{"id":2132,"date":"2026-10-03T14:37:33","date_gmt":"2026-10-03T14:37:33","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=2132"},"modified":"2026-10-03T14:37:37","modified_gmt":"2026-10-03T14:37:37","slug":"wave-particle-duality","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/modern-physics\/wave-particle-duality\/","title":{"rendered":"Wave-Particle Duality: One Electron, Two Slits"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nWave particle duality is the principle that light and matter each behave as a spread-out wave in some experiments and as a localised particle in others. A single relation ties the two faces together: an object of momentum p has a wavelength equal to the Planck constant h divided by p.\n\n<\/p><\/div>\n\n<p>Fire electrons at a barrier with two narrow openings in it, one electron at a time, and every single one arrives at the far screen as a dot. Nothing smeared, nothing half-landed. Leave it running and those dots stack up into a striped pattern that only a wave could have drawn.<\/p>\n\n<p>Each electron crossed the apparatus alone, so there was nothing for it to interfere with except itself. That is wave-particle duality in one experiment, and the rest of this page takes it apart: the relation linking momentum to wavelength, what the stripes are worth in micrometres, and what becomes of them the moment you try to find out which opening was used.<\/p>\n\n<h2>What Is Wave-Particle Duality?<\/h2>\n\n<p>Wave-particle duality is the rule that a quantum object carries both descriptions at once: it travels like a wave and it is detected like a particle, and which face you meet depends on what the apparatus asks of it.<\/p>\n\n<p>The word doing the work is <em>and<\/em>. Textbooks that say light &#8220;sometimes behaves as a wave and sometimes as a particle&#8221; make it sound like a costume change, as though the electron decides en route which to be. It does not. One description governs the journey, the other governs the arrival, and both are in force for every electron.<\/p>\n\n<p>Detection is where this stops being philosophy. An electron is never half-detected: the screen records one localised hit, at one position, carrying the whole of the electron\u2019s charge. The wave does not arrive; it sets the odds on where the hit lands.<\/p>\n\n<p>So the honest sentence is this one. The wave tells you where the particle is likely to turn up, and only a great many arrivals make it visible \u2014 a single dot tells you almost nothing, while ten thousand dots draw the wave in full.<\/p>\n\n<p>None of this is special to electrons. Photons, neutrons and whole atoms have all been sent through two-path interferometers and have all produced fringes, and one piece of arithmetic covers every one of them. Our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/modern-physics\/quantum-mechanics\/\">quantum mechanics<\/a> places duality alongside the other ideas it belongs with.<\/p>\n\n<h2>The Evidence on Both Sides: Quanta of Light, Waves of Matter<\/h2>\n\n<p>Duality earns its name because the evidence runs in both directions: light, modelled as a wave for a century, delivers its energy in indivisible lumps, and matter, modelled as particles for rather longer, diffracts.<\/p>\n\n<h3>Light arrives in quanta<\/h3>\n\n<p>Shine light on a clean metal surface and electrons come off \u2014 but whether any come off at all depends on the colour of the light, not on how bright it is. A wave of any frequency should eventually deliver enough energy if you wait long enough. It does not work that way.<\/p>\n\n<p>Light hands over its energy in fixed packets, and a packet either does the job or it does not. The experiment, its threshold and its arithmetic belong to our <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/modern-physics\/photoelectric-effect\/\">photoelectric effect<\/a> guide; all this page needs from it is the conclusion that light arrives in quanta.<\/p>\n\n<h3>Matter diffracts<\/h3>\n\n<p>The other direction took longer to test, because the wavelengths involved are minute. Send a beam of electrons at a crystal, though, and it emerges in sharp spots and rings \u2014 the signature of a wave meeting a regular array of scatterers rather than of pellets bouncing off it.<\/p>\n\n<p>The <a href=\"https:\/\/www.nobelprize.org\/prizes\/physics\/1937\/summary\/\" target=\"_blank\" rel=\"noopener\">Nobel Prize in Physics 1937<\/a> was awarded jointly to Clinton Joseph Davisson and George Paget Thomson for their experimental discovery of the diffraction of electrons by crystals. Our guide to the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/modern-physics\/de-broglie-wavelength\/\">de Broglie wavelength<\/a> works through the matter-wave relation and its arithmetic in detail; this page takes it as given and builds an experiment on top of it.