ARCHETYPE: de Broglie Wavelength Numericals

Give a paper-setter one electron and a voltage, and they can mint a dozen questions from it. What is its de Broglie wavelength after falling through 100 volts? Through 2000 volts? What if it is a proton instead, at the same voltage? What if you are told its speed, or its kinetic energy in electron-volts, rather than the voltage? What wavelength would a photon of the same energy carry, and why is it a hundred times longer? Every one of these is the same single idea — de Broglie's $\lambda = h/p$ — wearing a different costume. Learn to strip the costume off and the whole family collapses to one relation and a couple of ready-made shortcuts. This is a numerical gym, so the setup is the article. At its heart is a single equation you already met in the concept lesson: a particle of momentum $p$ has a wave of wavelength $\lambda = h/p$ . From it flow three working forms — $\lambda = h/mv$ when you know the speed, $\lambda = h/\sqrt{2mK}$ when you know the kinetic energy, and the exam favourite $\lambda = 12.27/\sqrt{V}$ ångström for an electron accelerated through $V$ volts (and $0.286/\sqrt{V}$ ångström for a proton). Memorise those four faces of one formula and every de Broglie numerical on a CBSE, ICSE, JEE or NEET paper becomes fill-in-the-blanks. The trap that catches half the class is a hidden square root: wavelength falls as $1/\sqrt{V}$ and as $1/\sqrt{K}$ , not as $1/V$ — quadruple the energy and you only halve the wavelength. Keep the square root, carry four significant figures, and match the mass to the particle. This article drills the master recipe and the four variants examiners actually pose, with every number computed to the last digit.

Key takeaways

The engine of every de Broglie numerical is one relation: a particle of momentum $p$ carries a wave of wavelength $\lambda = h/p$ , with $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ .
Three working forms: $\lambda = h/mv$ (speed given), $\lambda = h/\sqrt{2mK}$ (kinetic energy given), and for a charge accelerated through $V$ volts $\lambda = h/\sqrt{2mqV}$ .
Electron shortcut: $\lambda = 12.27/\sqrt{V}$ Å; proton shortcut: $\lambda = 0.286/\sqrt{V}$ Å — both for a particle accelerated from rest through $V$ volts.
Wavelength falls as $1/\sqrt{V}$ and $1/\sqrt{K}$ , never as $1/V$ : quadrupling the energy only halves the wavelength.
At the same energy an electron's wavelength dwarfs a proton's by $\sqrt{m_p/m_e}\approx 43$ , and a photon's dwarfs the electron's — matching the mass to the particle is the whole game.
Key idea: one relation, $\lambda = h/p$ , recast as $\lambda = h/\sqrt{2mK}$ and the shortcuts $12.27/\sqrt{V}$ Å (electron) and $0.286/\sqrt{V}$ Å (proton), solves the entire de Broglie numerical family — provided you keep the square root and match the mass to the particle.

The core insight

Every matter-wave numerical reduces to $\lambda = h/p$ ; writing the momentum as $p = mv = \sqrt{2mK} = \sqrt{2mqV}$ turns it into $\lambda = h/\sqrt{2mqV}$ , which for an electron is $12.27/\sqrt{V}$ Å and for a proton $0.286/\sqrt{V}$ Å, so the whole family is one formula and its costumes — remember only that $\lambda\propto 1/\sqrt{V}$ and that the mass must match the particle.

Where this lives today

The arithmetic you just drilled is the operating manual for a whole industry of matter-wave instruments. The electron microscope exists because a $100\ \mathrm{V}$ electron has a $0.123\ \mathrm{nm}$ wavelength — thousands of times shorter than visible light — so it resolves structures far below any optical microscope; crank the voltage into the hundreds of kilovolts and modern transmission electron microscopes image individual atomic columns, their resolution set by exactly the $\lambda = h/\sqrt{2meV}$ you computed (with relativistic corrections at those energies). Neutron diffraction, whose thermal-neutron wavelength of $\sim0.15\ \mathrm{nm}$ you found in Example 5, maps crystal and magnetic structure at neutron sources worldwide and earned the 1994 Nobel Prize. Low-energy electron diffraction (LEED) reads surface atomic arrangements for the semiconductor industry, and atom interferometers — matter waves of whole atoms — now make gravimeters and inertial sensors of extraordinary sensitivity. Every one of these begins with the same one-line calculation: turn an energy or a voltage into a momentum, and divide $h$ by it.

