Atomic Models: the Bohr Atom

Beginner

An atom has a tiny, heavy, positive nucleus with light, negative electrons around it. Picture it as a miniature solar system, as people did around 1911, and you hit a problem: an electron circling the nucleus should keep giving off light, lose energy and spiral inwards in a split second. Yet atoms do not collapse.

Step 1. In 1913 Niels Bohr proposed that an electron may only use certain allowed orbits, like the rungs of a ladder, and gives off no light while it stays on one. Each rung has its own energy: the lowest, nearest the nucleus, has the least.

Step 2. Atoms climb to higher rungs when something hands them energy: the heat of a flame, or collisions with fast electrons in the glowing gas of a neon sign.

Step 3. An excited electron soon drops back. The energy it loses leaves as one packet of light, a photon. In the animation the electron on the third ring drops to the second and a red photon flies off. Later an identical red photon arrives, the electron swallows it and climbs back up; light of a slightly different colour would pass by untouched.

Step 4. The gaps between rungs are fixed, so each drop always gives the same colour. Hydrogen glows in a few sharp lines rather than a rainbow, and every element has its own rungs, so its lines are a fingerprint. Spread a star's light through a prism and you can read what it is made of.

The rings are only a picture: the electron is really a fuzzy cloud, not a tiny planet, but the fixed energy levels are real.

Check yourself: an electron sits on rung 4. Which single drop gives the photon with the most energy? (Straight down to rung 1.)

Intermediate

Atoms, light and the electronvolt

Spread through a prism, the light of glowing hydrogen is not a full rainbow but a few separate bright lines. Alpha-particle scattering showed in 1911 that an atom is mostly empty: a nucleus only $10^{-15}$ to $10^{-14}\mathrm{m}$ across holds all the positive charge and almost all the mass, while the electrons fill a region about $10^{-10}\mathrm{m}$ across. For hydrogen, one proton and one electron, two facts need explaining: its atoms are stable, and hot hydrogen glows only at a few sharp wavelengths, such as a red line at 656 nm, and it absorbs light at those same wavelengths.

Atomic energies are measured in electronvolts, $1\mathrm{eV}=1.602\times 10^{-19}\mathrm{J}$, the energy an electron gains across one volt. The zero is an electron at rest far from the proton, so a bound electron has a negative energy. A photon's energy and wavelength are linked by $E=hc/\lambda$, with $hc\approx 1240\mathrm{eV}\cdot \mathrm{nm}$.

Bohr's three rules

A miniature solar system cannot work: an orbiting electron accelerates, so it should radiate, spiral into the nucleus in about $10^{-11}\mathrm{s}$ and emit a smear of colours rather than sharp lines. In 1913 Niels Bohr kept the orbit but added rules that classical physics does not contain.

The first rule is classical: the electron circles the proton, held by the Coulomb force. The second selects the orbits: the angular momentum $mvr$ must be a whole multiple of the reduced Planck constant $\hslash =h/2\pi \approx 1.055\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$,
$mvr=n\hslash$
where $n=1,2,3,\dots$ is the principal quantum number.

The third rule is the boldest: on an allowed orbit the atom is in a stationary state, and the electron does not radiate at all. Light is emitted or absorbed only in a jump between allowed orbits, as a single photon carrying the energy difference.

Radii, energies and photons

With the force balance, the angular-momentum rule fixes the radius and energy of each orbit, and a jump from a level $i$ down to a lower level $f$ gives one photon:

• $r_n=n^2{a}_0$
• $E_n=-\frac{13.6\mathrm{eV}}{n^2}$
• $hf=E_i-E_f$
• $\lambda =\frac{hc}{E_i-E_f}$

Here $a_0\approx 0.0529\mathrm{nm}$ is the Bohr radius. The radii grow as $n^2$, in the ratio 1 : 4 : 9 : 16, while the speed falls as $v_n=v_1/n$, with $v_1\approx 2.19\times 10^{6}m/s$. The subscript $f$ marks the final level; $f$ alone is the photon's frequency.