<\/p>\n\n<h2>The Wave-Particle Duality Formula: Wavelength from Momentum<\/h2>\n\n<p>One equation carries the whole idea. The wavelength of any object is the Planck constant divided by that object\u2019s momentum.<\/p>\n\n<div class=\"pf-formula\">\u03bb = h \/ p<\/div>\n\n<ul>\n<li><strong>\u03bb<\/strong> \u2014 the wavelength, in metres (m). Electron wavelengths on this page are quoted in picometres, and 1 pm = 10<sup>\u221212<\/sup> m.<\/li>\n<li><strong>h<\/strong> \u2014 the <a href=\"https:\/\/physics.nist.gov\/cgi-bin\/cuu\/Value?h\" target=\"_blank\" rel=\"noopener\">Planck constant<\/a>, 6.62607015 \u00d7 10<sup>\u221234<\/sup> J\u00b7s, exact by definition since the 2019 revision of the SI.<\/li>\n<li><strong>p<\/strong> \u2014 the momentum, in kilogram metres per second (kg m\/s). Well below the speed of light, p = mv.<\/li>\n<\/ul>\n\n<p>Read the equation again and notice what is missing from it. Nothing on either side asks what the object is made of, whether it has mass, or whether it has charge \u2014 momentum goes in, wavelength comes out. Problem 4 below gives a photon and an electron the same momentum and gets the same wavelength from both. If you would rather have the number than the arithmetic, our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/de-broglie-wavelength\">De Broglie Wavelength Calculator<\/a> takes a mass and a speed and returns the wavelength.<\/p>\n\n<p>For an electron gun the momentum arrives as a voltage rather than a speed. Accelerate an electron from rest through a potential difference V and all the work done on it, eV, ends up as kinetic energy p<sup>2<\/sup> \/ (2 m<sub>e<\/sub>), which rearranges to a single square root.<\/p>\n\n<div class=\"pf-formula\">p = sqrt(2 m<sub>e<\/sub> e V)<\/div>\n\n<ul>\n<li><strong>V<\/strong> \u2014 the accelerating potential difference, in volts (V).<\/li>\n<li><strong>e<\/strong> \u2014 the elementary charge, 1.602176634 \u00d7 10<sup>\u221219<\/sup> C, also exact since 2019.<\/li>\n<li><strong>m<sub>e<\/sub><\/strong> \u2014 the electron mass, taken here as 9.109 \u00d7 10<sup>\u221231<\/sup> kg so that every figure on this page matches the calculator linked above.<\/li>\n<\/ul>\n\n<p>Put 100 V through that and the electron comes out with a wavelength of 122.6 pm. Problem 1 does the arithmetic line by line. Hold on to the magnitude: a tenth of a nanometre is the same order as the spacing between neighbouring atoms in a solid, which is the single fact that makes electron diffraction a laboratory technique rather than a curiosity.<\/p>\n\n<p>The lab below runs the whole experiment. Choose the accelerating voltage, the gap between the openings, how wide they are and how far away the screen sits, then let the stream run and watch the pattern assemble itself out of single detections \u2014 or send one electron at a time and watch it land as a single dot.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Wave-Particle Duality Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/wave-particle-duality.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<p>If you would rather take the controls one at a time, with worked settings to copy, the <a href=\"https:\/\/physicsfundamentalsinfo.com\/labs\/wave-particle-duality-simulator\">wave-particle duality simulator<\/a> has a page of its own.<\/p>\n\n<h2>The Double Slit, One Electron at a Time<\/h2>\n\n<p>Two openings, one particle in the apparatus at a time, and a pattern that needs both openings to exist: that is the experiment, and it is why this page is not simply an essay about a formula.<\/p>\n\n<p>Three lengths set everything. The openings are a distance d apart, each one is a width a across, and the screen sits a distance L beyond the barrier. Bright fringes land wherever the two paths differ by a whole number of wavelengths, which is the condition d sin\u03b8 = m\u03bb for integer m.<\/p>\n\n<p>At the angles involved here, sin\u03b8 and tan\u03b8 agree to better than one part in a million, so the fringes come out evenly spaced and their separation has a one-line formula.<\/p>\n\n<div class=\"pf-formula\">w = \u03bb L \/ d<\/div>\n\n<ul>\n<li><strong>w<\/strong> \u2014 the fringe spacing, the distance on the screen between neighbouring bright fringes, in metres (m).<\/li>\n<li><strong>L<\/strong> \u2014 the distance from the barrier to the screen, in metres (m).<\/li>\n<li><strong>d<\/strong> \u2014 the separation of the two openings, centre to centre, in metres (m).<\/li>\n<\/ul>\n\n<p>Those 122.6 pm electrons through openings 500 nm apart, onto a screen half a metre away, give a fringe spacing of 122.65 \u00b5m \u2014 about an eighth of a millimetre. Problem 2 works it through, and problem 3 runs it backwards to recover d from a measured spacing.