 10-20 J
= 0.376963 eV ≈ 0.377 eV
| n | λn / nm | En / eV |
|---|---|---|
| 1 | 2.00 | 0.377 |
| 2 | 1.00 | 1.51 |
| 3 | 0.667 | 3.39 |
The corresponding unrounded energy calculations for n = 2 and 3 give 2.41256 × 10-19 J and 5.42827 × 10-19 J. The ratio 1:4:9 is exact in the model; displayed energies are rounded.
E3 - E2 = 5E1
The increasing gaps are not equally spaced. At fixed n and mass, doubling L quarters En. At fixed n and L, doubling m halves it.
The nonzero ground energy is consistent with the momentum spread needed for confinement. It does not mean that a stationary density graph depicts a ball travelling back and forth at a definite speed. The infinite well is also not the exact potential or energy sequence of an atom.
Check your understanding
For the same particle mass, compare n = 2 in an infinite well of width 2L with n = 1 in a well of width L. How do their energies compare?
Show the answer
They are equal.
E_n is proportional to n2/L2 for fixed mass. Here 22/(2L)2 = 1/L2, the same ratio as the original ground state.
07
Atomic transitions and spectra
Make a guess. You are not marked.
Hydrogen atoms are excited to their n = 3 level. How many different photon energies can they emit as they return to n = 1?
Show the answer
Three: 3 to 1, 3 to 2 and 2 to 1.
An electron can drop straight from n = 3 to n = 1, or go 3 to 2 and then 2 to 1. Each drop between a different pair of levels emits a photon of a different energy, giving three lines.
An isolated atom has discrete electronic energy levels. Emission and absorption involve differences between those levels, with the actual initial state determining which transitions are available.
Atomic hydrogen is a real example of discrete bound electronic states with spatial wavefunctions. Their energies need not be equally spaced or follow the infinite well's n2 pattern. The box is a model of confinement, not an exact model of hydrogen.
For a transition that emits a photon, the photon carries the decrease in atomic energy. Absorption requires a photon that supplies an available upward gap. Photon energy is positive:
= hf = hc/λ
Read a supplied three-level atom
Consider levels at -6.00, -3.00 and -1.00 eV relative to a zero ionisation reference. All three pairwise transitions are permitted in this supplied model. It is not a labelled hydrogen model. The following comparison considers only these bound-bound transitions.
Use the actual starting level and the allowed energy gap
The three downward transitions can emit photons when their upper states are occupied. A ground-state absorber can use the upward 3 eV or 5 eV transition; it cannot use the 2 eV gap from an unoccupied excited state. Zero is the ionisation reference, not another bound level.
Worked emitted photon
Subtract the levels before using hf
For emission from -1.00 to -3.00 eV, the atom loses 2.00 eV. Using 1 eV = 1.60 × 10-19 J, h = 6.63 × 10-34 J s and c = 3.00 × 108 m/s:
f = Ephoton/h = 4.82655 × 1014 Hz
λ = c/f = 6.215625 × 10-7 m
≈ 622 nm
| Levels / eV | Photon energy / eV | λ / nm |
|---|---|---|
| -1 to -3 | 2.00 | 622 |
| -3 to -6 | 3.00 | 414 |
| -1 to -6 | 5.00 | 249 |
For the 3.00 and 5.00 eV photons, the frequencies are about 7.24 × 1014 and 1.21 × 1015 Hz. With the supplied constants, their wavelength calculations give 414.375 and 248.625 nm before rounding.
A direct upper-to-ground transition emits one 5.00 eV photon. A two-step cascade emits one 2.00 eV photon and one 3.00 eV photon, with the same total energy. It does not emit one photon at an average of their wavelengths.
Distinguish emission and absorption spectra
An excited low-density gas can produce separate emission lines against a dark background. Their wavelengths correspond to permitted downward gaps, with suitable excited-state populations. The level diagram alone does not determine each line's strength.
A cooler gas in front of a continuous source can remove selected wavelengths from the transmitted beam, producing absorption lines. The relevant lower levels must be occupied. Absorbed energy has not disappeared; later emission can send it in other directions.
An excited population can show all three emission lines
An appropriately excited population can supply the 2 eV, 3 eV and 5 eV downward transitions. The numerical axis includes ultraviolet. Equal marker height or width does not mean equal emitted intensity.