The setup

Physics first, because in a numerical gym the physics is the setup. In 1924 Louis de Broglie proposed that the wave–particle duality already known for light extends to matter: any particle of momentum $p$ has an associated wavelength

\lambda = \frac{h}{p}. \qquad (1)

where: $\lambda$ — the de Broglie wavelength; $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ — Planck's constant; $p$ — the particle's momentum. The physical reading is stark: the heavier or faster the particle, the shorter its wave. This is exactly the same $p = h/\lambda$ that a photon obeys, now demanded of electrons, protons, atoms — everything. The whole numerical craft is turning whatever the question gives you (a speed, a kinetic energy, an accelerating voltage) into a momentum, then dividing $h$ by it. There are three costumes the momentum wears, and recognising which one a question hands you is 80% of the battle. If you are given the speed $v$ , then $p = mv$ . If you are given the kinetic energy $K$ , use $K = p^2/2m$ so that $p = \sqrt{2mK}$ . If you are given the accelerating voltage $V$ through which a charge $q$ falls from rest, then the work done becomes kinetic energy, $K = qV$ , so $p = \sqrt{2mqV}$ . Feed each into equation (1):

\lambda = \frac{h}{mv}, \qquad \lambda = \frac{h}{\sqrt{2mK}}, \qquad \lambda = \frac{h}{\sqrt{2mqV}}. \qquad (2)

where: $m$ — the particle's mass ( $m_e = 9.11\times10^{-31}\ \mathrm{kg}$ for an electron, $m_p = 1.67\times10^{-27}\ \mathrm{kg}$ for a proton); $q$ — its charge ( $e = 1.60\times10^{-19}\ \mathrm{C}$ for both electron and proton). The physical reading: because $K$ and $V$ sit under a square root, the wavelength shrinks only as their square root — a fourfold rise in energy buys a mere halving of $\lambda$ . This single fact defeats more students than any other in the topic.

Figure

An electron gun accelerates an electron through V volts so it emerges as a matter wave of wavelength lambda equal to h over p equal to 12.27 over root V angstroms, then diffracts into rings off a crystal

The last costume in equation (2) is worth a name of its own because examiners lean on it hardest. Put the electron's mass and charge into $\lambda = h/\sqrt{2meV}$ and all the constants collapse into a single number, leaving nothing but the voltage under a root:

\lambda_{\text{electron}} = \frac{12.27}{\sqrt{V}}\ \text{Å}, \qquad \lambda_{\text{proton}} = \frac{0.286}{\sqrt{V}}\ \text{Å}. \qquad (3)

where: $V$ is the accelerating voltage in volts and $\lambda$ comes out in ångström ( $1\ \text{Å} = 10^{-10}\ \mathrm{m} = 0.1\ \mathrm{nm}$ ). The proton constant is smaller by exactly $\sqrt{m_p/m_e}$ because the only thing that changed is the mass under the root. These two shortcuts, plus the awareness that $\lambda\propto 1/\sqrt{V}$ , dispatch the majority of exam questions in under a minute. A worked landmark to memorise: an electron through $100\ \mathrm{V}$ has $\lambda = 12.27/\sqrt{100} = 1.23\ \text{Å} = 0.123\ \mathrm{nm}$ — an atomic-scale wave, which is precisely why electron beams diffract off crystals.

Figure

A straight-line graph of de Broglie wavelength in angstrom against one over root V, steep for the electron with slope 12.27 and shallow for the proton with slope 0.286, both through the origin

The master relations, in one table

Every question is one of these forms; the skill is spotting which line applies before you touch the calculator.

\lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}. \qquad (4)

where: the four expressions are the same wavelength written for the four things a question can give — momentum, speed, kinetic energy, accelerating voltage. Reading them left to right: momentum is the fundamental input; speed feeds it through $p=mv$ ; kinetic energy through $p=\sqrt{2mK}$ (valid while the particle is non-relativistic, which every board question is); voltage through $K = qV$ . Two consequences you should carry into every problem. First, $\lambda\propto 1/\sqrt{K}\propto 1/\sqrt{V}$ — the square root is not optional. Second, for a given energy $\lambda\propto 1/\sqrt{m}$ , so a light particle has a long wave and a heavy one a short wave. An electron and a proton dropped through the same voltage differ in wavelength by $\sqrt{m_p/m_e} = \sqrt{1833}\approx 42.8$ — the electron's wave is about 43 times longer.