The energies are negative because the electron is bound: freeing it from the ground state takes 13.6 eV, the ionisation energy of hydrogen. The levels $-13.6$, $-3.40$, $-1.51$ and $-0.85\mathrm{eV}$ crowd together and approach zero as $n\to \infty$. The force balance $m{v}^2/r=k{e}^2/r^2$ also makes the kinetic energy $K=\frac{1}{2}m{v}^2$ half the size of the potential energy $U=-k{e}^2/r$, so that $E=K+U=-K$.

Worked example: the red line in the animation

The electron drops from $n=3$ to $n=2$:
$E_3=-\frac{13.6}{9}=-1.51\mathrm{eV},E_2=-\frac{13.6}{4}=-3.40\mathrm{eV}$
$hf=E_3-E_2=-1.51-(-3.40)=1.89\mathrm{eV}$
$\lambda =\frac{hc}{E_3-E_2}=\frac{1240\mathrm{eV}\cdot \mathrm{nm}}{1.89\mathrm{eV}}\approx 656\mathrm{nm}$
This is red light, the Hα line, of frequency $f=1.89\mathrm{eV}/(4.14\times 10^{-15}\mathrm{eV}\cdot \mathrm{s})\approx 4.57\times 10^{14}\mathrm{Hz}$.

These are the animation's numbers. The $n$ readout switches from 3 to 2 and the $E$ readout from $-1.51$ to $-3.40\mathrm{eV}$ just as the red photon labelled $hf$ leaves, and the line below them shows $hf=1.89\mathrm{eV}$ and $\lambda =656\mathrm{nm}$. When an identical photon arrives, the electron absorbs it and the readouts return to 3 and $-1.51\mathrm{eV}$.

The ring shrinks from $r_3=9a_0=0.476\mathrm{nm}$ to $r_2=4a_0=0.212\mathrm{nm}$, and the electron speeds up by a factor 3/2. Its kinetic energy rises by 1.89 eV while its potential energy falls by twice as much, 3.78 eV; the net loss is the photon. One turn takes a time proportional to $n^3$: on screen 3.0 s on $n=3$ against 0.9 s on $n=2$, in reality $1.2\times 10^{-15}\mathrm{s}$ on $n=2$.

Worked example: lighting a hydrogen lamp

In a discharge tube, fast electrons strike hydrogen atoms in their ground state. A collision can pass on any part of the electron's energy, but the atom accepts only exact gaps. What is the least energy that makes the tube glow red, and which lines then appear?

The red line starts on $n=3$, so the atom must first be lifted there:
$E_3-E_1=-1.51-(-13.6)=12.09\mathrm{eV}$
An electron with 11 eV can only raise the atom to $n=2$ (10.2 eV), and the atom falls straight back, emitting ultraviolet at $1240/10.2\approx 122\mathrm{nm}$: nothing visible.

From $n=3$ there are two ways down: straight to $n=1$ (12.09 eV, 103 nm), or $3\to 2$ (1.89 eV, 656 nm) followed by $2\to 1$ (10.20 eV, 122 nm). Both routes release $1.89+10.20=12.09\mathrm{eV}$, and of the three lines only the red one is visible. Faster electrons reach higher levels: an atom raised to level $n$ can emit $n(n-1)/2$ different lines, six from $n=4$, including the blue-green one at 486 nm.

Reading an energy-level diagram

An energy-level diagram draws each level as a horizontal line at its energy, from $n=1$ up to the ionisation limit $E=0$; the expert layer of the animation shows one to scale. A downward arrow is an emission, an upward one an absorption, and the arrow's length is the photon energy: the longer the arrow, the shorter the wavelength. The rings of the atom show the opposite trend: the orbits spread apart while the levels crowd together.

Jumps ending on the same level form a series: Lyman (to $n=1$, ultraviolet, 122 to 91 nm), Balmer (to $n=2$, visible lines at 656, 486, 434 and 410 nm, converging on 365 nm) and Paschen (to $n=3$, infrared). One rule, found from measurements before Bohr explained it, covers them all:
$\frac{1}{\lambda }=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$
with $R_H\approx 1.097\times 10^7{\mathrm{m}}^{-1}$. For Hα, $R_H(1/4-1/9)=1.52\times 10^6{\mathrm{m}}^{-1}$, which gives 656 nm once more.