Ground-state absorption removes only the available gaps
For this all-ground-state bound-transition model, 249 nm and 414 nm are removed from the beam; 622 nm is not. Absorbed energy is not destroyed, and subsequent emission can be in other directions. A different lower-state population changes the possible absorption lines.
If all atoms start at -6.00 eV, photons of 3.00 or 5.00 eV can excite the listed upper states. A 2.00 eV photon cannot make either of those ground-state transitions. It could produce the -3.00 to -1.00 eV transition if that lower excited state were occupied.
These photons are below the 6.00 eV ground-state ionisation threshold. Start from the actual occupied level and the transitions permitted in the question. A mismatch with these specified gaps is not a universal claim that the photon cannot interact with matter in any other way.
Check your understanding
A supplied atom has levels -6.00, -3.00 and -1.00 eV, with the three pairwise transitions permitted. It starts in the -6.00 eV ground state. Can a 2.00 eV photon produce one of these bound-bound absorptions?
Show the answer
No; neither upward transition from its actual initial state has a 2.00 eV gap.
Start from the occupied -6.00 eV level. The supplied bound-bound absorptions require 3.00 or 5.00 eV. This conclusion is limited to the stated transitions, not every possible light-matter interaction.
Revision
Quantum physics at a glance
Identify the evidence and the quantity being predicted. A photon energy, a probability density, a momentum spread and an atomic energy gap require different calculations.
Choose the model before the equation
| Task | Keep this distinction |
|---|---|
| Light evidence | Threshold frequency supports quantised energy exchange; interference and diffraction support wave behaviour. Frequency changes energy per photon, while intensity at fixed frequency changes photon rate. |
| Momentum and wavelength | A photon has p = E/c = h/λ despite zero rest mass. A non-relativistic electron uses p = mv and λ = h/p. Do not use E = pc for its kinetic energy. |
| Position probability | Amplitude can be signed; density |ψ|2 is nonnegative. An interval probability is density area, not density alone or the square of amplitude area. |
| Coherent alternatives | Add amplitudes before squaring. Relative signs or phases affect interference. A global sign reversal of one complete real state does not change its density. |
| Localisation | A narrower position distribution needs a broader momentum distribution. The width relation concerns spread, not a required mean momentum or apparatus error. |
| Confinement | Boundary conditions select standing shapes and quantised energies. A wavefunction plot is not a particle trajectory, and the box's level sequence is not an atom's exact sequence. |
| Spectral lines | Use an allowed gap from the actual occupied level. Photon energy is the magnitude of the energy change, not the absolute value of one level. |
Photon and matter-wave calculations
Threshold: Φwork = hf0
Matter wave: λ = h/p
Non-relativistic massive particle: p = mv
For the 6.00 × 1014 Hz photon, λ = 500 nm, E ≈ 2.49 eV and p ≈ 1.33 × 10-27 kg m/s. The 1.50 × 106 m/s electron instead has λ ≈ 0.485 nm. Doubling its speed halves wavelength while quadrupling kinetic energy.
Normalisation and area
For a profile that vanishes outside 0 < x < L:
C2L = 1, so C = 1/√L
Sine profile ψ = C sin(πx/L):
C2L/2 = 1, so C = √(2/L)
Choose a global sign without changing probabilities. At L = 4.00 nm, the positive coefficients are 0.500 and 0.707107 nm-1/2. The square profile gives probability 0.375 over 1.00-2.50 nm. The sine profile's central 1.00-3.00 nm probability is about 0.818310; a coarse trapezium estimate need not equal that smooth-model area.
The coordinate and density units must match: nm-1 times nm gives a dimensionless probability. For coherent real contributions, (a + b)2 includes 2ab. Do not replace it with a2 + b2. Detection-bin frequencies estimate finite-interval probabilities and do not establish a classical path or exact point probability.
Width scales, box states and atomic gaps
Infinite well: ψn = √(2/L) sin(nπx/L)
λn = 2L/n
En = h2n2/(8mL2)
The width relation is an order-of-magnitude convention, not an exact equality. Infinite-well states have n = 1, 2, 3, ... and n - 1 interior nodes. Their first energies are in the ratio 1:4:9. Doubling width quarters energies at fixed n and mass; n = 2 at width 2L has the same energy as n = 1 at width L.
