Topic 2 of 7
Matter waves and detections
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.
Optional check An electron at 1.50 x 10^6 m/s has de Broglie wavelength 0.485 nm. Its speed doubles and the non-relativistic approximation remains valid. What is the new wavelength?
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.
Count-data exercise
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.