Dual nature of matter — de Broglie λ = h - p: Question bank

The one rule everything here tests: $λ = h / p = h / m v$ — Planck's constant on top, momentum below. Wave nature is loud only when $λ$ is comparable to the size of whatever the particle meets.

True or false — justify

Every moving particle, even a resting-then-pushed dust speck, has a de Broglie wavelength.

True — the instant it has non-zero momentum

p = m v

it has

λ = h / p

. Only a truly stationary object (

v = 0

) has no finite wavelength (

λ \to \infty

A photon has zero rest mass, so it cannot have a de Broglie wavelength.

False — the relation needs momentum, not rest mass. A photon carries

p = E / c

, so

λ = h / p = c / ν

, its ordinary wavelength.

Doubling a particle's speed doubles its de Broglie wavelength.

False — speed sits in the denominator, so doubling

v

halves

λ

. Faster ⇒ shorter wave.

Two particles with the same kinetic energy always have the same de Broglie wavelength.

False —

λ = h / 2 m K E

also depends on mass, so a heavier particle at equal

K E

has a shorter wavelength.

A proton and an electron accelerated through the same voltage get the same wavelength.

False — both gain the same

K E = q V

(same

q

), but

λ = h / 2 m q V

; the heavier proton has the much shorter wavelength.

The matter wave physically wiggles the particle up and down like a vibrating string.

False — it is a probability/guiding wave; its amplitude squared tells you where the particle is likely to be, not a mechanical vibration of the particle itself.

A cricket ball has no de Broglie wavelength at all.

False — it has one (~

1 0^{- 34}

m), it is just so absurdly small that no slit or crystal can diffract it, so we never observe waviness.

If a particle slows to a stop its de Broglie wavelength shrinks to zero.

False — as

v \to 0

p \to 0

, so

λ = h / p \to \infty

(grows without bound). Slow means long, not short, wavelength.

Spot the error

"For an electron I'll use

λ = h / m c

since

c

appeared in the derivation."

Wrong:

c

belonged to the photon (which alone travels at light speed). For matter use the particle's own speed

v

λ = h / m v

"The electron gains

50

eV, so I plug

K E = 50

straight into

λ = h / 2 m K E

Error: eV must be converted to joules first (

\times 1.6 \times 1 0^{- 19}

). Mixing eV with SI mass and

h

gives nonsense units.

"Heavier objects carry more matter, so they must have longer matter waves."

Error: mass is in the denominator of

λ = h / m v

. More mass ⇒ shorter wavelength — which is precisely why big objects show no wave behaviour.

"Since

λ = h / p

was derived using

E = h ν

, matter waves must travel at the speed of light."

Error: only the derivation's analogy borrowed a photon. A matter wave is associated with a slow particle; nothing forces it to move at

c

"An electron microscope beats a light microscope because electrons are smaller than photons."

Error: size isn't the reason. It's wavelength — an accelerated electron's

λ \sim 0.1

nm is far shorter than visible light's

\sim 500

nm, and shorter

λ

means finer resolution.

E = p c

can be used to find the kinetic energy of a slow electron."

Error:

E = p c

holds only for massless particles (photons). For matter use

K E = \frac{1}{2} m v^{2}

or the full

E^{2} = (p c)^{2} + (m_{0} c^{2})^{2}

Edge cases

What is the de Broglie wavelength of a genuinely stationary particle (

v = 0

It is undefined/infinite:

p = 0

makes

λ = h / p

blow up, meaning no localised wave — the concept only applies to moving particles.

What happens to

λ

as a particle's speed approaches the speed of light?

Momentum

p

grows relativistically without bound, so

λ \to 0

— the wave becomes vanishingly short and the particle looks purely particle-like.

For a neutral particle like a neutron, does the accelerating-voltage formula apply?

No —

λ = h / 2 m q V

needs a charge

q

to gain energy from a field. A neutron has

q = 0

, so you must know its

K E

or speed directly.

Compare the wavelength of a slow electron and a fast electron of equal mass.

The slow one has the longer wavelength, since

λ \propto 1/ v

; slower momentum means a bigger, more spread-out wave.

At the same speed, which is wavier — an electron or a proton?

The electron, because it is ~1800× lighter; smaller

m

gives larger

λ

, so its wave behaviour is far easier to observe.

Is there any object too heavy to have a matter wave?

No object is exempt — every moving mass has one; heavy objects simply have wavelengths so tiny they can never be measured, so "no wave" really means "immeasurably small wave."

Recall Quick self-test before you move on

Cover these and answer aloud. If I halve the mass but keep speed fixed, $λ$ does what? ::: Doubles — mass is in the denominator. Why can't a football diffract through a doorway? ::: Its $λ \sim 1 0^{- 34}$ m is unimaginably smaller than the doorway, so no bending occurs. Photon momentum with zero rest mass — contradiction? ::: No: momentum comes from energy via $p = E / c$ , not from rest mass.

Connections

Planck's Quantum Theory — supplies $E = h ν$ , one leg of the derivation these traps rely on.
Photoelectric Effect — the particle-nature-of-light mirror image of matter waves.
Heisenberg Uncertainty Principle — the direct consequence tested in the "why" section.
Schrödinger Wave Equation — where the probability wave becomes $ψ$ .
Davisson–Germer Experiment — the experiment answering "why crystals?".
Bohr Model of Atom — standing-wave orbits, $2 π r = nλ$ .

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