<\/p>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/wave-particle-duality-diagram-01-768x524.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/wave-particle-duality-diagram-01-768x524.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/wave-particle-duality-diagram-01.jpg 1400w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1400\" height=\"956\" alt=\"A double-slit experiment drawn from the side, with the pattern it writes on the screen. On the left a filled block is the electron gun. A dashed horizontal optic axis runs from the gun to the centre of a screen on the right. Between them stands a vertical barrier drawn as three filled blocks, leaving two openings; a dimension line to the left of the barrier marks the separation d between the two openings and a short bracket marks the opening width a. Two solid rays leave the two openings and meet at one point P on the screen, above the centre. An arc between the axis and the direction of P marks the angle theta. Close to the lower opening a thick wine segment is the extra distance the lower path covers, labelled as the path difference d sin theta, with a dashed perpendicular dropped from the upper opening onto the lower ray. To the right of the screen a gold trace is the relative intensity against position on the screen, drawn on the same vertical scale as the screen itself; a dimension line beside it marks w, the gap of 122.65 micrometres between the central maximum and the first maximum. Labels beneath name the electron gun, the double slit with d equal to 500 nanometres and a equal to 100 nanometres, the screen and the intensity trace, and a dimension line gives the slit-to-screen distance L of 0.500 metres. Footnotes state that one pixel of height is 5.65 micrometres of screen position, that the horizontal scale is compressed about 286 times relative to the vertical, and that the angle to P is therefore drawn as about 16 degrees instead of its real 0.981 milliradians.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">The geometry and the pattern on one figure. The vertical scale is shared between the screen and the trace, but the barrier is schematic and its openings are drawn much further apart than that scale allows, while the horizontal scale is compressed hundreds of times, which is the only reason the angle to P is visible at all.<\/figcaption>\n<\/figure>\n\n<h3>Why the fringes are not all equally bright<\/h3>\n\n<p>Each opening on its own spreads the beam, which is ordinary <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/diffraction-physics\/\">diffraction<\/a>, and that spreading sets an envelope the fringes have to live inside. The envelope first falls to zero where a sin\u03b8 = \u03bb, and the fringes fade as they climb towards it.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;white-space:nowrap;\">Order m<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;white-space:nowrap;\">Position from the centre (\u00b5m)<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;white-space:nowrap;\">Relative intensity<\/th><th style=\"text-align:left;padding:8px 10px;border-bottom:2px solid #C8932A;font-weight:700;white-space:nowrap;\">What it is<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0.00<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">1.000<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">central maximum, the brightest fringe<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">1<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">122.65<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0.875<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">first fringe, one fringe spacing out<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">2<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">245.29<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0.573<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">already visibly dimmer<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">3<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">367.94<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0.255<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">a quarter of the centre<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">4<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">490.58<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0.055<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">the faintest fringe inside the envelope<\/td><\/tr>\n<tr><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">5<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">613.23<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">0.000<\/td><td style=\"padding:8px 10px;border-bottom:1px solid #D9CFB8;\">missing order: it lands exactly on the envelope\u2019s first zero<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>Every figure in that table is for 100 V electrons with d = 500 nm, a = 100 nm and L = 0.500 m, and every position appears on both sides of the centre. Four bright fringes survive on each side before the envelope closes them out.