De Broglie hypothesis — matter waves λ = h - p: Worked examples

Before anything, let us pin down every symbol we will lean on, so nothing sneaks in undefined.

Definition The cast of symbols

$λ$ (Greek "lambda") — the wavelength: the length of one full ripple of the matter wave, measured in metres. Picture the distance from one wave-crest to the next.
$p$ — momentum: "how much motion" a thing carries. For slow objects $p = m v$ (mass times velocity). Units: $kg\cdotm/s$ .
$m$ — mass in kilograms; $v$ — speed in metres per second.
$K$ — kinetic energy, the energy of motion, in joules (J).
$E$ — total energy carried by a particle, also in joules (J). For a photon (light particle) all of its energy is this $E$ ; there is no separate "rest" part because it has no mass. We meet $E$ in the photon example (Cell H).
$c = 3 \times 1 0^{8} m/s$ — the speed of light in a vacuum, the fixed top speed of the universe. It appears whenever light or relativity enters (Cells H and I).
$h = 6.626 \times 1 0^{- 34} J\cdots$ — Planck's constant, the fixed number that scales the whole quantum world.
$q$ or $e$ — electric charge; for an electron $e = 1.6 \times 1 0^{- 19} C$ (coulombs).
$V$ — accelerating voltage in volts.
$γ$ (Greek "gamma") — the relativistic factor $γ = 1/ 1 - v^{2} / c^{2}$ , always $\geq 1$ ; it measures how much fast motion (near $c$ ) inflates momentum. Built in Cell I.
$1 \overset{˚}{A}$ ("angstrom") $= 1 0^{- 10} m$ — a handy tiny ruler, roughly the size of an atom.

We will use exactly four "faces" of the same one formula. Which face you pick depends on what the problem hands you.

The scenario matrix

Every de Broglie problem falls into one of these cells. The rest of the page fills each one.

Cell	What's given	Which face	Watch out for	Example
A	mass + speed, small particle	$h / m v$	keep SI units	Ex 1
B	mass + speed, macroscopic object	$h / m v$	answer absurdly tiny — that's correct	Ex 2
C	electron + voltage $V$	$12.27/ V$ Å	this shortcut is electrons only	Ex 3
D	kinetic energy $K$ (any particle)	$h / 2 m K$	never use $h / m v$ if you only know $K$	Ex 4
E	temperature $T$ (thermal particle)	$K = \frac{3}{2} k_{B} T$ then face D	$K$ is average, not exact	Ex 5
F	ratio / comparison, same $K$	$λ \propto 1/ m$	which quantity is held fixed?	Ex 6
G	ratio / comparison, same $v$	$λ \propto 1/ m$	different from cell F!	Ex 7
H	photon (massless) — degenerate case	$λ = h / p$ , $p = E / c$	$h / m v$ is forbidden ( $m = 0$ )	Ex 8
I	relativistic (speed near light) — limiting case	$p = γ m v$	$h / m v$ fails, must correct	Ex 9
J	exam twist: $λ$ given, find $V$ or $K$	invert a face	algebra runs backward	Ex 10

The two "danger cells" are H (mass is zero, so the $h / m v$ face literally divides by zero) and I (speed so high that $p \neq = m v$ ). We will look those two straight in the eye.

Cell A — small particle, speed given

Worked example Ex 1 · An electron moving at

$2 \times 1 0^{6} m/s$ Find its de Broglie wavelength. Electron mass $m_{e} = 9.11 \times 1 0^{- 31} kg$ .

Forecast: electrons are famously "wavy." Guess: will $λ$ come out around an atom's size ( $\sim 1 \overset{˚}{A} = 1 0^{- 10} m$ ), or vanishingly small? Hold your guess.

