Questionnaire 1.1.5 — Matter Waves

Four questions on the wave nature of matter: why "size" comparisons between quantum objects are ill-posed, Bohr's angular-momentum quantization, the de Broglie relation, and judging when an electron can be treated non-relativistically.

Key Ideas

The de Broglie wavelength assigns a wavelength to every particle of momentum $p$ :

\lambda = \frac{h}{p} = \frac{h}{mv} \quad (\text{non-relativistic})

Wavelength and momentum are inversely related — a faster particle has a shorter wavelength, never a longer one.

Bohr quantization is the postulate that in the Bohr model of hydrogen the electron's orbital angular momentum can only take values

L = n\hbar = n\,\frac{h}{2\pi}, \qquad n = 1, 2, 3, \ldots

so the quantum of angular momentum is the reduced Planck constant $\hbar = h/2\pi$ , not $h$ itself. The factor of $2\pi$ comes from fitting a whole number of de Broglie wavelengths around the circumference of the orbit: $2\pi r = n\lambda$ together with $\lambda = h/p$ gives $L = pr = nh/2\pi$ .

Quantum objects have no classical intrinsic size. An electron or a photon is described by a quantum state, and the spatial extent of that state (the wave packet) depends on how the particle was prepared, not on what kind of particle it is. Asking whether "an electron is larger than a photon" treats both as little classical balls with diameters — the question is ill-posed as stated.

A particle is safely non-relativistic when its speed satisfies $v \ll c$ , or equivalently when its momentum satisfies $pc \ll mc^2$ (rest energy $m_e c^2 \approx 511\ \text{keV}$ for an electron, and $hc \approx 1240\ \text{eV·nm}$ is the handy shortcut for computing $pc$ from a wavelength).

Exercises

E1 (medium). According to quantum mechanics,

a) an electron is smaller than a photon b) an electron is larger than a photon c) the question of whether an electron is larger than a photon is meaningless as stated

Solution

Answer: c) the question of whether an electron is larger than a photon is meaningless as stated.

Why, step by step:

"Size" in the everyday sense is a classical notion — a rigid boundary enclosing the object. Quantum objects are not little balls; each is described by a quantum state.
The spatial extent of that state is the extent of its wave packet, which depends entirely on preparation: the same electron can be localized to a fraction of a nanometre in an atom or spread over micrometres in a beam. The same is true of a photon.
Since neither particle carries an intrinsic diameter, comparing their "sizes" has no defined meaning — the question is ill-posed as stated.

Why the tempting options fail:

a) trades on the electron being "point-like" in scattering experiments, but point-like refers to the absence of substructure, not to a comparable "size" that could be ranked against a photon's.
b) trades on the photon being "just light" and therefore small; a photon's mode can be metres long or atoms wide — it has no fixed size either.

See it: the slider below reshapes the same electron — its spatial extent is a property of the prepared state, not of the particle.

E2 (easy). In the Bohr model of the hydrogen atom, the angular momentum is quantized in units of

a) Planck's constant b) Planck's constant divided by $2\pi$ c) Planck's constant divided by the wavelength of the electron

Solution

Answer: b) Planck's constant divided by $2\pi$ .

Why, step by step:

Bohr's postulate is $L = n\hbar$ with $\hbar = h/2\pi$ , so the quantum (step size) of angular momentum is $h/2\pi$ .
De Broglie's standing-wave argument shows where the $2\pi$ comes from: a stable orbit must fit a whole number of wavelengths around its circumference, $2\pi r = n\lambda$ .
Substituting $\lambda = h/p$ gives $2\pi r = nh/p$ , hence $L = pr = n\,\dfrac{h}{2\pi} = n\hbar$ .

Why the tempting options fail:

a) $h$ has the right dimensions (J·s) but is off by the geometric factor $2\pi$ that comes from wrapping the wave around a full circle.
c) $h/\lambda = p$ is the linear momentum, which doesn't even have the dimensions of angular momentum.

See it: drag $n$ off an integer and watch the de Broglie wave miss itself — only whole numbers of wavelengths survive, and each closure is an allowed orbit with $L = n\hbar$ .

E3 (easy). True or false: an electron with a higher speed has a correspondingly longer wavelength.

Solution

Answer: False.

Why, step by step:

The de Broglie relation is $\lambda = h/p = h/(mv)$ .
Speed appears in the denominator: doubling $v$ halves $\lambda$ .
So a faster electron has a shorter wavelength — the statement inverts the relationship.

Why the tempting reading fails:

It's easy to reason "more speed → more energy → bigger wave," but wave amplitude or frequency growing with energy is a different statement; the spatial wavelength shrinks as momentum grows ( $f = E/h$ rises while $\lambda = h/p$ falls).

See it: push the electron's speed up and watch $\lambda = h/(mv)$ shrink in real time.

E4 (medium). An electron has a $1\ \text{nm}$ wavelength. Is it reasonable to treat this electron as an approximately non-relativistic particle (i.e. traveling much slower than the speed of light)?

Solution

Answer: Yes.

Why, step by step:

The cleanest test compares the momentum energy scale $pc$ to the rest energy $m_e c^2$ : the particle is non-relativistic when $pc \ll m_e c^2$ .
From the de Broglie relation, $pc = \dfrac{hc}{\lambda} = \dfrac{1240\ \text{eV·nm}}{1\ \text{nm}} = 1240\ \text{eV}$ .
The electron rest energy is $m_e c^2 \approx 511{,}000\ \text{eV}$ , so $pc/m_e c^2 \approx 2.4\times 10^{-3} \ll 1$ . Equivalently $v \approx \dfrac{h}{m_e \lambda} \approx 7.3\times 10^{5}\ \text{m/s} \approx 0.24\%$ of $c$ .
The kinetic energy is $E = \dfrac{p^2}{2m_e} = \dfrac{(pc)^2}{2m_e c^2} = \dfrac{(1240)^2}{2\times 511{,}000}\ \text{eV} \approx 1.5\ \text{eV}$ — tiny compared to the rest energy, so the non-relativistic treatment is excellent.

Why the tempting answer fails:

No would follow from intuiting that a "nanometre-scale quantum electron" must be exotic and fast; but $1\ \text{nm}$ is a long matter wavelength (atomic-bond scale), which means low momentum. Relativity only kicks in for electron wavelengths around the Compton wavelength, $h/m_e c \approx 2.4\times 10^{-3}\ \text{nm}$ — some 400× shorter.

The reusable decision procedure for any "is it relativistic?" wavelength problem:

flowchart TD
    lam["given λ"] --> pc["pc = hc/λ<br/>(hc ≈ 1240 eV·nm)"]
    pc --> cmp{"pc ≪ mc²?<br/>(electron: mc² ≈ 511 keV)"}
    cmp -->|"yes — e.g. λ = 1 nm → pc = 1240 eV"| nonrel["non-relativistic ✓<br/>E = p²/2m"]
    cmp -->|"no — λ near h/mc ≈ 0.0024 nm"| rel["relativistic —<br/>use E² = (pc)² + (mc²)²"]