In the supplied atom, levels -6, -3 and -1 eV give permitted gaps of 3, 2 and 5 eV. The emitted wavelengths are about 414, 622 and 249 nm. A ground-state absorber can use the 3 and 5 eV upward gaps; the 2 eV absorption requires the -3 eV state to be occupied. The 249 nm line is ultraviolet.
Quantities and units
| Quantity and symbol | Unit | Meaning |
|---|---|---|
| Energy E, En | J or eV | 1 eV = 1.60 × 10-19 J. |
| Work function Φwork | J | Minimum surface-removal energy, distinct from magnetic flux. |
| Planck constant h | J s | 6.63 × 10-34 J s in these examples. |
| Momentum p | kg m/s = N s | Photon and massive-particle relations have different conditions. |
| Frequency f | Hz | Sets energy per photon. |
| Vacuum light speed c | m/s | 3.00 × 108 m/s. |
| Wavelength λ; width L; position x | m or nm | 1 nm = 10-9 m. Match density units to the position coordinate. |
| Mass m, me | kg | me = 9.11 × 10-31 kg. |
| Elementary charge e | C | 1.60 × 10-19 C; eV itself is an energy unit. |
| One-dimensional amplitude ψ | m-1/2 or nm-1/2 | Its modulus squared is the corresponding inverse-length density. |
| Probability; quantum number n | No unit | Probability is between 0 and 1. Here n labels a box state, not amount of substance. |
| Box potential energy V | J or eV | The local symbol V here does not mean electric potential. |
A three-dimensional electron cloud instead uses probability per volume, with SI density unit m-3. Always identify what the graph's vertical axis represents before reading a height or calculating an area.
Practise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
Define the work function of a metal and the threshold frequency. [2]
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Exam-style question
Ultraviolet light of wavelength 250 nm falls on a metal with a work function of 2.30 eV. Calculate the maximum kinetic energy of the emitted electrons, in eV. Take h = 6.63 x 10-34 J s, c = 3.00 x 108 m s-1 and e = 1.60 x 10-19 C. [4]
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Exam-style question
Explain how the photoelectric effect gives evidence that light consists of photons. [3]
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Exam-style question
A graph of maximum kinetic energy of photoelectrons (in eV) against frequency of the light is a straight line. It crosses the frequency axis at 4.8 x 1014 Hz and passes through (8.0 x 1014 Hz, 1.32 eV). Use the graph to find the Planck constant and the work function in eV. Take e = 1.60 x 10-19 C. [3]
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Exam-style question
Data-based question: Measuring the Planck constant with a photocell [20]
Light of a single wavelength shines on a clean metal surface inside an evacuated photocell. A reverse potential difference is increased until the photocurrent just falls to zero; this is the stopping potential Vs. The experiment is repeated for five wavelengths. Take c = 3.00 x 108 m s-1 and e = 1.60 x 10-19 C.
| wavelength / nm | 400 | 450 | 500 | 550 | 600 |
|---|---|---|---|---|---|
| Vs / V | 1.11 | 0.76 | 0.49 | 0.26 | 0.07 |
- (a) Explain what the stopping potential tells you about the emitted electrons. [2]
- (b) Write down the photoelectric equation and show that eVs = hf - Φ, where Φ is the work function. [2]
- (c) Calculate the frequency of the light for each wavelength. [1]
- (d) Use the data to find the gradient of a graph of Vs against f, and hence a value for the Planck constant h. [3]
- (e) Find the threshold frequency, and the work function of the metal in eV. [3]
- (f) A student says: "Red light of wavelength 700 nm will release electrons from this metal if it is bright enough." Evaluate this claim. [2]
- (g) The intensity of the 400 nm light is doubled. State and explain the effect on the stopping potential and on the photocurrent before the reverse p.d. is applied. [3]
- (h) The 450 nm beam delivers 2.0 mW to the surface. Calculate the number of photons arriving each second and the photocurrent if 5.0% of them each release an electron. [3]
- (i) State one assumption you made in (h). [1]
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Quick recall
A surface has threshold frequency 5.00 x 10^14 Hz. In the ordinary single-photon model, what happens if the intensity at 4.00 x 10^14 Hz increases while frequency and other conditions stay fixed?