Derivation AThe electron shortcut λ = 12.27/√V Å from λ = h/p

The goal is to compress $\lambda = h/\sqrt{2m_e eV}$ — the wavelength of an electron accelerated from rest through $V$ volts — into the single ready-to-use number $12.27/\sqrt{V}$ ångström, so that any accelerated-electron question becomes one division. The strategy is to collect every constant ( $h$ , $m_e$ , $e$ ) into one lumped coefficient, leaving only the voltage as a variable. This is pure algebra with a square root: no calculus enters, because kinetic energy relates to momentum through the plain identity $K = p^2/2m$ .

Step 1. Turn voltage into kinetic energy. An electron of charge $e$ falling from rest through a potential difference $V$ gains kinetic energy $K = eV$ (work done by the field). There is no wavelength yet — just energy bookkeeping.

Step 2. Turn kinetic energy into momentum. Non-relativistically $K = p^2/2m_e$ , so $p = \sqrt{2m_e K} = \sqrt{2m_e eV}$ . This is the only physics step; everything else is arithmetic.

Step 3. Insert into $\lambda = h/p$ . This gives $\lambda = h/\sqrt{2m_e eV}$ . Now lump the constants: $\lambda = \big(h/\sqrt{2m_e e}\big)\big/\sqrt{V}$ .

Step 4. Evaluate the coefficient. With $h = 6.63\times10^{-34}$ , $m_e = 9.11\times10^{-31}$ , $e = 1.60\times10^{-19}$ , the group $h/\sqrt{2m_e e} = 6.63\times10^{-34}/\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})} = 1.227\times10^{-9}\ \mathrm{m\cdot\sqrt V}$ . In ångström that is $12.27$ , so $\lambda = 12.27/\sqrt{V}$ Å. Check: $V = 100$ gives $12.27/10 = 1.23\ \text{Å} = 0.123\ \mathrm{nm}$ , and $V = 54$ (the Davisson–Germer value) gives $12.27/\sqrt{54} = 1.67\ \text{Å}$ . Verified: the coefficient $12.27$ Å·√V and the landmark $\lambda(100\ \mathrm{V}) = 0.123$ nm, $\lambda(54\ \mathrm{V}) = 1.67$ Å are recomputed independently in verify_hs2-de-broglie-numericals.py.

Go deeper · Derivation B · 1 minThe proton shortcut 0.286/√V Å, and why it is 43 times smaller

Derivation BThe proton shortcut 0.286/√V Å, and why it is 43 times smaller

The goal is the companion coefficient for a proton — $\lambda = 0.286/\sqrt{V}$ Å — and, more usefully, the clean ratio that lets you convert an electron answer into a proton one without recomputing. The strategy is identical to the electron case with one swap: the proton's mass $m_p$ replaces $m_e$ under the root, while its charge is still $+e$ . The whole difference is therefore a single square root of a mass ratio.

Step 1. Repeat the accelerated-particle result for a proton: $\lambda_p = h/\sqrt{2m_p eV}$ , exactly equation (3) with $m_p$ in place of $m_e$ . The charge is the same $e$ , so only the mass changed.

Step 2. Form the coefficient. $h/\sqrt{2m_p e} = 6.63\times10^{-34}/\sqrt{2(1.67\times10^{-27})(1.60\times10^{-19})} = 2.86\times10^{-11}\ \mathrm{m\cdot\sqrt V} = 0.286\ \text{Å}\cdot\sqrt{\mathrm V}$ . So $\lambda_p = 0.286/\sqrt{V}$ Å, and at $V = 100$ , $\lambda_p = 0.0286\ \text{Å}$ — a wavelength far below atomic spacing.

Step 3. Take the ratio to the electron. Dividing the two shortcuts, $\lambda_e/\lambda_p = \sqrt{m_p/m_e}$ , independent of the voltage. With $m_p/m_e = 1.67\times10^{-27}/9.11\times10^{-31} = 1833$ , this is $\sqrt{1833} = 42.8$ . So at any common voltage the electron's wave is about 43 times the proton's — a fact worth quoting directly to save a full recomputation.

Step 4. Sanity-check against the coefficients themselves: $12.27/0.286 = 42.9$ , matching $\sqrt{m_p/m_e} = 42.8$ to rounding. The mass ratio and the coefficient ratio are the same number reached by two roads. Verified: the proton coefficient $0.286$ Å·√V, $\lambda_p(100\ \mathrm{V}) = 0.0286$ Å, and the ratio $\sqrt{m_p/m_e} = 42.8$ are recomputed in verify_hs2-de-broglie-numericals.py.

