Question bank — Davisson-Germer experiment — electron diffraction

False. A sharp intensity bump at one special angle needs constructive interference of overlapping waves; billiard-ball particles give a smooth featureless spread, so the peak is a wave-only signature.

False. Higher $V$ means larger momentum $p=\sqrt{2meV}$, and $\lambda =h/p$, so $\lambda$ decreases. Faster particles have shorter waves.

False. It proves the opposite direction of duality — that matter (electrons) behaves as a wave. Light-as-particle is the job of the Photoelectric effect.

True. A regular lattice gives one well-defined atomic spacing, so scattered waves reinforce at sharp predictable angles; a disordered lump smears many spacings together and washes the peak out.

False. The kinetic energy $54$ eV is minuscule next to the electron rest energy $511,000$ eV, so $v\ll c$ and the non-relativistic $p=\sqrt{2meV}$ is essentially exact.

False. $\varphi$ is measured from the incident beam, $\theta$ from the crystal plane; they are linked by $\theta =90^{\circ }-\varphi /2$, not by equality.

True in the observed sense — real electrons arrive as discrete detected charges (particle side of Wave-particle duality) whose statistical distribution follows the wave pattern. A pure wave with no quantisation would not give individual clicks.

The number $0.91$ Å is not the real Ni(111) plane spacing ($d\approx 2.03$ Å); the clean $1.65$ Å actually comes from the surface-grating relation $a\sin \varphi =n\lambda$ with row spacing $a\approx 2.15$ Å, not from Bragg with a fudged $d$.

For grazing scatter off a surface row the extra path is $a\sin \varphi$, not $a\cos \varphi$. Using cosine would predict the peak at the wrong angle.

Energy gained is $eV$ (charge times voltage), which then equals $\frac{1}{2}m{v}^2$. Confusing the voltage $V$ with the speed $v$ mixes two completely different quantities.

It is the whole point, not a coincidence: two independent routes to $\lambda$ agreeing is exactly what confirms de Broglie hypothesis for matter.

It measures scattered current (how many electrons) versus angle. The wavelength is inferred afterwards from the geometry of where the peak sits.

Electron–electron repulsion is negligible here; the pattern comes from each electron's own wave interfering with itself after scattering off the ordered lattice.

Diffraction only produces observable, well-separated maxima when the grating spacing is of the same order as the wavelength; a grating far larger than $\lambda$ gives peaks crammed together near the forward direction and invisibly weak features.

One road (theory, $\lambda =h/p$) assumes de Broglie is true; the other (experiment, from diffraction geometry) does not. Their agreement is the test — a single road could never confirm the hypothesis.

Heating a metal boils electrons off its surface (thermionic emission), providing a steady controllable supply of free electrons to accelerate.

Electrons speed up slightly as they enter the crystal's inner potential, shortening their wavelength just inside the surface — a small refraction correction not in the vacuum formula.

A narrow monoenergetic beam gives one sharp wavelength and one incidence direction, so all scattered waves share the same geometry and the interference peak stays sharp instead of smearing.

They are two consistent geometric descriptions of the same ordered lattice; each just references its angle to a different feature (surface rows vs. internal planes), so both must encode the same physics.

$\lambda$ keeps shrinking; the first-order peak moves toward smaller $\varphi$ and, once $\lambda$ drops below what that order can satisfy, that peak can vanish entirely while higher-order ones may appear.

$\lambda =12.27/\sqrt{V}$ grows without bound as $V\to 0$; the wavelength becomes far larger than the atomic spacing, so no diffraction maximum forms — the geometry can no longer be satisfied.

No. Its momentum is enormous, so $\lambda =h/p$ is unimaginably small compared to atomic spacing, and the wave nature is utterly unobservable — this is why Wave-particle duality shows up only for tiny masses.

With no single repeating spacing there is no fixed path-difference condition, so the sharp peak dissolves into a broad diffuse halo — order in the lattice is essential to a sharp maximum.

That is the undeflected transmitted/reflected direction where path difference is zero for every atom; it's always bright but carries no wavelength information, so it isn't the diagnostic diffraction peak.

No — Bragg's law $2d\sin \theta =n\lambda$ depends only on $\lambda$ and spacing, not on what kind of wave it is; that equivalence is precisely why matter-wave diffraction mirrors X-ray diffraction.