<\/p>\n\n<p>Order 5 is the interesting row. Because d is exactly five times a here, the fifth fringe would arrive precisely where the single-opening envelope has already fallen to zero, so it never appears at all \u2014 a missing order. Change the ratio d\/a and a different order goes missing; problem 5 works out which.<\/p>\n\n<h3>Watching the pattern assemble<\/h3>\n\n<p>Start the stream and the first few dozen detections look like nothing at all: scattered dots, no structure, no sign of a fringe. The pattern is a statement about probability, and probability needs a sample.<\/p>\n\n<p>By a few hundred the stripes begin to show through. By ten thousand the histogram of dot positions traces the intensity curve closely enough that you could read the fringe spacing off it with a ruler. Nothing changed about any individual electron on the way \u2014 only the number of them.<\/p>\n\n<p>One caveat about any simulation of this, including the one above. The arrival positions come from a seeded pseudo-random generator: it is a model of quantum randomness, not a sample from a quantum source, and no classical computer can supply the real thing.<\/p>\n\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/a-real-single-electron-pattern-building-up.webp\"\n       alt=\"Single electrons arriving one at a time on a detector, the scattered dots building into the interference fringes of wave-particle duality\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" width=\"1280\" height=\"640\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Each dot is one electron arriving whole. The fringes are not in any single dot; they are in how the dots are distributed.<\/figcaption>\n<\/figure>\n\n<h2>What Happens When You Ask Which Slit?<\/h2>\n\n<p>The fringes disappear. Any measurement that records which opening the electron went through destroys the interference pattern and leaves only the broad single-opening spread behind.<\/p>\n\n<p>This is the result that makes duality more than a turn of phrase. The pattern is not a property of the electron; it is a property of the experiment. Leave both paths open and unexamined and you get fringes. Learn the path and you get a smooth hump.<\/p>\n\n<p>The standard way to put a number on it is fringe visibility, which compares the brightest and darkest points of the pattern.<\/p>\n\n<div class=\"pf-formula\">visibility = (I<sub>max<\/sub> \u2212 I<sub>min<\/sub>) \/ (I<sub>max<\/sub> + I<sub>min<\/sub>)<\/div>\n\n<p>Fringes at full contrast, with genuinely dark gaps between them, give a visibility of 1.00. No fringes at all gives 0.00. In the idealised model used on this page and in the lab above, switching on a perfect which-way measurement moves the visibility from one of those values straight to the other.<\/p>\n\n<p>The halfway point between two bright fringes makes the contrast concrete. With both paths open that position is a dark line \u2014 the two paths cancel there. Measure the path and the same position reads 0.968 of the central intensity: almost as bright as the middle of the pattern.<\/p>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/wave-particle-duality-diagram-02-768x524.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/wave-particle-duality-diagram-02-768x524.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/10\/wave-particle-duality-diagram-02.jpg 1400w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1400\" height=\"956\" alt=\"Relative intensity plotted against position on the screen, from minus 700 to plus 700 micrometres, on two linear axes. The gold curve is the pattern with both paths open: a row of maxima at the centre and at multiples of the fringe spacing, each one lower than the last. The grey dashed curve is the pattern once the path has been measured: the same smooth single-slit envelope with no maxima of its own, and the gold curve touches it at every maximum. A wine dashed rule stands at plus 61.32 micrometres, half a fringe spacing from the centre, where the gold curve is at zero and the grey curve is at 0.968. A second wine dashed rule stands at plus 613.23 micrometres, where both curves fall to zero because the single-slit envelope does. A legend names the two curves. Footnotes give the two readings at half a fringe spacing, the relative heights of orders zero to four, and the equation and settings the curves are computed from.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">The same envelope with and without the interference inside it. Measuring the path does not dim the pattern or narrow it \u2014 it fills in the dark lines between the fringes, which is a different thing entirely. The envelope\u2019s own zero, marked by the second rule, stays dark on both curves.