Question bank — Hydrogen energy levels Eₙ = −13.6 - n² eV

True — but "more" means less negative, i.e. closer to $0$. Since $E_n=-13.6/n^2$, larger $n$ shrinks the fraction, so the electron is higher up the well (more weakly bound), not lower.

True — $n=1$ gives $-13.6$ eV, the deepest shelf. No allowed state sits below it; this is the ground state and why hydrogen is stable.

False — any positive total energy means the electron is free (already escaped) with leftover kinetic energy. Bound states are strictly negative; the formula only produces negative values.

False — because the spacing follows $1/n^2$, the bottom gap ($1\to 2$) is huge ($10.2$ eV) and gaps shrink toward zero as $n$ grows. Levels crowd together near $E=0$ (see the well diagram above).

True — $n$ runs over all positive integers, and $-13.6/n^2$ gets ever closer to $0$ without reaching it, so infinitely many levels pile up just below the escape line.

False — that constant is for hydrogen ($Z=1$). Neutral helium has two electrons and a different structure; only a one-electron ion like He${}^{+}$ obeys $E_n=-13.6Z^2/n^2$ with $Z=2$.

True — from the force balance $m{v}^2=k{e}^2/r$ (with $k=1/4\pi {\epsilon }_0$), kinetic energy is $\frac{1}{2}k{e}^2/r$ and potential is $-k{e}^2/r$, so their sum is $-\frac{1}{2}k{e}^2/r=-\text{KE}$. This is why bound states come out negative automatically.

True — the depth (how far below zero) is $13.6/n^2$; replacing $n$ by $2n$ multiplies by $1/4$. So $n=2$ is one quarter as deep as $n=1$.

False — the ground state is the lowest shelf, so there is nowhere lower to fall to. Emission requires dropping to a lower level, which the ground state lacks.

Error: multiplied instead of dividing. The rule is $-13.6/n^2$, so $E_3=-13.6/9\approx -1.51$ eV. Higher shelves are shallower, not deeper.

Error: a photon carries a difference of two levels, not a single level's value. Use $\Delta E=E_3-E_2=(-1.51)-(-3.40)=1.89$ eV; also energy released is positive.

Error: kinetic energy (KE) is indeed positive, but total energy also includes the negative potential energy (PE $=-k{e}^2/r$), which wins. Net total is negative for any bound electron.

Error: $13.6$ eV frees it from the ground state. From $n=2$ you only climb from $-3.40$ eV to $0$, needing $3.40$ eV. Ionization energy depends on the starting shelf.

Error: it doesn't stop — it moves slower ($v\propto 1/n$) but never zero for finite $n$. And "stopped and free" is the $n\to \infty$ limit, where $E\to 0$, not a real finite orbit. (Here $a_0=0.529$ Å is the Bohr radius.)

Error: the sign has nothing to do with direction. It encodes that the state lies below the free-electron reference $E=0$ — a statement about binding, not motion.

Error: closely-spaced high levels give tiny photon energies. The biggest photons come from the widest gap — transitions ending at $n=1$ (Lyman series).

Because "electron infinitely far, at rest" is the natural zero of potential energy in Coulomb's law. It makes bound states negative and "energy to free it" simply equal to $-E_n$.

The force balance $m{v}^2=k{e}^2/r$ ties kinetic energy to potential: KE $=\frac{1}{2}|\text{PE}|$. Adding KE and PE gives half the (negative) potential. This is a general result for inverse-square attraction (the virial theorem).

Because only discrete energy shelves exist, only discrete energy differences exist, and each difference maps to one photon energy via $E=hc/\lambda$. Discrete gaps force discrete wavelengths — see Hydrogen spectral series.

Classically any orbit is allowed and an accelerating electron would radiate and spiral in. Quantizing angular momentum ($\hslash =h/2\pi$ being its natural unit) was an unexplained input, later justified by wave nature, that selects stable orbits — see Quantum numbers.

It is the Rydberg energy, built only from fundamental constants $m,e,{\epsilon }_0,h$. The $n$-dependence lives entirely in the $1/n^2$ factor outside it — see Rydberg constant and formula.

Falling means going to a more negative (lower) level, so the starting level is higher; the higher-minus-lower difference is positive. That positive amount becomes the emitted photon's energy.

Because only quantized orbits are allowed; there is no shelf below $n=1$. The ground state has nonzero radius $a_0=0.529$ Å and cannot lose more energy, so it is stable.

$13.6/n^2\to 0$, so $E_n\to 0^{-}$: the electron is at the very brink of freedom, infinitely weakly bound, at infinite radius. This limit is the ionization threshold.

No — $n$ starts at $1$. Setting $n=0$ would divide by zero (infinite depth) and also violate $mvr=n\hslash$, which would give zero angular momentum. The lowest real shelf is $n=1$.

The electron escapes and keeps the surplus as kinetic energy. Above $E=0$ energies are continuous, so any excess is allowed — unlike the discrete bound region below.

No — a bound-to-bound jump needs an energy that exactly matches a gap. A mismatched photon passes through untouched (unless it exceeds the full $13.6$ eV and ionizes).

$E_1=-13.6\times 2^2=-54.4$ eV, four times deeper than hydrogen because the doubled nuclear charge pulls the single electron in much harder ($E\propto Z^2$).

Ending at $n=1$ crosses the widest gaps, giving the most energetic (ultraviolet, Lyman) photons; ending at $n=2$ gives visible Balmer lines. The final shelf sets the series and rough colour band.