Compute momentum $p = m_{e} v = (9.11 \times 1 0^{- 31}) (2 \times 1 0^{6}) = 1.822 \times 1 0^{- 24} kg\cdotm/s$ . Why this step? We know mass and speed, and the speed is small compared to light, so cell A's face $h / m v$ applies — and its heart is $p = m v$ .
Divide $λ = \frac{h}{p} = \frac{6.626 \times 1 0 ^{- 34}}{1.822 \times 1 0 ^{- 24}} = 3.64 \times 1 0^{- 10} m$ . Why this step? This is $λ = h / p$ itself — the master face.
In angstroms: $3.64 \times 1 0^{- 10} m = 3.64 \overset{˚}{A}$ .

Verify: units check — $\frac{J\cdots}{kg\cdotm/s} = \frac{kg\cdotm ^{2} / s ^{2} \cdot s}{kg\cdotm/s} = m$ . ✓ And $3.6 \overset{˚}{A}$ is a few atom-widths — exactly the "wavy on the atomic scale" our forecast hoped for. ✓

Cell B — macroscopic object

Cell C — electron accelerated through a voltage

Cell D — kinetic energy given (any particle)

Worked example Ex 4 · A proton with

$K = 1.0 MeV$ Find $λ$ . Proton mass $m_{p} = 1.673 \times 1 0^{- 27} kg$ ; $1 MeV = 1.6 \times 1 0^{- 13} J$ .

Forecast: protons are ~1836× heavier than electrons, so at a given energy they carry more momentum and should be less wavy than an electron. Expect $λ$ well below $1 \overset{˚}{A}$ .

Convert energy $K = 1.0 MeV = 1.6 \times 1 0^{- 13} J$ . Why this step? The formula demands joules; MeV is just a nickname.
Momentum $p = 2 m_{p} K = 2 (1.673 \times 1 0^{- 27}) (1.6 \times 1 0^{- 13})$ . Inside: $2 (1.673 \times 1 0^{- 27}) (1.6 \times 1 0^{- 13}) = 5.35 \times 1 0^{- 40}$ , so $p = 5.35 \times 1 0^{- 40} = 2.31 \times 1 0^{- 20} kg\cdotm/s$ . Why this step? Cell D's face — energy in, momentum out, no detour through $v$ .
Wavelength $λ = \frac{6.626 \times 1 0 ^{- 34}}{2.31 \times 1 0 ^{- 20}} = 2.87 \times 1 0^{- 14} m$ .

Verify: $2.87 \times 1 0^{- 14} m = 2.87 \times 1 0^{- 4} \overset{˚}{A}$ — thousands of times smaller than an atom, as forecast for a heavy, energetic proton. ✓ Is $1 MeV ≪ m_{p} c^{2} \approx 938 MeV$ ? Yes, so the non-relativistic $2 m K$ is valid here (contrast cell I). ✓

Cell E — thermal particle (temperature given)

Worked example Ex 5 · A thermal neutron at

$T = 300 K$ (room temperature) Find $λ$ . Neutron mass $m_{n} = 1.675 \times 1 0^{- 27} kg$ .

Forecast: neutron diffraction is a real lab technique, which only works if $λ \approx$ crystal spacing $\approx 1 \overset{˚}{A}$ . So we expect an answer near $1 \overset{˚}{A}$ .

Average KE $K = \frac{3}{2} k_{B} T = 1.5 (1.38 \times 1 0^{- 23}) (300) = 6.21 \times 1 0^{- 21} J$ . Why this step? Convert the temperature into the kinetic energy the formula wants.
Momentum $p = 2 m_{n} K = 2 (1.675 \times 1 0^{- 27}) (6.21 \times 1 0^{- 21})$ . Inside: $2.08 \times 1 0^{- 47}$ , so $p = 4.56 \times 1 0^{- 24} kg\cdotm/s$ . Why this step? Now it's exactly cell D.
Wavelength $λ = \frac{6.626 \times 1 0 ^{- 34}}{4.56 \times 1 0 ^{- 24}} = 1.45 \times 1 0^{- 10} m = 1.45 \overset{˚}{A}$ .