<\/figcaption>\n<\/figure>\n\n<p>There is a reason the trade is so unforgiving. Working out which opening was used means localising the electron at the barrier, and localising it there costs you knowledge of its sideways momentum, which is exactly what fixed the fringe positions. Our guide to the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/modern-physics\/heisenberg-uncertainty-principle\/\">Heisenberg uncertainty principle<\/a> handles that exchange in general.<\/p>\n\n<h2>Why Everyday Objects Show No Fringes<\/h2>\n\n<p>They obey the same relation; their fringes are simply far too fine for anything to resolve. Momentum sits in the denominator of \u03bb = h \/ p, so an everyday momentum drives the wavelength, and with it the fringe spacing, down past any length that means anything.<\/p>\n\n<p>Problem 7 runs the numbers for a thrown ball through a pair of metre-wide openings. The fringe spacing comes out around 10<sup>\u221233<\/sup> m \u2014 smaller than an atomic nucleus by a factor of roughly 10<sup>18<\/sup>.<\/p>\n\n<p>So the line between quantum and classical is not a size. It is a comparison: the wavelength against the scale of the apparatus. An electron at 122.6 pm meeting openings 500 nm apart has a wavelength worth about a four-thousandth of the gap, and that is already enough to make fringes you can measure in micrometres.<\/p>\n\n<p>Push the wavelength down far enough relative to the apparatus and the fringes crowd together until they blur into the smooth, featureless distribution that classical physics predicts. Nothing switches off. The pattern just stops being resolvable.<\/p>\n\n<h2>Where Wave-Particle Duality Turns Up in Real Experiments<\/h2>\n\n<p>Wherever a wavelength of about 100 pm is useful, matter waves are the practical way to get one. Four places where that pays off:<\/p>\n\n<h3>Electron microscopy<\/h3>\n\n<p>Resolution is limited by wavelength, and visible light runs out at a few hundred nanometres. A 100 V electron at 122.6452 pm has a wavelength about 4,500 times shorter than green light at 550 nm, which is why an electron beam reaches detail an optical microscope can never touch. Real instruments use far higher voltages and shorter wavelengths still.<\/p>\n\n<h3>Electron diffraction from crystals<\/h3>\n\n<p>Atoms in a solid sit a few hundred picometres apart, and a wave only diffracts usefully from a structure comparable to its own wavelength. Electrons at tens to hundreds of volts land in exactly that range, so a crystal acts on them as a three-dimensional grating \u2014 the effect the 1937 Nobel Prize recognised.<\/p>\n\n<h3>Surface science<\/h3>\n\n<p>Low-energy electrons scatter strongly and so penetrate only the top few atomic layers. That makes low-energy electron diffraction a tool for the surface specifically rather than the bulk, which is a hard thing to get at any other way.<\/p>\n\n<h3>Neutron diffraction<\/h3>\n\n<p>Neutrons are uncharged, so they pass deep into a sample instead of stopping at its surface, and they scatter from nuclei rather than from electron clouds. Facilities such as the <a href=\"https:\/\/www.ncnr.nist.gov\/\" target=\"_blank\" rel=\"noopener\">NIST Center for Neutron Research<\/a> exist to supply beams with the right wavelengths for the job.<\/p>\n\n<h2>Common Misconceptions About Wave-Particle Duality<\/h2>\n\n<h3>&#8220;The electron travels as a wave and turns into a particle at the screen&#8221;<\/h3>\n\n<p>Nothing turns into anything. There is one object with one quantum description, and that description predicts both the spreading and the single localised detection. The two-stage story is a mental crutch that fails the moment you ask when the change is supposed to happen.<\/p>\n\n<h3>&#8220;Observing it means a person has to look&#8221;<\/h3>\n\n<p>No consciousness is involved. What destroys the fringes is any interaction that leaves a record of the path \u2014 a stray photon scattering off the electron, a detector coupled to one opening, a gas molecule in a poor vacuum. Whether a human ever reads that record makes no difference at all.<\/p>\n\n<h3>&#8220;The fringes come from electrons bumping into each other&#8221;<\/h3>\n\n<p>This was a reasonable guess, and it is straightforward to rule out. Send the electrons through so sparingly that only one is ever in the apparatus, and the same fringes appear, just more slowly. Each electron interferes with itself.<\/p>\n\n<h3>&#8220;Duality is really only about light&#8221;<\/h3>\n\n<p>It cuts both ways, and the matter half is the half that surprised people. Light behaving as particles was startling enough; electrons, neutrons and atoms behaving as waves is the same statement read backwards, and it is the half that turned into working laboratory equipment.