Question bank — Bohr atomic model and electron shells

A lower (more negative) energy level means the electron is more tightly bound. ::: True. More negative energy means you must add more energy to reach $E=0$ (freedom), so the electron is harder to pull off — the ground state $n=1$ is the most tightly bound. As $n$ increases, the energy levels get more negative. ::: False. $E_n=-13.6/n^2$ climbs toward $0$ as $n$ grows (less negative), so higher shells are higher in energy, not lower. An electron in a stationary state is standing still. ::: False. "Stationary state" means the orbit is stable and radiates no energy — the electron is still moving in a circle; only its energy is constant. The energy levels get farther apart as $n$ increases. ::: False. They bunch together: since $E_n\propto -1/n^2$, spacing shrinks and the levels crowd toward $0$ as $n\to \infty$ (see figure s01). Doubling the shell number doubles the orbit radius. ::: False. $r_n=a_0{n}^2$ scales with $n$ squared, so doubling $n$ makes the radius four times larger, not two. An atom absorbing a photon must gain exactly the energy gap between two allowed levels. ::: True. Only jumps between allowed levels exist; a photon whose energy doesn't match some gap $E_{n_f}-E_{n_i}$ simply passes through unabsorbed. A free (ionized) electron has zero energy in this model. ::: True (at rest). $E\to 0$ as $n\to \infty$ marks the boundary of being bound; a just-freed electron sits at $E=0$, and any extra energy becomes its kinetic energy above zero. The kinetic energy of the orbiting electron is half the magnitude of its potential energy. ::: True. Force balance gives $KE=\frac{1}{2}k{e}^2/r$ while $|PE|=k{e}^2/r$, so $KE=\frac{1}{2}|PE|$ — the virial relation for a $1/r$ force.

"The Coulomb force pushes the electron outward, balancing gravity." ::: Two errors: Coulomb attraction pulls the electron inward (opposite charges), and gravity is negligible here — it's the inward Coulomb force that plays the role of centripetal force. "Bohr allowed any circular orbit as long as the force balanced." ::: The force-balance equation alone allows a continuous range of radii; Bohr's extra rule $mvr=n\hslash$ (angular momentum quantization) is what selects the discrete allowed orbits. "In the Bohr model of hydrogen, shell $n=3$ holds 18 electrons." ::: The Bohr hydrogen atom has only one electron, so it occupies just one shell at a time; the $2n^2=18$ capacity is a rule for multi-electron atoms, a separate framework from the single-electron Bohr picture — don't mix the two. "The $2n^2$ capacity guarantees shell $n=3$ is completely full before shell $n=4$ begins filling." ::: In multi-electron atoms $2n^2$ is only a maximum; actual filling follows sub-shell energy ordering, so the $n=4$ shell can start before $n=3$ reaches 18. (This is a fact about multi-electron atoms, beyond the single-electron Bohr derivation.) "Silicon conducts well because it has 4 valence electrons." ::: Four valence electrons make it a semiconductor, not a good conductor — they're bound tightly enough to insulate cold, only partly freed with added energy. Good conductors (copper) have few loosely held valence electrons. "When an electron falls to a lower shell it absorbs a photon." ::: Falling to a lower (more negative) level releases energy, so the atom emits a photon; absorption is what lifts an electron to a higher shell. "The emitted photon energy equals $E_{n_f}$, the final level's energy." ::: The photon carries the difference $E_{n_i}-E_{n_f}$ between the two levels, not the final level's energy by itself. "Because the electron accelerates in its circle, it continuously radiates light." ::: That's the classical prediction (figure s02) Bohr rejected; postulate 2 declares stationary orbits radiate nothing — light appears only on a jump between orbits.

Why is the total energy of a bound electron negative? ::: Because $E=KE+PE=\frac{1}{2}k{e}^2/r-k{e}^2/r=-\frac{1}{2}k{e}^2/r$: the positive kinetic energy is only half the magnitude of the negative potential energy (the virial theorem for a $1/r$ force), so their sum is negative — you must add energy to reach the free state $E=0$. Why did Bohr forbid the smooth inward spiral classical physics predicted? ::: Because an accelerating (orbiting) charge radiates by Maxwell's laws, draining energy so fast that hydrogen would spiral in within ~$10^{-11}$ s (figure s02), yet real atoms are stable and emit only sharp spectral lines. Why does heated hydrogen glow in specific colors rather than a full rainbow? ::: Each color is one fixed energy gap between two allowed levels, so only the discrete jump-energies appear as light — a continuum would require a continuum of levels, which are forbidden. Why does the number of valence electrons decide electrical conductivity? ::: Conduction needs electrons that can be freed from atoms; the outermost (valence) electrons are the loosest, so how many there are and how tightly they're held sets how easily current flows. Why does the energy formula have $n^2$ in the denominator (not $n$)? ::: It falls out of substituting the quantized radius $r_n=a_0{n}^2$ into $E\propto -1/r$, so $E\propto -1/n^2$. Why must the Coulomb attraction exactly equal $m{v}^2/r$ for a circular orbit? ::: A circle requires a constant inward (centripetal) pull of size $m{v}^2/r$, and the only inward force available is the Coulomb attraction, so they must match — this balance is what fixes $KE$ at half of $|PE|$.

What happens to the orbit radius as $n\to \infty$? ::: $r_n=a_0{n}^2$ grows without bound, so the electron drifts arbitrarily far out — the limit of becoming unbound (ionized). What is the energy of the $n\to \infty$ state, and what does it physically mean? ::: $E_n\to 0$: the electron is just barely free, sitting at the top of the well — the ionization threshold (figure s01). Can an electron sit at an energy between two allowed levels, say halfway between $n=1$ and $n=2$? ::: No — only $n=1,2,3,\dots$ exist; there is no allowed state in the gap, which is the whole point of quantization. Is there a level below $n=1$ (an $n=0$ or lower)? ::: No — $n=1$ is the lowest allowed shell and the most tightly bound state; smaller $n$ isn't defined, so the electron cannot fall further and crash in. Does the $2n^2$ capacity rule mean a multi-electron atom always fills shells fully before starting the next? ::: No — $2n^2$ is only an upper bound; real filling follows sub-shell energy order, so an outer shell can begin before an inner one is completely full. What does the model predict for a photon whose energy doesn't match any level gap? ::: The atom can't absorb it (no matching jump exists), so that photon simply passes through unabsorbed. For a hydrogen atom already in the ground state ($n=1$), can it emit a photon spontaneously? ::: No — there's no lower level to fall to, so it stays put; emission needs an available lower shell to jump into.