De Broglie hypothesis — matter waves λ = h - p: Question bank

Intuition How to use this bank

These are the questions that catch people who "know the formula" but haven't felt the idea. Every answer below is a short chain of reasoning — read the reveal, then re-derive it in your head. If a claim feels obviously true, that's usually the trap. Master note: parent topic.

Key symbols you must already own before reading:

$λ$ = wavelength (crest-to-crest distance of the associated wave).
$v$ = speed of the particle (metres per second).
$m$ = mass of the particle; for the electron specifically we write $m_{e}$ (electron rest mass, $9.11 \times 1 0^{- 31} kg$ ).
$p$ = momentum ("push"; $p = m v$ for slow bodies).
$h$ = Planck's constant ( $6.626 \times 1 0^{- 34} J\cdots$ ).
$K$ = kinetic energy.
$q$ = electric charge of the particle; the elementary charge is $e = 1.6 \times 1 0^{- 19} C$ .
$V$ = accelerating voltage (in volts); a charge $q$ falling through $V$ gains $K = q V$ .
$c$ = speed of light ( $3.0 \times 1 0^{8} m/s$ ).
$k_{B}$ = Boltzmann's constant ( $1.38 \times 1 0^{- 23} J/K$ ); $T$ = absolute temperature (in kelvin).

If any of these are fuzzy, go back to the parent note first.

True or false — justify

Each line: a statement, then ::: the verdict with the reason. Cover the reason, guess T/F and why, then check.

A heavier particle at the same speed has a shorter matter wavelength.

True.

λ = h / (m v)

; at fixed

v

, larger

m

means larger

p

, so smaller

λ

. Mass sits in the denominator.

A heavier particle at the same kinetic energy also has a shorter wavelength.

True, but for a different reason:

λ = h / 2 m K

, so at fixed

K

larger

m

still gives larger

p = 2 m K

, hence smaller

λ

. Both "same speed" and "same

K

" favour the lighter particle being wavier.

A photon has no rest mass, so the de Broglie relation cannot apply to it.

False. The master form is

λ = h / p

, which never mentions mass. A photon has finite momentum

p = E / c

, so

λ = h / p

works perfectly — it was literally the seed of the whole idea.

If a particle is at rest, its de Broglie wavelength is zero.

False. At rest

p = 0

, so

λ = h / p \to \infty

— the wavelength is infinite, not zero. A "wave" with no momentum has no spatial oscillation at all, which shows the picture breaks down for a truly stationary particle.

The matter wave is the particle spread out into a smeared blob of stuff.

False. It is a probability amplitude (Born): the particle is still detected as one localised dot. The wave tells you where the dot is likely to land, not that matter itself is jelly. See Wave–particle duality.

Doubling an electron's speed doubles its wavelength.

False — it halves it.

λ \propto 1/ p \propto 1/ v

(non-relativistic), so more speed means shorter wave. People confuse "more of the motion quantity" with "more wavelength".

A cricket ball has no de Broglie wavelength.

False. It has one — about

1 0^{- 34}

m. It's just so absurdly smaller than the ball or any slit that no diffraction is ever observable, so the ball looks purely classical.

λ = h / (m v)

is valid at any speed, including near light-speed.

False.

p = m v

only holds non-relativistically. Near

c

p = γ m v

, so you must use

λ = h / p

with the correct relativistic

p

. Starting from the mass form here gives a wrong (too-long) wavelength.

Two particles with the same wavelength must have the same momentum.

True.

λ = h / p

is a one-to-one link:

p = h / λ

. Same

λ \Rightarrow

same

p

— regardless of mass, charge, or what the particle is.

λ = h / p

was rigorously derived for matter from Newton's laws.

False. For matter it is a hypothesis (postulated by symmetry with light), not a derivation. Its truth was decided by experiment — the Davisson–Germer experiment saw electron diffraction in 1927.

Spot the error

Each line states a flawed reasoning; the reveal names the flaw and fixes it.

"An electron and a proton accelerated through the same voltage have the same wavelength, since they get the same energy."

Error: same

K = q V

but different mass.

λ = h / 2 m K

, so the lighter electron has the longer wave. Equal energy

\neq =

equal wavelength.

"To make the electron's wave shorter, slow it down."

Error: backwards. Shorter

λ

needs larger

p

, i.e. faster.

λ \propto 1/ v

, so slowing down lengthens the wave.

"For a

100 V

electron I'll use

λ = 12.27/ V = 0.123 \overset{˚}{A}

Error: the formula is

λ = 12.27/ V

, not

12.27/ V

. Correct is

12.27/ 100 = 1.227 \overset{˚}{A}

. The square root comes from

p = 2 m_{e} e V

"A neutron is neutral, so we can't use

λ = h / 2 m q V

or any de Broglie relation for it."

Category error: the voltage form assumes a charged particle (

q \neq = 0

) accelerated by an electric field, so it genuinely cannot apply to a neutral neutron. But that is a limit of one special-case formula, not of de Broglie itself —

λ = h / p

and

λ = h / 2 m K

apply fine, and thermal neutrons diffract off crystals at

\sim 1 \overset{˚}{A}

"Since

λ = h / p

and

h

is tiny, all matter wavelengths are undetectably small."