Common mistakes

• The photon carries the difference between two levels: $3\to 2$ gives 1.89 eV, not 1.51 or 3.40 eV, nor their sum, 4.91 eV.
• The energy goes as $1/n^2$, not as $1/n$: the $n=2$ level lies at $-3.40\mathrm{eV}$, not at $-6.8\mathrm{eV}$.
• The radius grows as $n^2$: the $n=3$ orbit is 9 times wider than the $n=1$ orbit, not 3 times; 27 is the ratio of their periods.
• A higher orbit is slower and yet has more energy: on $n=3$ the kinetic energy is 1.89 eV lower than on $n=2$, but the potential energy is 3.78 eV higher.
• A photon cannot be partly absorbed: a 2.00 eV photon passes an atom in $n=2$ untouched, because no level lies 2.00 eV higher. Only above 3.40 eV, where the electron is set free, can a surplus go into its kinetic energy; a colliding electron can give up just part of its energy.

Where you meet it

Every element has its own levels, so its lines are a fingerprint. Cooler gas in the Sun's outer layers absorbs its own wavelengths: hydrogen's Hα is the dark line that Fraunhofer labelled C, at 656 nm, and helium was discovered in sunlight in 1868, before anyone found it on Earth. Nebulae glow red with Hα, sodium lamps shine yellow at 589 nm, and lasers work by jumps between levels.

The photon rule comes from Photoelectric Effect; Wave–Particle Duality explains why only some orbits are allowed, and Schrödinger's Equation and Tunnelling replaces orbits by waves while keeping these energies. In Semiconductors, Diodes and LEDs the levels of many atoms merge into bands, and Radioactivity and Half-Life turns to the nucleus, whose levels lie thousands to millions of times further apart.

Advanced

Why a classical atom cannot exist

Before quantum ideas, physicists tried to build the hydrogen atom from parts they understood: a light electron of charge $-e$ and mass $m$, a heavy proton of charge $+e$, and the Coulomb attraction between them, of size $k{e}^2/r^2$ at distance $r$, with $k=1/(4\pi {\epsilon }_0)$. From these parts, classical physics cannot make a stable atom, for three separate reasons. Each failure points to something the new physics has to supply.

First, the charges cannot sit still. Wherever there is no charge, the electrostatic potential $\varphi$ obeys Laplace's equation ${\nabla }^2\varphi =0$. A function that obeys it has no minimum and no maximum in that region: wherever it curves up in one direction, it curves down in another. So an electron at rest among fixed charges is never at the bottom of a potential well, and there is always a direction in which it can slide away. This is Earnshaw's theorem, and it means the electron must move.

Second, a moving electron cannot keep its orbit. Circular motion is accelerated motion, and an accelerating charge radiates. With centripetal acceleration $v^2/r$, the electron loses energy at the rate (Larmor's formula)
$P=\frac{2}{3}\frac{k{e}^2(v^2/r{)}^2}{c^3}$
where $c$ is the speed of light. At the observed size of the atom, 0.053 nm, the acceleration $v^2/r$ is about $9\times 10^{22}m/{\mathrm{s}}^2$. Losing energy at this rate, the electron would spiral into the proton in about $1.6\times 10^{-11}\mathrm{s}$. On the way in its orbital frequency would rise steadily, so the atom would give out a rising sweep of light, a chirp, instead of the sharp spectral lines that are observed.

Third, classical physics cannot even say how big an atom should be. A size has to be a length built from the constants of the problem. From $k{e}^2$ (in J m) and $m$ (in kg) no length can be formed. Bringing in $c$ gives one, $k{e}^2/(m{c}^2)\approx 2.8\times 10^{-15}\mathrm{m}$, but it is about 19 000 times smaller than an atom. A new constant is needed, and Planck's reduced constant $\hslash =h/(2\pi )$ supplies it.

Uncertainty sets the size of the atom

One quantum rule alone, the uncertainty principle, already explains why an atom has a definite size and roughly what that size is. The idea is a competition between two energies, and Figure 1 shows it.

