Dual nature of matter — de Broglie λ = h - p: Worked examples

Before anything, three constants we will reuse. Each is just a number with units — memorise the units, not the digits.

The scenario matrix

Every de Broglie problem is one of these cells. The Formula to reach for column tells you which face of the master result to use.

#	Case class	What you're given	Formula to reach for	Example
A	Direct: speed known	$m$ , $v$	$λ = \frac{h}{m v}$	Ex 1
B	Kinetic energy known	$m$ , $K E$	$λ = \frac{h}{2 m K E}$	Ex 2
C	Accelerating voltage	charge $q$ , $V$	$λ = \frac{h}{2 m q V}$	Ex 3
D	Photon (limiting case, $m_{0} = 0$ )	wavelength or energy	$λ = \frac{h}{p} = \frac{h c}{E}$	Ex 4
E	Degenerate: macroscopic / slow	big $m$ , ordinary $v$	$λ = \frac{h}{m v}$ (see it vanish)	Ex 5
F	Unit trap (eV, kV, grams)	mixed units	convert first, then B or C	Ex 6
G	Comparison / ratio twist	two particles' $m$ , $K E$	take ratio $\frac{λ _{1}}{λ _{2}}$	Ex 7
H	Real-world word problem	thermal / everyday	pick the right cell	Ex 8

We cover A–H below, each labelled with its cell.

Figure — Dual nature of matter — de Broglie λ = h - p

The picture above is the whole page in one glance: as momentum $p$ climbs (heavier or faster), the wavelength $λ$ falls along that curve. Every example is just a dot on this curve. Watch where each lands.

Worked examples

Worked example Ex 2 — Cell B (kinetic energy known)

An electron has kinetic energy $K E = 50 eV$ . Find its de Broglie wavelength.

Forecast: $50 eV$ is a modest energy. Will $λ$ land near the atomic (~0.1 nm) scale? Guess yes.

Convert energy to joules: $50 eV \times 1.602 \times 1 0^{- 19} = 8.01 \times 1 0^{- 18} J$ . Why this step? The formula uses SI joules; eV is a disguise for joules.
We don't know $v$ directly, so use the KE form. Why this form? Because $p = 2 m K E$ lets us skip solving for $v$ . $λ = \frac{h}{2 m _{e} K E}$
Compute the denominator: $2 (9.11 \times 1 0^{- 31}) (8.01 \times 1 0^{- 18}) = 1.459 \times 1 0^{- 47} = 3.82 \times 1 0^{- 24}$ . Why this step? This is $p$ in kg m/s.
Divide: $λ = \frac{6.626 \times 1 0 ^{- 34}}{3.82 \times 1 0 ^{- 24}} = 1.73 \times 1 0^{- 10} m = 0.173 nm$ .

Verify: Cross-check with the voltage shortcut. $50 eV$ means an electron dropped through $50 V$ , so $λ \approx \frac{1.226}{50} = 0.173 nm$ ✓ — the two roads meet.

Worked example Ex 4 — Cell D (photon: the limiting case)

Green light has wavelength $500 nm$ . (a) What is the photon's momentum? (b) Confirm $λ = h / p$ still holds.

Forecast: A photon is the origin of the formula. So $λ = h / p$ must hold exactly — it's not an approximation for light.

For a photon, $p = \frac{h}{λ}$ — just the master formula rearranged. Why this step? Photons are the one case where the formula was proved, not postulated. $p = \frac{6.626 \times 1 0 ^{- 34}}{500 \times 1 0 ^{- 9}} = 1.325 \times 1 0^{- 27} kg m/s$
Now feed that $p$ back: $λ = h / p = 500 nm$ . Why this step? Sanity loop — going forward and backward must return the input.

Verify: Photon energy $E = p c = (1.325 \times 1 0^{- 27}) (3.0 \times 1 0^{8}) = 3.98 \times 1 0^{- 19} J = 2.48 eV$ , the familiar energy of green light ✓. Note: here we used $c$ , not $v$ — because a photon is the massless particle that travels at $c$ . This is the exact case where the "use $v$ not $c$ " mistake does not apply.

Worked example Ex 6 — Cell F (unit trap)

A $2.0 g$ bullet moves at $500 m/s$ , and separately an electron has $K E = 1.5 keV$ . Two traps at once. Find both wavelengths.

Forecast: Both traps are about not plugging in raw numbers. Guess: bullet gives a Cell-E-tiny $λ$ ; electron gives a sub-atomic $λ$ .

