Question bank — Rutherford's gold-foil experiment — nuclear model

FALSE. It proves the positive charge is not spread out in their path; it is concentrated in a tiny non-pudding core, so most trajectories simply never come near it.

FALSE. It never makes contact — Coulomb repulsion stops and reverses it at the distance of closest approach $r_0$, still well outside the actual nuclear surface.

FALSE. It's thin so each α undergoes essentially one deflecting interaction; a thick foil would give many overlapping scatterings and blur the clean single-nucleus result.

FALSE. Electrons occupy the vast empty space around the tiny nucleus; the nucleus holds only the positive charge and nearly all the mass.

FALSE. A projectile heavier than its target cannot rebound backward in a collision — it plows forward. Backscatter requires the nucleus to out-mass the α, which is exactly why it implies a heavy core.

FALSE. The orbiting electron accelerates, radiates energy, and spirals in within $\sim 10^{-11}$–$10^{-10}$ s — the model predicts collapse, contradicting real stable atoms.

FALSE. $r_0$ is only an upper bound on the nuclear radius — the α stops before touching, so the true nucleus is smaller than $r_0$.

FALSE. From $r_0=\frac{1}{4\pi {\epsilon }_0}\frac{4Z{e}^2}{m{v}^2}$, more KE ($\frac{1}{2}m{v}^2$) means the α pushes closer before stopping, so $r_0$ gets smaller.

He discovered the nucleus. The electron was found by J. J. Thomson in 1897; Rutherford's experiment says nothing new about electrons.

The nucleus is positive ($+Ze$). Like charges repel, so a positive nucleus repels the positive α — that repulsion is precisely what causes backscatter (see Coulomb's Law).

It does not. A spiraling electron emits a continuous range of frequencies, but real atoms show discrete lines. Explaining line spectra required Bohr's quantized orbits.

There is no friction; it's an elastic Coulomb interaction in vacuum. The KE is temporarily stored as electrostatic PE, then fully returned as the α reverses.

Centripetal force points inward, not outward. It's the Coulomb attraction (pointing inward) that supplies the centripetal requirement; nothing pushes the electron out.

It proves the nucleus is dense and massive, not large. Large deflection needs a strong, concentrated charge in a tiny region, and a scatterer heavier than the α.

$Z=79$ is the number of protons (nuclear charge $+79e$), discovered via the proton. It has nothing to do with α-particles, which are the incoming projectiles.

No single point in a diffuse cloud is strong enough to fully reverse a fast α; the gentle, distributed repulsion can only nudge it slightly, so the pudding model predicts no large-angle bounces.

To turn a heavy, fast α completely around, the scatterer must out-mass it. A light-but-charged target would recoil and let the α through, so backscatter forces the core to be massive.

Air molecules would scatter and slow the α-particles before they reach the foil, smearing the results; vacuum ensures each α interacts only with the gold.

An electron is a charge undergoing centripetal acceleration, and accelerating charges emit electromagnetic waves. Planets are neutral masses; their gravitational-wave emission is negligibly tiny.

We only care about the turning point where speed hits zero, and energy conservation links the known starting KE directly to the PE there — no need to track the complicated force along the whole curved path.

That tiny fraction is the ratio of the nucleus's shadow disc (its scattering cross-section) to the whole atom's face; only ~1 in 20,000 α's land on the disc, so the deflecting target occupies a minuscule area — a tiny nucleus in mostly empty space.

It feels only a weak, distant Coulomb push and deflects through a small angle — this is the "some swerve" middle group, neither straight-through nor backscattered.

It decelerates straight in, stops at $r_0$, and retraces its path back at $180^{\circ }$ — the extreme backscatter case, and the one used to define the closest approach.

Weaker nuclear charge means less repulsion and a smaller $r_0$; fewer and smaller-angle deflections occur, so backscatter becomes even rarer.

KE is momentarily zero (it has stopped), but the acceleration is maximum — the repulsive Coulomb force is strongest there because $r$ is smallest, and it's about to fling the α back out.

Then the α is no longer stopped purely by Coulomb repulsion — it penetrates to the nuclear surface, and the simple KE = PE estimate breaks down because short-range nuclear forces and possible reactions take over. Rutherford's clean scattering formula holds only below this Coulomb-barrier energy.

About $10^{-11}$–$10^{-10}$ s before the electron spirals into the nucleus. It matters because matter plainly is stable, so this collapse prediction is the decisive failure that motivated the Bohr model.