<\/p>\n\n<h2>Where This Model Stops Being Exact<\/h2>\n\n<p>Everything above is correct for the regime it describes, and the regime has edges. Five of them are worth stating plainly.<\/p>\n\n<p><strong>The which-way switch is idealised.<\/strong> Visibility 1.00 with both paths open and 0.00 with the path measured are the two limits of the intensity equation, not readings from any apparatus. Partial which-way information gives partial decoherence and an intermediate visibility, and quantifying that for a real apparatus is a measurement, not an application of this page.<\/p>\n\n<p><strong>The electrons here are treated non-relativistically.<\/strong> That is justified rather than assumed: from 20 V to 600 V, v\/c runs from 0.0088 to 0.0485, so the Lorentz factor exceeds 1 by only 1.18 \u00d7 10<sup>\u22123<\/sup> at the top of that range. The relativistic correction to \u03bb is about eV \/ (4 m<sub>e<\/sub> c<sup>2<\/sup>), which is 0.03 per cent at 600 V \u2014 below the precision anything on this page is quoted to.<\/p>\n\n<p><strong>The small-angle fringe formula is safe here, not everywhere.<\/strong> Across every setting in the lab above, the angle to the first maximum stays under 0.914 mrad, so w = \u03bbL\/d and the exact L tan\u03b8 agree to better than one part in a million. Open the geometry up to larger angles and that stops being true; the small-angle form is a convenience of this apparatus, not a general licence.<\/p>\n\n<p><strong>Simulated randomness is not quantum randomness.<\/strong> The dots in the lab above are placed by a seeded pseudo-random generator chosen to reproduce the intensity curve. It models quantum randomness faithfully in distribution and is not a quantum random source.<\/p>\n\n<p><strong>Plenty is left out.<\/strong> There is no space charge between electrons, no detector inefficiency, no scattering off the edges of the openings, no stray field and no vibration. Real electron interferometers also magnify the pattern with electron optics, because fringes tens to hundreds of micrometres apart are not something you look at directly.<\/p>\n\n<h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">An electron is accelerated from rest through a potential difference of 100 V. Find its de Broglie wavelength. Take h = 6.62607015e-34 J s, the electron mass as 9.109e-31 kg and the elementary charge as 1.602176634e-19 C.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: All the work done on the electron becomes kinetic energy, so eV = p<sup>2<\/sup> \/ (2 m<sub>e<\/sub>), which rearranges to p = sqrt(2 m<sub>e<\/sub> e V).\n\nStep 2: Substitute with units. 2 \u00d7 9.109 \u00d7 10<sup>\u221231<\/sup> kg \u00d7 1.602176634 \u00d7 10<sup>\u221219<\/sup> C \u00d7 100 V = 2.918845 \u00d7 10<sup>\u221247<\/sup> kg<sup>2<\/sup> m<sup>2<\/sup> s<sup>\u22122<\/sup>.\n\nStep 3: Take the root, then divide. p = sqrt(2.918845 \u00d7 10<sup>\u221247<\/sup>) = 5.402634 \u00d7 10<sup>\u221224<\/sup> kg m\/s, and \u03bb = h \/ p = 6.62607015 \u00d7 10<sup>\u221234<\/sup> \/ 5.402634 \u00d7 10<sup>\u221224<\/sup> = 1.226452 \u00d7 10<sup>\u221210<\/sup> m.\n\n<strong>Answer: 122.6 pm, or 122.6452 pm before rounding<\/strong>\n\nThat is the same order of magnitude as the gap between neighbouring atoms in a solid, which is exactly why a 100 V electron beam is a useful probe of crystal structure.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">Those same 100 V electrons pass through two openings 500 nm apart and land on a screen 0.500 m beyond the barrier. Find the fringe spacing. Use the wavelength from problem 1, 122.6452 pm.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: At these angles the fringes are evenly spaced, with w = \u03bb L \/ d.\n\nStep 2: Substitute with units. \u03bbL = 1.226452 \u00d7 10<sup>\u221210<\/sup> m \u00d7 0.500 m = 6.13226 \u00d7 10<sup>\u221211<\/sup> m<sup>2<\/sup>.\n\nStep 3: Divide by the separation, d = 500 nm = 5.00 \u00d7 10<sup>\u22127<\/sup> m. w = 6.13226 \u00d7 10<sup>\u221211<\/sup> \/ 5.00 \u00d7 10<sup>\u22127<\/sup> = 1.226452 \u00d7 10<sup>\u22124<\/sup> m.\n\n<strong>Answer: 122.65 \u00b5m, about an eighth of a millimetre<\/strong>\n\nNote how far the scale has shifted. A wavelength of 10<sup>\u221210<\/sup> m produced a pattern at 10<sup>\u22124<\/sup> m; the ratio L\/d is a magnifier, and it is worth a factor of a million here.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A different pair of openings gives a fringe spacing of 61.32 micrometres, with the same 122.6452 pm electrons and the same 0.500 m screen distance. How far apart are the openings?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Rearrange w = \u03bb L \/ d for the unknown: d = \u03bb L \/ w.