Bullet: convert grams to kilograms first — $2.0 g = 2.0 \times 1 0^{- 3} kg$ . Why this step? SI demands kilograms; grams would inflate $λ$ by 1000×. $p = (2.0 \times 1 0^{- 3}) (500) = 1.0 kg m/s, λ = \frac{6.626 \times 1 0 ^{- 34}}{1.0} = 6.63 \times 1 0^{- 34} m$
Electron: convert $1.5 keV = 1500 eV = 1500 \times 1.602 \times 1 0^{- 19} = 2.403 \times 1 0^{- 16} J$ . Why this step? "k" means ×1000, then eV→J. Two conversions stacked — the classic trap.
$λ = \frac{h}{2 m _{e} K E} = \frac{6.626 \times 1 0 ^{- 34}}{2 ( 9.11 \times 1 0 ^{- 31} ) ( 2.403 \times 1 0 ^{- 16} )}$ . Denominator $= 4.378 \times 1 0^{- 46} = 2.092 \times 1 0^{- 23}$ , so $λ = 3.17 \times 1 0^{- 11} m = 0.0317 nm$ .

Verify: Voltage check for the electron: $1500 V$ ⇒ $λ \approx \frac{1.226}{1500} = 0.0317 nm$ ✓. Bullet $λ$ is macroscopic-tiny as forecast.

Worked example Ex 7 — Cell G (comparison / ratio twist)

A proton and an electron are accelerated through the same voltage $V$ . Which has the longer wavelength, and by what factor? ( $m_{p} / m_{e} = 1836$ .)

Forecast: Same charge magnitude, same $V$ , so same $K E$ . Lighter particle ⇒ smaller $p$ ⇒ longer $λ$ . Guess the electron wins.

Both gain $K E = q V$ with the same $q$ (charge magnitude $e$ ) and same $V$ . Why this step? Equal $K E$ is the whole trick — it makes the ratio clean.
$λ = \frac{h}{2 m K E}$ , so with $K E$ equal, $λ \propto \frac{1}{m}$ . Why this step? Everything except $m$ cancels in the ratio. $\frac{λ _{e}}{λ _{p}} = \frac{m _{p}}{m _{e}} = 1836 = 42.8$

Verify: The electron's wavelength is ~ $42.8 \times$ longer than the proton's ✓. Sanity: lighter ⇒ longer, matches the forecast and the parent's "mass in the denominator" rule. Notice the exam-style beauty: no numbers for $h$ , $V$ , or $q$ were needed — the ratio killed them all.

Worked example Ex 8 — Cell H (real-world word problem)

"Thermal" neutrons at room temperature ( $T = 300 K$ ) are used in materials research. Their average kinetic energy is $K E = \frac{3}{2} k_{B} T$ with $k_{B} = 1.38 \times 1 0^{- 23} J/K$ . Neutron mass $m_{n} = 1.675 \times 1 0^{- 27} kg$ . Find $λ$ — and explain why neutrons "see" crystals.

Forecast: Neutrons are heavy but very slow (thermal). Guess $λ$ lands near atomic spacing (~0.1 nm), which is why they're used for crystal studies.

Kinetic energy: $K E = \frac{3}{2} (1.38 \times 1 0^{- 23}) (300) = 6.21 \times 1 0^{- 21} J$ . Why this step? The word "thermal at temperature $T$ " is code for this energy — recognise the phrase.
Use Cell B (we have $m$ and $K E$ , not $v$ ): $λ = \frac{h}{2 m _{n} K E} = \frac{6.626 \times 1 0 ^{- 34}}{2 ( 1.675 \times 1 0 ^{- 27} ) ( 6.21 \times 1 0 ^{- 21} )}$ Denominator $= 2.080 \times 1 0^{- 47} = 4.560 \times 1 0^{- 24}$ .
$λ = \frac{6.626 \times 1 0 ^{- 34}}{4.560 \times 1 0 ^{- 24}} = 1.45 \times 1 0^{- 10} m = 0.145 nm$ .

Verify: Crystal atomic spacings are ~ $0.1$ – $0.3$ nm. Our $λ = 0.145 nm$ sits right in that window ✓ — so thermal neutrons diffract off crystals just like the electrons in the Davisson–Germer Experiment. Real, useful, forecast confirmed.

Recall Which formula for which cell?

Given speed ::: use $λ = h / m v$ (Cell A). Given kinetic energy ::: use $λ = h / 2 m K E$ (Cell B). Given accelerating voltage $V$ for a charge $q$ ::: use $λ = h / 2 m q V$ , or $1.226/ V$ nm for electrons (Cell C). A photon ::: use $λ = h / p = h c / E$ and speed $c$ , never $v$ (Cell D). Two particles, same voltage ::: $λ \propto 1/ m$ , take the ratio (Cell G). "Thermal at temperature $T$ " ::: means $K E = \frac{3}{2} k_{B} T$ , then Cell B (Cell H).

Connections

Parent topic — the derivation these examples exercise.
Davisson–Germer Experiment — Ex 2/3/8 wavelengths are exactly the atomic scale that made electron/neutron diffraction possible.
Photoelectric Effect — Ex 4's photon energy $E = p c$ links back to the light-as-particle result.
Heisenberg Uncertainty Principle — why the tiny $λ$ of Ex 5 keeps macroscopic objects "classical".
Schrödinger Wave Equation — every $λ$ here becomes a wavefunction wiggle in $ψ$ .
Bohr Model of Atom — electron wavelengths from Cell C set the standing-wave orbit condition $2 π r = nλ$ .