Legend

$a$	Size of the region that holds the electron: how close it is kept to the proton.	$\mathrm{m}$
$E$	Energy of the electron.	$\mathrm{J}$
$K$	Kinetic energy forced on the electron by confinement, at least about ${\hslash }^2/(2m{a}^2)$.	$\mathrm{J}$
$U$	Potential energy of the attraction, about $-k{e}^2/a$.	$\mathrm{J}$
$K+U$	Total energy $E(a)$ of the electron: kinetic plus potential.	$\mathrm{J}$
$a_0$	Size at which the total energy is lowest: the Bohr radius, about $0.053\mathrm{nm}$.	$\mathrm{m}$
$E(a_0)$	Lowest value of the total energy, $-k{e}^2/(2a_0)\approx -13.6\mathrm{eV}$.	$\mathrm{J}$

Suppose the electron is kept within a distance $a$ of the proton; $a$ runs along the horizontal axis of Figure 1. The uncertainty principle says that a particle confined to a region of size $a$ has a momentum spread of at least about $\hslash /a$. Its momentum therefore cannot be zero, and its kinetic energy $K$ is at least about $(\hslash /a{)}^2/(2m)={\hslash }^2/(2m{a}^2)$. This is the blue curve: it climbs as $1/a^2$ when $a$ shrinks.

The potential energy of the attraction, $U$, is about $-k{e}^2/a$: the red curve. It also grows in size as $a$ shrinks, but only as $1/a$, more slowly than $K$.

Now follow the purple curve, the total $K+U$, from right to left. At large $a$ both terms are small and the total is close to zero. As $a$ shrinks the attraction wins at first, so the total falls. At small $a$ the $1/a^2$ rise of $K$ overtakes the $1/a$ fall of $U$, and the total climbs steeply. In between lies a lowest point, at $a=a_0$, where the total is $E(a_0)$. The electron settles there: that lowest point is the ground state.

To locate it, write the total as
$E(a)=\frac{{\hslash }^2}{2m{a}^2}-\frac{k{e}^2}{a}$
At the lowest point the slope is zero:
$\frac{dE}{da}=-\frac{{\hslash }^2}{m{a}^3}+\frac{k{e}^2}{a^2}=0$
Multiplying by $a^3$ and solving gives the size; putting it back into $E(a)$ gives the energy:
$a_0=\frac{{\hslash }^2}{mk{e}^2}\approx 0.0529\mathrm{nm},E(a_0)=-\frac{k{e}^2}{2a_0}\approx -13.6\mathrm{eV}$
At $a_0$ the kinetic term ${\hslash }^2/(2m{a}_0^2)$ equals $k{e}^2/(2a_0)$, half the size of the potential term, which is why the total is $-k{e}^2/(2a_0)$.

An argument that uses $\hslash /a$ only as an order of magnitude cannot be exact, and landing on both the true size and the true binding energy is partly luck. The physical point is sound, though. An atom is as large as it is because squeezing the electron into less space would cost more kinetic energy than the extra attraction would pay back. The same trade-off is why solids and liquids resist compression: pushing atoms closer together would confine their electrons more tightly.

Bohr's orbits and their energies

In 1913 Niels Bohr reached the same numbers by another route: he kept classical circular orbits but allowed only some of them. His radii, speeds and energies, derived below, reproduce hydrogen's spectrum. Figure 2 shows one allowed orbit with the force and velocity of the electron.

Legend

$e$	Size of the elementary charge: the nucleus carries $+e$, the electron $-e$.	$\mathrm{C}$
$\vec{F}$	Coulomb force on the electron, pointing at the nucleus.	$\mathrm{N}$
$\vec{v}$	Velocity of the electron, along the circle.	$m/s$
$r_n$	Radius of the orbit: the distance from the nucleus to the electron.	$\mathrm{m}$
$\lambda$	Wavelength of the electron's wave; one wavelength is the arc from one crest to the next.	$\mathrm{m}$
$n$	Number of the orbit; $n$ wavelengths fit round it, here $n=3$.

Figure 2 shows the orbit $n=3$. The electron, charge $-e$, circles the nucleus, charge $+e$, at radius $r_n$, and its velocity $\vec{v}$ is tangent to the circle. The only force on it is the Coulomb attraction $\vec{F}$, which points along the radius towards the nucleus. A force towards the centre is exactly what circular motion requires, so $\vec{F}$ alone supplies the centripetal force. The wavy line and $\lambda$ come in further down.