Limiting cases and the trap table

Because this is a gym, the "limiting cases" double as the pitfalls examiners exploit. Read the table as both.

pitfall or limiting case	what actually happens	the fix
$\lambda\propto 1/V$ ?	no — $\lambda\propto 1/\sqrt{V}$ ; quadruple $V$ , halve $\lambda$	keep the square root on $V$ and on $K$
electron mass used for a proton	$\lambda$ comes out $\sqrt{m_p/m_e}\approx 43$ times too big	match the mass to the named particle
$V$ or $K$ given, but you divide by $v$	wrong; convert to momentum first via $\sqrt{2mK}$	pick the right costume from equation (2)
photon and electron treated alike	photon $\lambda = hc/E$ ; electron $\lambda = h/\sqrt{2mK}$	a photon has no mass — different formula
macroscopic object, $m$ large	$\lambda\sim10^{-34}\ \mathrm{m}$ , unmeasurable	correct — this is why balls show no waves
$V\to$ large (tens of kV)	non-relativistic shortcut slightly over-estimates $\lambda$	fine for board voltages; note the ceiling
units: $K$ in eV vs joules	mixing them scrambles the root	convert eV to joules ( $\times1.6\times10^{-19}$ ) first

Common trapWavelength falls as one over the square root of the energy, not the energy

The single costliest habit in this gym is dropping the square root. The de Broglie wavelength of a particle with kinetic energy $K$ is $\lambda = h/\sqrt{2mK}$ , so it scales as $1/\sqrt{K}$ — and for an accelerated charge as $1/\sqrt{V}$ . Double the accelerating voltage and the wavelength shrinks not to a half but to $1/\sqrt{2} = 0.707$ of its value; quadruple the voltage and it halves. Students who "cancel" the square root and write $\lambda\propto 1/V$ get every multi-voltage question wrong by a compounding factor. The second half of the same trap is the mass: at a given energy $\lambda\propto 1/\sqrt{m}$ , so the moment a question switches from electron to proton (or to an alpha particle, or a neutron) you must swap the mass under the root, which rescales the answer by the square root of the mass ratio. Write $\lambda = h/\sqrt{2mK}$ in full, name $m$ , and the square root will look after itself.

Worked examples

1 · Board level — an electron accelerated through 100 V (CBSE and NEET pattern)

An electron, initially at rest, is accelerated through a potential difference of $100\ \mathrm{V}$ . Find its de Broglie wavelength.

Step 1. Identify the costume: you are given a voltage, so use $\lambda = h/\sqrt{2m_e eV}$ (or the shortcut). Here $V = 100\ \mathrm{V}$ , $m_e = 9.11\times10^{-31}\ \mathrm{kg}$ , $e = 1.60\times10^{-19}\ \mathrm{C}$ .

Step 2. Shortcut route: $\lambda = 12.27/\sqrt{V} = 12.27/\sqrt{100} = 12.27/10 = 1.23\ \text{Å}$ .

Step 3. Full route as a check: $\lambda = 6.63\times10^{-34}/\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})(100)} = 6.63\times10^{-34}/(5.40\times10^{-24}) = 1.23\times10^{-10}\ \mathrm{m}$ .

Step 4. Convert and sense-check: $\lambda = \boxed{1.23\ \text{Å} = 0.123\ \mathrm{nm}}$ . This is comparable to atomic spacings, which is exactly why $100\ \mathrm{V}$ electrons diffract off crystals — the basis of the electron microscope.

2 · Board level — from a given speed, and why a cricket ball shows no waves (ICSE pattern)

Part (a): find the de Broglie wavelength of an electron moving at $1.0\times10^{6}\ \mathrm{m/s}$ . Part (b): find it for a $0.15\ \mathrm{kg}$ cricket ball travelling at $40\ \mathrm{m/s}$ , and comment.

Step 1. You are given a speed, so use $\lambda = h/mv$ . Part (a): $m = m_e = 9.11\times10^{-31}\ \mathrm{kg}$ , $v = 1.0\times10^{6}\ \mathrm{m/s}$ .

Step 2. Compute (a): $\lambda = 6.63\times10^{-34}/(9.11\times10^{-31}\times1.0\times10^{6}) = 6.63\times10^{-34}/(9.11\times10^{-25}) = 7.28\times10^{-10}\ \mathrm{m} = \boxed{0.728\ \mathrm{nm}}$ — again atomic-scale, and measurable.