Verify: $\approx 1.5 \overset{˚}{A}$ — right at atomic-spacing scale, exactly why neutron diffraction is done at room temperature. ✓ Matches parent note's $\approx 1.4 \overset{˚}{A}$ (rounding). ✓

Cells F & G — comparisons (this is where students mix things up)

The single most common exam trap: is the energy the same, or the speed the same? The proportionality flips.

Figure — De Broglie hypothesis — matter waves λ = h - p

Worked example Figure — reading the two panels

The blueprint figure above draws the actual wave-ripples of an electron (cyan) and a proton (amber), stacked so you can compare crest-to-crest spacing directly.

Left panel ("Same kinetic energy $K$ "): both particles carry equal energy. The electron's cyan wave is drawn with a clearly longer crest-to-crest distance than the proton's amber wave. The double-headed white arrows mark one full wavelength of each — the electron's arrow is far wider. This visualizes $λ \propto 1/ m$ : since $m_{p} / m_{e} = 1836 \approx 42.8$ , the electron's ripple is about 43× longer.
Right panel ("Same speed $v$ "): now the speed is locked instead. The proton's amber wave is squeezed even tighter than in the left panel, because here $λ \propto 1/ m$ (no square root), giving a factor of $1836$ . The electron's cyan wave is unchanged, so the gap between the two waves is dramatically bigger. The takeaway you should read off the picture: the electron is wavier in both cases, but how much wavier depends entirely on what is held fixed — the right panel's gap is far larger than the left's. That difference is the whole lesson of Cells F and G.

Cell H — the degenerate case: a massless photon

Worked example Ex 8 · A photon of energy

$E = 2.0 eV$ (visible light) Find its wavelength. $1 eV = 1.6 \times 1 0^{- 19} J$ ; $c = 3 \times 1 0^{8} m/s$ .

Forecast: $2 eV$ is roughly red-orange light, so expect $λ$ around $600 nm = 6000 \overset{˚}{A}$ — enormously bigger than any matter wave above, because photons carry so little momentum.

Energy in joules $E = 2.0 (1.6 \times 1 0^{- 19}) = 3.2 \times 1 0^{- 19} J$ .
Photon momentum $p = \frac{E}{c} = \frac{3.2 \times 1 0 ^{- 19}}{3 \times 1 0 ^{8}} = 1.067 \times 1 0^{- 27} kg\cdotm/s$ . Why this step? $m = 0$ blocks the mass face; $p = E / c$ is the photon's proper momentum.
Wavelength $λ = \frac{h}{p} = \frac{6.626 \times 1 0 ^{- 34}}{1.067 \times 1 0 ^{- 27}} = 6.21 \times 1 0^{- 7} m = 621 nm$ .

Verify: $621 nm$ is orange visible light — exactly what $2 eV$ photons are. ✓ And it dwarfs every matter wavelength above, because tiny $p$ ⇒ long $λ$ . The mass face would have crashed; the momentum face sailed through. ✓

Cell I — the limiting case: relativistic speed

Worked example Ex 9 · An electron at

$v = 0.9 c$ Find $λ$ , and compare with the wrong (naive) answer. Electron mass $m_{e} = 9.11 \times 1 0^{- 31} kg$ .

Forecast: the naive $h / m v$ underestimates $p$ (it drops the $γ$ stretch), so it overestimates $λ$ . The correct $λ$ should therefore be shorter than the naive one — by roughly the factor $γ$ .