\n\nStep 2: The numerator is the same as in problem 2. \u03bbL = 1.226452 \u00d7 10<sup>\u221210<\/sup> m \u00d7 0.500 m = 6.13226 \u00d7 10<sup>\u221211<\/sup> m<sup>2<\/sup>.\n\nStep 3: Divide by the measured spacing, w = 61.32 \u00b5m = 6.132 \u00d7 10<sup>\u22125<\/sup> m. d = 6.13226 \u00d7 10<sup>\u221211<\/sup> \/ 6.132 \u00d7 10<sup>\u22125<\/sup> = 1.0000424 \u00d7 10<sup>\u22126<\/sup> m.\n\n<strong>Answer: 1.00 \u00b5m, which is 1000 nm to three significant figures<\/strong>\n\nDoubling the separation halved the spacing, exactly as w = \u03bbL\/d requires. This is how slit separations are measured in practice: the pattern is far easier to get a ruler on than the barrier is.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A photon is given exactly the same momentum as the 100 V electron of problem 1, namely 5.402634e-24 kg m\/s. Find the photon energy in kiloelectronvolts, and its wavelength. Take c = 299792458 m\/s.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: A photon of energy E carries momentum p = E \/ c, so E = pc. Its wavelength still comes from the same relation, \u03bb = h \/ p.\n\nStep 2: Substitute with units. E = 5.402634 \u00d7 10<sup>\u221224<\/sup> kg m\/s \u00d7 2.99792458 \u00d7 10<sup>8<\/sup> m\/s = 1.61967 \u00d7 10<sup>\u221215<\/sup> J.\n\nStep 3: Convert to electronvolts by dividing by the elementary charge. 1.61967 \u00d7 10<sup>\u221215<\/sup> \/ 1.602176634 \u00d7 10<sup>\u221219<\/sup> = 10,109 eV. The wavelength needs no new work: \u03bb = h \/ p is unchanged at 122.6452 pm.\n\n<strong>Answer: 10.11 keV, and the same 122.6 pm wavelength as the electron<\/strong>\n\nSame momentum, same wavelength, and the formula never asked which object it was handed. What differs is the bill: the electron reached that wavelength on 100 eV of kinetic energy, while the photon needed 10,109 eV for it, about 101 times as much.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">Two openings 500 nm apart are each 100 nm wide. Which interference order is missing from the pattern, and how many bright fringes survive on each side of the centre inside the central envelope?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Write down the two conditions. Bright fringes sit where d sin\u03b8 = m\u03bb. The single-opening envelope first falls to zero where a sin\u03b8 = \u03bb.\n\nStep 2: Set them against each other. Order m lands on that zero when m\u03bb\/d = \u03bb\/a, which is simply m = d\/a. Here d\/a = 500 nm \/ 100 nm = 5.\n\nStep 3: So order 5 arrives exactly where the envelope has already gone to zero and is wiped out. The surviving orders on each side are the whole numbers below 5: 1, 2, 3 and 4.\n\n<strong>Answer: order 5 is the missing order, and four bright fringes survive on each side of the central maximum<\/strong>\n\nThe rule generalises: whenever d\/a is a whole number, that order goes missing. Make the openings narrower and the envelope widens, letting more orders through.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">For the same openings, 500 nm apart and 100 nm wide, how bright is the second-order fringe compared with the central maximum?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: At any maximum the two-path factor cos<sup>2<\/sup>(\u03c0 d sin\u03b8 \/ \u03bb) equals 1, so the ratio is set entirely by the single-opening envelope, sinc<sup>2<\/sup>(\u03c0 a sin\u03b8 \/ \u03bb), where sinc(x) = sin(x)\/x.\n\nStep 2: At order m the fringe condition gives sin\u03b8 = m\u03bb\/d, so the argument collapses to \u03c0am\/d, with no wavelength left in it. For m = 2: \u03c0 \u00d7 100 nm \u00d7 2 \/ 500 nm = 0.4\u03c0 = 1.256637 rad.\n\nStep 3: Evaluate. sin(1.256637) = 0.9510565, so sinc = 0.9510565 \/ 1.256637 = 0.756827, and the relative intensity is 0.756827<sup>2<\/sup> = 0.572787.\n\n<strong>Answer: 0.573, so the second fringe is about 57 per cent as bright as the centre<\/strong>\n\nBecause the wavelength cancelled in step 2, this answer depends only on the ratio d\/a. Change the voltage and the whole pattern rescales, but the brightness of order 2 relative to the centre does not move.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A 0.145 kg ball is thrown at 30.0 m\/s through a fence with two openings 1.00 m apart, onto a wall 10.0 m beyond it. Find the fringe spacing.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Momentum first. p = mv = 0.145 kg \u00d7 30.0 m\/s = 4.35 kg m\/s.\n\nStep 2: Combine the two relations instead of computing a wavelength: substituting \u03bb = h\/p into w = \u03bbL\/d gives w = hL \/ (pd).\n\nStep 3: Substitute with units. w = (6.62607015 \u00d7 10<sup>\u221234<\/sup> J\u00b7s \u00d7 10.0 m) \/ (4.35 kg m\/s \u00d7 1.00 m) = 1.523235 \u00d7 10<sup>\u221233<\/sup> m.