Error:

h

tiny is only half the story —

p

can be tiny too. For a slow electron

p

is small enough that

λ \sim

atomic size. It's the ratio

h / p

that matters, not

h

alone.

λ = h / p

means as

p \to 0

the wavelength vanishes."

Error:

p

in the denominator means

λ \to \infty

p \to 0

, not

\to 0

. Low momentum

=

long wave.

"Photons and electrons both obey

λ = h / p

, so a photon and electron of the same wavelength have the same energy."

Error: same

λ \Rightarrow

same

p

, but energy relates differently. Photon:

E = p c

. Electron (slow):

K = p^{2} /2 m

. Same

p

gives very different energies.

Why questions

Deeper "why", one or two sentences of real reasoning.

Why does wave behaviour disappear for everyday objects?

Their momentum

p = m v

is enormous, so

λ = h / p \sim 1 0^{- 34} m

— smaller than any obstacle or slit. Diffraction only shows when

λ

is comparable to the aperture size, which never happens for macroscopic bodies.

Why did de Broglie use the photon's

λ = h / p

as his springboard rather than any matter formula?

Because

λ = h / p

contains no property unique to light — no

c

, no "massless". If the relation carries nothing light-specific, symmetry suggests it should hold for all matter too.

Why do we prefer the momentum form

λ = h / p

over the mass form

λ = h / (m v)

The momentum form is universal — it works for massless photons (

p = E / c

) and for relativistic particles (

p = γ m v

). The mass form is only the slow-speed special case and breaks for both.

Why is

λ = h / 2 m K

safer than

λ = h / (m v)

in most exam problems?

Problems usually give energy or voltage, not speed.

p = 2 m K

gets momentum directly from

K

, avoiding an extra step and the temptation to misuse

p = m v

at high speed.

Why can electrons be used to image atoms but visible light cannot resolve them?

100 V

electron has

λ \approx 1 \overset{˚}{A}

, matching atomic spacing, so it diffracts off the lattice; visible light has

λ \sim 5000 \overset{˚}{A}

, far too coarse to resolve atomic-scale detail.

Why does the de Broglie idea naturally lead to the uncertainty principle?

A wave with a definite wavelength (

\Rightarrow

definite

p

) is spread over all space (

\Rightarrow

undefined position). Localising the particle needs a mix of wavelengths, blurring

p

. That trade-off is the Heisenberg Uncertainty Principle.

Why does the Bohr orbit condition

2 π r = nλ

make sense as a matter wave?

An electron's wave must join up smoothly around the orbit; only whole numbers of wavelengths fit without destructive cancellation, giving allowed (quantised) orbits. See Bohr model.

Edge cases

Boundary and degenerate scenarios the formula invites.

A particle at exactly zero momentum: what is

λ

λ = h /0 \to \infty

. Physically the wave picture degenerates — an infinitely long wave has no localisation, consistent with a perfectly stationary (hence position-uncertain) particle.

Relativistic electron with

K

comparable to

m_{e} c^{2}

: which formula for

λ

Not

λ = h / (m v)

nor

h / 2 m K

. Use

λ = h / p

with

p = \frac{1}{c} K^{2} + 2 K m_{e} c^{2}

. This comes straight from the relativistic identity

E^{2} = (p c)^{2} + (m_{e} c^{2})^{2}

: write total energy as

E = K + m_{e} c^{2}

, substitute, expand

(K + m_{e} c^{2})^{2} = K^{2} + 2 K m_{e} c^{2} + (m_{e} c^{2})^{2}

, cancel the

(m_{e} c^{2})^{2}

terms, and solve for

p

Neutral particle (neutron) — can the voltage formula

λ = 12.27/ V

ever apply?

No, twice over: that formula is electron-specific (it bakes in

m_{e}

and

e

), and a neutral particle has

q = 0

so it can't be electrically accelerated at all. Use thermal energy

K = \frac{3}{2} k_{B} T

instead.

Two identical electrons, one moving twice as fast — compare wavelengths.

λ \propto 1/ v

, so the faster one has half the wavelength. Same particle, different momentum, different wave.

As voltage

V \to \infty

for an accelerated electron, what happens to

λ

(ignoring relativity)?

λ = 12.27/ V \to 0

. Higher energy means shorter wave — but before

λ

truly vanishes, relativistic

p

takes over and the simple formula fails.

Photon vs electron carrying equal energy

E

: which has the longer wavelength?

The photon. Photon:

p = E / c

. Slow electron:

p = 2 m E

, which is much larger for the same

E

. Larger

p \Rightarrow

shorter

λ

, so the electron is less wavy at equal energy.

Recall One-line summary of every trap here

Almost every mistake comes from forgetting that $λ = h / p$ is about momentum in the denominator: more push $\Rightarrow$ shorter wave, zero push $\Rightarrow$ infinite wave, and mass only matters through $p$ .

Connections

Wave–particle duality — the umbrella idea behind all these traps.
Davisson–Germer experiment — why the hypothesis is believed.
Heisenberg Uncertainty Principle — grows directly from the wave nature.
Bohr model — standing matter waves fixing orbits.
Photoelectric effect — where photon momentum $p = E / c$ comes from.
Compton effect — independent proof of photon momentum.