Step 3. Part (b): $\lambda = h/mv = 6.63\times10^{-34}/(0.15\times40) = 6.63\times10^{-34}/6.0 = \boxed{1.1\times10^{-34}\ \mathrm{m}}$ .

Step 4. Comment: the ball's wavelength is about $10^{-34}\ \mathrm{m}$ — twenty-odd powers of ten smaller than an atomic nucleus, and utterly unmeasurable. This is the honest answer to "why don't everyday objects show wave behaviour?": they do, but $\lambda = h/mv$ makes the wave absurdly small once $m$ is macroscopic.

3 · Competitive level — proton versus electron through the same voltage (JEE-Main and NEET pattern)

An electron and a proton are each accelerated from rest through the same potential difference of $100\ \mathrm{V}$ . Find the ratio of their de Broglie wavelengths, and the proton's actual wavelength.

Step 1. Both use $\lambda = h/\sqrt{2meV}$ with the same $V$ and same charge $e$ ; only the mass differs. So the ratio is $\lambda_e/\lambda_p = \sqrt{m_p/m_e}$ , with no need to evaluate either wavelength for the ratio.

Step 2. Compute the ratio: $m_p/m_e = 1.67\times10^{-27}/9.11\times10^{-31} = 1833$ , so $\lambda_e/\lambda_p = \sqrt{1833} = \boxed{42.8}$ — the electron's wave is about 43 times the proton's.

Step 3. Get the proton's wavelength from its shortcut: $\lambda_p = 0.286/\sqrt{V} = 0.286/\sqrt{100} = 0.286/10 = \boxed{0.0286\ \text{Å}}$ , i.e. $2.86\times10^{-12}\ \mathrm{m}$ .

Step 4. Cross-check: the electron at $100\ \mathrm{V}$ is $1.23\ \text{Å}$ (Example 1), and $1.23/0.0286 = 43.0$ , matching the mass-ratio route to rounding. Timing note: about 40 seconds if you quote $\lambda_e/\lambda_p = \sqrt{m_p/m_e}$ straight off — the ratio never needs the individual wavelengths. The trap is using the electron mass for the proton, which would make them equal.

4 · Competitive level — electron versus photon of the same energy (JEE-Advanced pattern)

A photon and an electron each carry an energy of $100\ \mathrm{eV}$ . Compare their wavelengths.

Step 1. Different beasts, different formulas. A photon is massless: $\lambda_{\text{ph}} = hc/E = 1240/E[\text{eV}]$ nm. An electron has mass: its $100\ \mathrm{eV}$ is kinetic energy, so $\lambda_e = h/\sqrt{2m_e K}$ , which for $K$ supplied as a voltage of $100\ \mathrm{V}$ is the shortcut.

Step 2. Photon: $\lambda_{\text{ph}} = 1240/100 = \boxed{12.4\ \mathrm{nm}}$ .

Step 3. Electron: $\lambda_e = 12.27/\sqrt{100} = 1.23\ \text{Å} = \boxed{0.123\ \mathrm{nm}}$ .

Step 4. Ratio: $\lambda_{\text{ph}}/\lambda_e = 12.4/0.123 = 101$ — the photon's wavelength is about a hundred times the electron's at the same energy. The reason is structural: the photon's energy is $E = pc$ (linear in $p$ ), the electron's is $E = p^2/2m$ (quadratic), so for the same $E$ the electron carries far more momentum and hence a much shorter wave. Timing note: about 60 seconds. Keep the two formulas apart — $hc/E$ for the photon, $h/\sqrt{2mK}$ for the electron; using one for the other is the classic full-mark loss here.

5 · Competitive level — the de Broglie wavelength of a thermal neutron (JEE-Advanced pattern)

Estimate the de Broglie wavelength of a neutron in thermal equilibrium at room temperature, $T = 300\ \mathrm{K}$ . Take the average kinetic energy as $K = \tfrac{3}{2}k_BT$ with $k_B = 1.38\times10^{-23}\ \mathrm{J/K}$ and $m_n \approx m_p = 1.67\times10^{-27}\ \mathrm{kg}$ .

Step 1. You are given a temperature, so first convert to kinetic energy: $K = \tfrac{3}{2}k_BT = 1.5(1.38\times10^{-23})(300) = 6.21\times10^{-21}\ \mathrm{J}$ , which is $6.21\times10^{-21}/1.6\times10^{-19} = 0.039\ \mathrm{eV}$ — the familiar "thermal" energy scale.

