Question bank — Derivation of Bohr's radii and energies from electrostatics + quantization

True — smaller $r$ means more negative $PE$ and, since $E=\frac{1}{2}PE$, a more negative total energy; the difference leaves as a light particle (photon), which is exactly what makes hydrogen's bright spectral lines.

False — radius scales as $n^2$, so doubling $n$ multiplies the radius by $4$, not $2$ ($r\propto n^2/Z$).

True — $E_1=-13.6Z^2$ eV is the deepest (most negative) level, so it takes the most energy to lift the electron out of it.

False — a larger $Z$ pulls harder, so $r\propto 1/Z$ shrinks the orbit; the electron is dragged in, not pushed out.

True — speed scales as $v\propto Z/n$, so the inner (smaller $n$) orbit is the faster one.

False — $E=0$ is the free electron with zero binding energy; it can still be moving, but it is no longer trapped by the nucleus.

False — it is a separate, added postulate (Quantization of Angular Momentum); electrostatics alone allows any radius, so the extra rule is what discretises the orbits.

False — $PE$ itself is negative, so $-\frac{1}{2}PE$ flips the sign and is positive; kinetic energy $\frac{1}{2}m{v}^2$ is always positive, as it must be for a real speed.

Error — this forgets the kinetic energy. By the Virial Theorem, $KE=-\frac{1}{2}PE$, so $E=\frac{1}{2}PE$ — half the value written.

Error — there is a second factor of $Z$ hidden in $r_n\propto 1/Z$; substituting it gives $E_n\propto Z^2/n^2$.

Error — "more energy" here means closer to zero since $E_n=-13.6/n^2$; larger $n$ is less bound and easier to remove, not more.

Error — the correct statement is Coulomb = centripetal force, the inward force required for circular motion; centrifugal is a fictitious outward pseudo-force, not what supplies the pull. See Centripetal Force and Circular Motion.

Error — $\hslash$ (h-bar) is Planck's constant $h$ divided by $2\pi$, so $\hslash =h/2\pi$ and the condition is $mvr=n\hslash =nh/2\pi$; dropping the $2\pi$ changes every result by a factor of about $6.28$.

Error — $a_0$ is only the $n=1,Z=1$ radius; a general orbit is $r_n=a_0{n}^2/Z$.

Error — cancelling a common factor of $r$ is algebraically exact; no information is lost, we simply combine it with quantization to fix $r$.

Because the electron is bound: energy is measured relative to the free electron at $E=0$, and it takes positive energy to free it, so the trapped state sits below zero.

The Coulomb energy carries one factor of $Z$, and the radius it depends on carries another factor of $1/Z$; combining them squares the charge dependence for energy while the radius keeps a single $1/Z$.

The force balance is one equation in two unknowns ($r$ and $v$); Quantization of Angular Momentum supplies the second equation that pins both down to discrete values.

A circling electron is an accelerating charge, and Maxwell's electromagnetism says an accelerating charge continuously radiates energy away as light (this is called Larmor radiation); losing energy, its orbit would shrink and it would crash into the nucleus in about $10^{-11}$ s. Bohr's answer is a bold postulate that the quantised orbits simply do not radiate, so the ground state $n=1$ is stable with no smaller orbit to fall to. See Bohr Model Postulates.

Ionization Energy is the work to move the electron from $n=1$ all the way to $n=\infty$; since the free electron has $E_{\infty }=0$, that work is $0-E_1=+13.6$ eV — the depth of the ground level is the escape cost.

The attraction binding electron to nucleus is electrical, between charges $-e$ and $+Ze$; gravity between them is roughly $10^{39}$ times weaker and utterly negligible, so Coulomb's Law is the correct force law.

Physically it says that for a $1/r$ attractive force a bound orbit stores exactly twice as much (negative) potential energy as it has (positive) kinetic energy, a balance forced by the force-balance equation itself. Practically, it lets you write $KE$, and hence $E=KE+PE=\frac{1}{2}PE$, purely in terms of $PE$ (or $r$) — eliminating $v$ so the whole energy drops out in one substitution instead of tracking speed separately.

Radius $r_n\propto n^2$ blows up to infinity while $E_n\propto -1/n^2$ climbs up to $0$ — the electron drifts infinitely far and becomes free.

No — $n=0$ would give $mvr=0$, meaning zero angular momentum, and $r_n\propto n^2=0$, a collapse onto the nucleus; the counting starts at $n=1$, the physical ground state.

The quantum condition $mvr=n\hslash$ sets angular momentum to whole multiples of the fixed chunk $\hslash$; this echoes de Broglie Wavelength and Standing Waves, where the electron's wave must close on itself around the orbit — only a whole number of wavelengths fits, so a fraction would leave the wave out of step and cancel itself out. Negative $n$ just relabels the same orbits (magnitude is what matters), so we keep $n=1,2,3,\dots$.

It gives $r=a_0=0.529$ Å and $E=-13.6$ eV — the ground state of ordinary hydrogen, the benchmark all other $r_n$ and $E_n$ are scaled from.

The radius $\propto 1/Z$ shrinks and the binding $\propto Z^2$ grows enormously; the electron would orbit extremely close and fast — a hint that relativistic corrections (ignored by Bohr) become important there.

Yes for energy — both give $E=-13.6$ eV since $E\propto Z^2/n^2$ — but their radii differ, because $r\propto n^2/Z$ is a different combination.

Every formula collapses: $r_n\to \infty$ and $E_n\to 0$ — with no charge there is no Coulomb pull, so no bound orbit exists at all, which is physically correct.