Bohr added one rule: the angular momentum $mvr$ can only be a whole multiple of $\hslash$. The two conditions are
$\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2},mvr=n\hslash ,n=1,2,3,\dots$
The first is Newton's second law: mass times centripetal acceleration equals the Coulomb force. Multiplying it by $r$ gives $m{v}^{2}r=k{e}^2$. Dividing this by the second condition leaves the speed, and the second condition, $r=n\hslash /(mv)$, then gives the radius:
$v_n=\frac{k{e}^2}{n\hslash }=\frac{\alpha c}{n},r_n=\frac{n^2{\hslash }^2}{mk{e}^2}=n^2{a}_0$
Here $\alpha =k{e}^2/(\hslash c)\approx 1/137$ is a pure number, the fine-structure constant. The radii grow as $n^2$, and the smallest, $r_1$, is exactly the $a_0$ found from the uncertainty argument. On $n=1$ the electron moves at $c/137\approx 2.19\times 10^{6}m/s$: fast, but well below the speed of light.

Bohr's angular-momentum rule looked arbitrary until de Broglie gave it a meaning in 1924, and this is the wavy line in Figure 2. An electron of momentum $mv$ behaves as a wave of wavelength $\lambda =h/(mv)$. Substituting $mv=h/\lambda$ into $mvr=n\hslash =nh/(2\pi )$ cancels $h$ and leaves
$2\pi r=n\lambda$
The circumference holds a whole number of wavelengths. Only then does the wave join smoothly onto itself after one trip round, like a standing wave on a closed loop.

In Figure 2, $n=3$: the wavy line has exactly three crests round the circle, and the arc marked $\lambda$ spans one wavelength. In numbers, $r_3=9a_0=0.476\mathrm{nm}$, so the circumference is $2\pi \cdot 0.476\mathrm{nm}\approx 2.99\mathrm{nm}$, and the electron's wavelength on this orbit is about 1.00 nm: three wavelengths fill it exactly.

Next, the energy. The force balance, written as $m{v}^2=k{e}^2/r$, gives the kinetic energy $K=\frac{1}{2}m{v}^2=\frac{1}{2}k{e}^2/r$. The potential energy is $U=-k{e}^2/r$, so $K=-\frac{1}{2}U$; this is the virial theorem for a $1/r$ potential. The total is therefore $E=K+U=-\frac{1}{2}k{e}^2/r$, and putting in $r_n$ gives
$E_n=-\frac{m(k{e}^2{)}^2}{2{\hslash }^2{n}^2}=-\frac{{\alpha }^{2}m{c}^2}{2n^2}=-\frac{13.6\mathrm{eV}}{n^2}$

The energy is negative because the electron is bound: $13.6\mathrm{eV}/n^2$ must be supplied to free it from level $n$. The middle form compares it with the electron's rest energy, $m{c}^2=511\mathrm{keV}$: the binding energy is only a fraction ${\alpha }^2/2\approx 2.7\times 10^{-5}$ of it. That is why non-relativistic mechanics works so well here.

When the electron jumps from a level $n_i$ down to a lower level $n_f$, it emits one photon that carries off the energy difference, $hc/\lambda =E_{n_i}-E_{n_f}$. With $E_n=-13.6\mathrm{eV}/n^2$ this becomes
$\frac{1}{\lambda }=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right),R_H=\frac{13.6\mathrm{eV}}{hc}\approx 1.097\times 10^7{\mathrm{m}}^{-1}$
This is the Rydberg formula. Balmer (1885) and Rydberg had found it by fitting measured lines, with $R_H$ as a number read off the data. Bohr calculated $R_H$ from $m$, $e$, $h$ and $c$, and it agreed with the measured value within the accuracy of those constants. That agreement made his rules impossible to dismiss.

Reading the level diagram and the spectrum strip

The energies $E_n$ are easiest to grasp as a ladder of levels, with each spectral line a jump between two rungs. Figure 3 draws that ladder to scale and shows which jumps make which series of lines; the expert layer of the animation uses the same layout.