Step 2. Momentum from energy: $p = \sqrt{2m_n K} = \sqrt{2(1.67\times10^{-27})(6.21\times10^{-21})} = \sqrt{2.07\times10^{-47}} = 4.55\times10^{-24}\ \mathrm{kg\,m/s}$ .

Step 3. Wavelength: $\lambda = h/p = 6.63\times10^{-34}/(4.55\times10^{-24}) = 1.46\times10^{-10}\ \mathrm{m} = \boxed{0.146\ \mathrm{nm} = 1.46\ \text{Å}}$ .

Step 4. Sense-check: a thermal neutron's wavelength lands right at atomic-lattice spacing, which is exactly why neutron diffraction is a workhorse tool for mapping crystal and magnetic structure. A slow, heavy neutron and a fast, light electron can share an atomic-scale wavelength because the neutron's small energy compensates its large mass under the root. Timing note: about 90 seconds — the extra step is the temperature-to-energy conversion $K = \tfrac{3}{2}k_BT$ . Do not plug $T$ into a wavelength formula directly; energy first, then momentum, then wavelength.

Exam traps

Dropping the square root. The wavelength is $\lambda = h/\sqrt{2mK} = 12.27/\sqrt{V}$ Å — it falls as $1/\sqrt{V}$ , not $1/V$ . Doubling the voltage cuts $\lambda$ to $0.707$ of its value, not to a half. This single slip wrecks every multi-voltage or "by what factor" question. Write the root explicitly.

Using the wrong mass. The electron shortcut $12.27/\sqrt{V}$ is only for electrons. A proton, deuteron, alpha particle or neutron needs its own mass under the root; the proton coefficient is $0.286$ , smaller by $\sqrt{m_p/m_e}\approx 43$ . When a question names a non-electron, go back to $\lambda = h/\sqrt{2mqV}$ and substitute.

Confusing photon and matter-wave formulas. A photon's wavelength is $\lambda = hc/E = 1240/E[\text{eV}]$ nm because $E = pc$ . A massive particle's is $\lambda = h/\sqrt{2mK}$ because $E = p^2/2m$ . At equal energy the electron's wave is $\sim100$ times shorter. Never feed an electron's kinetic energy into $hc/E$ .

Unit soup. Keep it SI before dividing: kinetic energy in joules (convert eV by $\times1.6\times10^{-19}$ ), mass in kilograms, and remember $1\ \text{Å} = 10^{-10}\ \mathrm{m} = 0.1\ \mathrm{nm}$ . A stray factor of $10^{-9}$ between nm and Å is an instant lost mark. Carry four significant figures and round only at the end.

Board scope. de Broglie numerical drills of this kind are firmly examined on CBSE, ICSE, JEE-Main, JEE-Advanced and NEET — the India-facing set. AP Physics 2 teaches $\lambda = h/p$ conceptually but is not a numerical-drill objective; A-Level 9702 does carry de Broglie wavelength calculations for accelerated electrons and electron diffraction, though the drill volume is lighter than the India-facing set; and IGCSE 0625 has no matter-wave content at all. Weight your practice to your board: if you are on CBSE or NEET, the accelerated-electron and electron-versus-proton problems above are core; if you are on AP, know the relation and one clean application; if you are on A-Level 9702, expect lighter accelerated-electron and electron-diffraction numericals.

Origins · 1 minGeorge Paget Thomson and how this was discovered

George Paget Thomson

1892–1975 · Aberdeen and London

George Paget Thomson gave matter-wave theory its most poetic confirmation. In 1927, at the University of Aberdeen, he fired a beam of fast electrons through very thin metal foils and caught, on a photographic plate beyond, a set of concentric diffraction rings — the unmistakable signature of a wave scattering off a crystal lattice, and precisely what de Broglie's $\lambda = h/p$ predicted. Working independently of Davisson and Germer's crystal-reflection experiment the same year, he confirmed the electron's wavelength by a different geometry, and the two shared the 1937 Nobel Prize in Physics for the discovery of electron diffraction. The result carries a delicious irony: his father, J. J. Thomson, had established the electron as a particle by measuring its charge-to-mass ratio in 1897 (and won the 1906 Nobel Prize for his investigations into the conduction of electricity by gases), while the son won his for showing the same electron is a wave. Father and son, particle and wave — the dual nature made a family affair.

