# 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:** 1. "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. 2. 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. 3. 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. ```qm-demo wave-packet-width ```

**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:** 1. Bohr's postulate is $L = n\hbar$ with $\hbar = h/2\pi$, so the quantum (step size) of angular momentum is $h/2\pi$. 2. 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, $2\pi r = n\lambda$. 3. 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 $L = n\hbar$. ```qm-demo bohr-standing-wave ```

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

Solution

**Answer: False.** **Why, step by step:** 1. The de Broglie relation is $\lambda = h/p = h/(mv)$. 2. Speed appears in the _denominator_: doubling $v$ halves $\lambda$. 3. 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. ```qm-demo de-broglie-speed ```

**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:** 1. The cleanest test compares the momentum energy scale $pc$ to the rest energy $m_e c^2$: the particle is non-relativistic when $pc \ll m_e c^2$. 2. From the de Broglie relation, $pc = \dfrac{hc}{\lambda} = \dfrac{1240\ \text{eV·nm}}{1\ \text{nm}} = 1240\ \text{eV}$. 3. 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 $c$. 4. 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: ```mermaid flowchart TD lam["given λ"] --> pc["pc = hc/λ
(hc ≈ 1240 eV·nm)"] pc --> cmp{"pc ≪ mc²?
(electron: mc² ≈ 511 keV)"} cmp -->|"yes — e.g. λ = 1 nm → pc = 1240 eV"| nonrel["non-relativistic ✓
E = p²/2m"] cmp -->|"no — λ near h/mc ≈ 0.0024 nm"| rel["relativistic —
use E² = (pc)² + (mc²)²"] ```