Gamma factor $γ = \frac{1}{1 - 0. 9 ^{2}} = \frac{1}{1 - 0.81} = \frac{1}{0.19} = 2.294$ . Why this step? At $0.9 c$ we are firmly relativistic; $γ$ measures how big the correction is before we touch momentum.
Speed in SI $v = 0.9 c = 0.9 \times 3 \times 1 0^{8} = 2.7 \times 1 0^{8} m/s$ . Why this step? The formula needs metres per second, not " $0.9 c$ ."
Correct momentum $p = γ m_{e} v = (2.294) (9.11 \times 1 0^{- 31}) (2.7 \times 1 0^{8}) = 5.64 \times 1 0^{- 22} kg\cdotm/s$ . Why this step? Cell I's face — insert the $γ$ stretch that the naive formula forgets.
Correct wavelength $λ = \frac{h}{p} = \frac{6.626 \times 1 0 ^{- 34}}{5.64 \times 1 0 ^{- 22}} = 1.17 \times 1 0^{- 12} m$ .
Naive (wrong) momentum $p_{naive} = m_{e} v = (9.11 \times 1 0^{- 31}) (2.7 \times 1 0^{8}) = 2.46 \times 1 0^{- 22}$ , giving $λ_{naive} = \frac{6.626 \times 1 0 ^{- 34}}{2.46 \times 1 0 ^{- 22}} = 2.69 \times 1 0^{- 12} m$ . Why this step? We compute the mistake on purpose, to see the size of the error.

Verify: the naive answer ( $2.69 pm$ ) is exactly $γ = 2.294$ times too big compared to the correct $1.17 pm$ : $2.69/1.17 = 2.30 \approx γ$ . ✓ Ignoring relativity would have doubled your wavelength — always check $K$ vs $m c^{2}$ first. ✓

Cell J — the inverse exam twist

Worked example Ex 10 · You measure an electron's wavelength as

$λ = 0.5 \overset{˚}{A}$ . Through what voltage $V$ was it accelerated?

Forecast: a short wavelength ( $0.5 \overset{˚}{A}$ is shorter than the $1.23 \overset{˚}{A}$ at $100 V$ ) means high momentum means high voltage. Expect $V$ well above $100 V$ .

Start from the electron face $λ = \frac{12.27}{V} \overset{˚}{A}$ and solve for $V$ : rearrange to $V = \frac{12.27}{λ}$ , then square: $V = (\frac{12.27}{λ})^{2}$ with $λ$ in Å. Why this step? The forward formula runs $V \to λ$ ; we simply reverse the algebra so $λ \to V$ .
Plug in $V = (\frac{12.27}{0.5})^{2} = (24.54)^{2} = 602 V$ . Why this step? Substitute the measured $λ$ into the inverted formula.

Verify: $602 V > 100 V$ , as the forecast demanded. And check it forward: $λ = 12.27/ 602 = 12.27/24.5 = 0.50 \overset{˚}{A}$ . ✓ The round trip closes — always plug your answer back into the original formula. ✓

Recall Which face do I use? (self-test)

Given only a temperature, which face and what's the first step? ::: Face D ( $h / 2 m K$ ); first compute $K = \frac{3}{2} k_{B} T$ . Why can't a photon use $λ = h / m v$ ? ::: Its mass is zero, so $h / (m v) = h /0$ is undefined; use $λ = h / p$ with $p = E / c$ . At equal kinetic energy, $λ \propto$ ? ::: $1/ m$ (so electron/proton ratio is $1836 \approx 42.8$ ). At equal speed, $λ \propto$ ? ::: $1/ m$ (ratio $1836$ , not $1836$ ). When does $p = m v$ fail? ::: Near light speed; then $p = γ m v$ with $γ = 1/ 1 - v^{2} / c^{2}$ . Given $λ$ , how do you find the electron voltage? ::: Invert the face: $V = (12.27/ λ)^{2}$ with $λ$ in Å.

Connections

Parent topic — the theory these examples exercise.
Davisson–Germer experiment — the $54 V$ electron of Ex 3 is exactly its setup.
Photoelectric effect — origin of the photon momentum $p = E / c$ used in Ex 8.
Compton effect — another test of photon momentum.
Heisenberg Uncertainty Principle — why these tiny wavelengths matter.
Bohr model — standing matter waves quantize orbits.
Wave–particle duality — the umbrella idea behind all ten cells.