History

1924

— Louis de Broglie, in his doctoral thesis, proposes that every particle of momentum p carries a wave of wavelength lambda equals h over p.

1927

— Clinton Davisson and Lester Germer confirm electron waves by diffraction from a nickel crystal at 54 volts.

1927

— George Paget Thomson diffracts electrons through thin metal foils, producing X-ray-like rings that clinch the matter wave.

1929

— de Broglie receives the Nobel Prize in Physics for the matter-wave hypothesis.

1937

— Davisson and G. P. Thomson share the Nobel Prize for the experimental discovery of electron diffraction.

1994

— Clifford Shull and Bertram Brockhouse win the Nobel Prize for neutron scattering, matter waves turned into a tool.

Going further

Texts. NCERT Physics Class 12, Ch. 11 ("Dual Nature of Radiation and Matter"), the de Broglie sections and exercises · H. C. Verma, Concepts of Physics Vol. 2, "Photoelectric Effect and Wave–Particle Duality," matter-wave problems · D. C. Pandey / Irodov, matter-wave numerical sets (accelerated particles, thermal particles, electron-vs-photon) · Cambridge A-Level 9702 (de Broglie relation and electron diffraction).

Provenance. Standard exam-archive problems. Recompute: electron coefficient $12.27$ Å·√V; $\lambda(100\ \mathrm{V}) = 1.23\ \text{Å} = 0.123\ \mathrm{nm}$ ; $\lambda(54\ \mathrm{V}) = 1.67\ \text{Å}$ ; electron at $v = 10^{6}\ \mathrm{m/s}$ gives $0.728\ \mathrm{nm}$ ; cricket ball ( $0.15\ \mathrm{kg}$ , $40\ \mathrm{m/s}$ ) gives $1.1\times10^{-34}\ \mathrm{m}$ ; proton coefficient $0.286$ Å·√V, $\lambda_p(100\ \mathrm{V}) = 0.0286\ \text{Å}$ ; ratio $\sqrt{m_p/m_e} = \sqrt{1833} = 42.8$ ; photon vs electron at $100\ \mathrm{eV}$ gives $12.4\ \mathrm{nm}$ vs $0.123\ \mathrm{nm}$ , ratio $101$ ; thermal neutron at $300\ \mathrm{K}$ , $K = \tfrac{3}{2}k_BT = 0.039\ \mathrm{eV}$ , $\lambda = 0.146\ \mathrm{nm}$ ; graph points $\lambda(400\ \mathrm{V}) = 0.61\ \text{Å}$ , $\lambda(900\ \mathrm{V}) = 0.41\ \text{Å}$ .

Prerequisites. de Broglie Matter Waves and the Dual Nature of Matter — the concept article that establishes $\lambda = h/p$ this gym uses. Leads to. the university seam de Broglie Matter Waves.

Go deeper (UG). The relativistic form of the matter wavelength, the full theory of electron and neutron diffraction, and the phase-versus-group-velocity structure of a matter wave packet are developed in de Broglie Matter Waves. Enrichment only — no board question needs them.

When do I use λ = h/mv, λ = h/√(2mK), and the 12.27/√V shortcut?

They are the same formula fed three different inputs, so let the question choose. If you are told the particle's speed $v$ , use $\lambda = h/mv$ directly. If you are told its kinetic energy $K$ (in joules, or in eV converted to joules), use $\lambda = h/\sqrt{2mK}$ . If you are told the accelerating voltage $V$ through which it fell from rest, the energy is $K = qV$ , so use $\lambda = h/\sqrt{2mqV}$ — and for an electron that collapses to $12.27/\sqrt{V}$ ångström, for a proton to $0.286/\sqrt{V}$ . The shortcuts only apply to a charge accelerated from rest through a stated voltage; if you are given a raw speed or energy, go back to the first two forms. Spotting which of the three a question hands you is most of the marks.

Where does the number 12.27 come from, and is it exact?

It is the lumped constant $h/\sqrt{2m_e e}$ evaluated in ångström-per-root-volt. Put in the HS values $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ , $m_e = 9.11\times10^{-31}\ \mathrm{kg}$ , $e = 1.60\times10^{-19}\ \mathrm{C}$ and you get $1.227\times10^{-9}\ \mathrm{m\cdot\sqrt V}$ , which is $12.27$ ångström. It is not a new constant — just $\lambda = h/\sqrt{2m_e eV}$ with everything except $V$ pre-computed, so an accelerated-electron question becomes a single division. It is not exact: it uses the rounded board constants, and it is non-relativistic. For board voltages (up to a few kilovolts) it is good to three figures; only when the voltage climbs into the tens of kilovolts does relativity begin to make it slightly over-estimate $\lambda$ . The proton's version, $0.286/\sqrt{V}$ , is the same group with $m_p$ swapped for $m_e$ .