\n\n<strong>Answer: 1.52 \u00d7 10<sup>\u221233<\/sup> m<\/strong>\n\nAn atomic nucleus is a few femtometres across, of order 10<sup>\u221215<\/sup> m, so these fringes are finer than a nucleus by a factor of about 10<sup>18<\/sup>. No screen, no detector and no amount of patience will ever bring them out. The relation has not failed; the pattern is simply unobservable, and the ball lands where Newton said it would.\n\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is wave-particle duality in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\n\nWave-particle duality means a quantum object spreads out like a wave while it travels and registers as a single particle when it is detected. An electron fired at two openings arrives as one dot, yet thousands of such dots pile up into interference fringes that need both openings to be open. The wave sets the odds; the particle is what the detector actually records.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does the double-slit experiment still work with one electron at a time?<\/summary><div class=\"pf-faq-item-answer\">\n\nYes, and that is the point of running it that way. Sending electrons through one at a time still builds the full interference pattern, provided nothing records which opening each one used. With only one electron in the apparatus, it cannot be interfering with its neighbours, so each electron is interfering with itself. This single result is what makes wave-particle duality impossible to argue away.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What happens if you measure which slit the electron goes through?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe interference pattern vanishes and only the broad single-opening spread is left. In the idealised model used here, the fringe visibility falls from 1.00 to 0.00 the moment a perfect which-way measurement is switched on. A real measurement is never that clean: it decoheres the state only partly, so the visibility settles somewhere between the two limits rather than at zero.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is an electron a wave or a particle?<\/summary><div class=\"pf-faq-item-answer\">\n\nNeither word is complete on its own. An electron propagates according to a wave equation and is always detected as one whole, localised object carrying one elementary charge. Calling it a wave cannot explain the single dot on the screen; calling it a particle cannot explain the fringes. The quantum description is one object with both behaviours built into it from the start.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why do everyday objects not behave like waves?<\/summary><div class=\"pf-faq-item-answer\">\n\nBecause their momentum is enormous compared with the Planck constant. Wavelength is h divided by momentum, so a thrown ball works out at roughly 10 to the minus 34 metres, and its fringes would be some 10 to the 18 times finer than an atomic nucleus. Nothing can resolve that. The relation still holds; its consequences are simply far below anything measurable.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the formula for wave-particle duality?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe relation is \u03bb = h \/ p: wavelength equals the Planck constant divided by momentum, and it applies to photons and to matter alike. For an electron accelerated from rest through a potential difference V, the momentum is sqrt(2 m<sub>e<\/sub> e V), which gives 122.6 pm at 100 V. The fringe spacing on a screen a distance L away then follows from w = \u03bbL\/d.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>Wave-particle duality is the principle that light and matter each travel as a wave and arrive as a particle, tied together by the relation wavelength equals the Planck constant divided by momentum. This page works through the evidence on both sides, the double slit run one electron at a time, what a which-slit measurement does to the fringes, and why everyday objects never show them. Seven worked problems, two original diagrams and an interactive lab.<\/p>\n","protected":false},"author":1,"featured_media":2135,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"class_list":["post-2132","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-modern-physics"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/2132","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=2132"}],"version-history":[{"count":2,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/2132\/revisions"}],"predecessor-version":[{"id":2147,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/2132\/revisions\/2147"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/2135"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=2132"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=2132"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=2132"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}