At the same kinetic energy, why is the electron's wavelength so much larger than the proton's?

Because for a given energy $\lambda\propto 1/\sqrt{m}$ , and the proton is far heavier. Start from $\lambda = h/\sqrt{2mK}$ : the mass sits under the square root, so a heavier particle carries more momentum for the same $K$ and therefore a shorter wave. The proton is $1833$ times the electron's mass, so at equal energy (or equal accelerating voltage) the proton's wavelength is smaller by $\sqrt{1833}\approx 42.8$ — the electron's wave is about 43 times longer. This is why electron beams, not proton beams, are the natural probe for crystal diffraction at ordinary voltages: at $100\ \mathrm{V}$ the electron sits at a convenient $1.23\ \text{Å}$ while the proton is a hard-to-use $0.0286\ \text{Å}$ . Always re-substitute the correct mass the instant the particle changes.

A photon and an electron have the same energy — why don't they have the same wavelength?

Because they store energy differently. A photon is massless and obeys $E = pc$ , so its momentum is $p = E/c$ and its wavelength is $\lambda = hc/E = 1240/E[\text{eV}]$ nm. A massive electron obeys $E = p^2/2m$ for its kinetic energy, so $p = \sqrt{2mK}$ and $\lambda = h/\sqrt{2mK}$ . Because the electron's energy is quadratic in momentum while the photon's is linear, at the same energy the electron carries far more momentum and hence a much shorter wave — about a hundred times shorter at $100\ \mathrm{eV}$ ( $12.4\ \mathrm{nm}$ for the photon versus $0.123\ \mathrm{nm}$ for the electron). The lesson: never use $hc/E$ for a massive particle, and never use $h/\sqrt{2mK}$ for a photon.

Do I ever need the relativistic formula for these problems?

Not for any standard board or entrance question. Every voltage you will meet — tens to a few thousand volts — keeps the electron slow compared with light, so the non-relativistic $\lambda = h/\sqrt{2meV}$ and its $12.27/\sqrt{V}$ shortcut are accurate to the three figures the exam wants. The approximation only starts to bite once the accelerating voltage reaches the tens of kilovolts (as in a real electron microscope), where the electron's speed becomes an appreciable fraction of $c$ and the shortcut begins to over-estimate the wavelength. At the HS level you should know that this ceiling exists and that the syllabus formula is the low-speed one; you are not asked to compute the relativistic correction. If a problem quotes an ordinary lab voltage, use the shortcut with confidence.

Why don't we notice the wave nature of everyday objects?

Because $\lambda = h/mv$ makes their wavelength preposterously small. Planck's constant is $6.63\times10^{-34}\ \mathrm{J\,s}$ , so dividing it by the momentum of anything macroscopic gives a wavelength around $10^{-34}\ \mathrm{m}$ — the cricket ball in Example 2 came out at $1.1\times10^{-34}\ \mathrm{m}$ , twenty powers of ten below the size of a nucleus. A wave that short can never diffract off any real obstacle, because diffraction only shows up when the wavelength is comparable to the gap or grating spacing, and nothing in the universe is that small. Matter waves are not switched off for big objects; they are simply crushed to invisibility by the largeness of $mv$ against the tininess of $h$ . Only for the lightest particles — electrons, neutrons, atoms — does $\lambda$ climb to the atomic scale where diffraction becomes observable.

Is this the same as the concept article on de Broglie matter waves?

It shares the physics but not the purpose. The concept article, de Broglie Matter Waves and the Dual Nature of Matter, explains why a particle has a wavelength — the analogy with the photon, the meaning of $\lambda = h/p$ , and how Davisson–Germer confirmed it. This article assumes all of that and drills using it fast and correctly under exam time: picking the right form for a speed, an energy or a voltage; applying the $12.27/\sqrt{V}$ and $0.286/\sqrt{V}$ shortcuts; comparing electron, proton and photon; and dodging the square-root and wrong-mass traps. Treat the concept article as the theory lecture and this as the problem class. If a formula here feels unmotivated, revisit the concept article; if you understand the idea but keep slipping on arithmetic and setup, this gym is where you fix it.