Legend

$E$	Energy of the atom, measured from the ionisation limit $E=0$ (dashed line).	$\mathrm{J}$
$n$	Number of the level: $n=1$ is the ground state.
$\to$	A jump between two levels, as in $3\to 2$: the electron drops from the first to the second and one photon leaves.

In Figure 3, energy $E$ increases upwards and each horizontal line is a level, labelled by $n$. The dashed line at the top is $E=0$, the ionisation limit: an electron with more energy than this is free. Measured from the ground state, level $n$ lies $E_n+13.6\mathrm{eV}=13.6\mathrm{eV}(1-1/n^2)$ higher, a fraction $1-1/n^2$ of the way up to the limit: 0, 3/4, 8/9 and 15/16 for $n=1$ to 4. So three quarters of the binding energy is already gone at $n=2$, and the higher levels crowd together just below $E=0$.

Each vertical arrow is one photon given out in a downward jump, labelled by the two levels it joins, as in $3\to 2$. Because the levels are to scale, the length of an arrow is the photon energy $E_i-E_f$, where $E_i$ and $E_f$ are the energies of the initial and final levels. Jumps that end on $n=1$ form the Lyman series (purple): $2,3,4\to 1$ give 10.2, 12.09 and 12.75 eV, or 122, 103 and 97 nm, all in the ultraviolet. Jumps that end on $n=2$ form the Balmer series: $3\to 2$ gives 1.89 eV (red, 656 nm) and $4\to 2$ gives 2.55 eV (blue-green, 486 nm).

Compare the arrows: every Lyman arrow is far longer than any Balmer arrow. Even the largest Balmer energy, a drop from the ionisation limit to $n=2$, is only 3.40 eV, far below the smallest Lyman energy, 10.2 eV, so the two series never overlap.

In the animation's diagram, a blue marker follows the electron from $n=3$ to $n=2$ and back. The spectrum strip above it covers the visible range, 380 to 700 nm, and shows the Balmer lines at 410, 434, 486 and 656 nm. They bunch towards the violet end because, as the starting level rises, $1/\lambda$ approaches the series limit $R_H/4$, at 365 nm, just off the strip. The red line, called Hα (the $3\to 2$ jump), lights up the moment the atom emits it.

What quantum mechanics keeps and changes

Bohr's model gets the energies right, but its picture of an electron on a sharp circular track is wrong. Quantum mechanics keeps the energies and replaces the track by a cloud of probability. Figure 4 shows that cloud for the ground state, sorted by distance from the nucleus.

Legend

$r$	Distance from the nucleus.	$\mathrm{m}$
$P$	Probability density: the chance per unit distance of finding the electron at distance $r$, $P=4r^2{e}^{-2r/a_0}/a_0^3$.	${\mathrm{m}}^{-1}$
$a_0$	Bohr radius: the distance at which $P$ is largest.	$\mathrm{m}$
$\bar{r}$	Mean distance of the electron from the nucleus, $1.5a_0$.	$\mathrm{m}$

The curve $P$ in Figure 4 is the probability per unit distance of finding the ground-state electron at distance $r$ from the nucleus: the chance of finding it between $r$ and $r+dr$ is $Pdr$. Three features stand out. $P$ is zero at $r=0$, because a thin shell of radius $r$ has a volume that shrinks to nothing as $r\to 0$. It peaks at exactly $a_0$, the radius Bohr gave. And it falls away to the right without ever reaching zero, so the cloud has no edge.

The mean distance, $\bar{r}=1.5a_0$, lies to the right of the peak, because the tail on the right is long while the rise on the left is short. Bohr's radius survives, but only as the most probable distance, not as a path.

Schrödinger's equation for the Coulomb potential (1926) gives exactly Bohr's energies $E_n$. What changes is the description of the states. The electron is a three-dimensional standing wave, and level $n$ contains $n^2$ different states with the same energy. They are labelled by the orbital angular-momentum number $l=0,1,\dots ,n-1$, and each $l$ comes in $2l+1$ orientations; adding them up, $1+3+5+\cdots +(2n-1)=n^2$. The ground state has $l=0$: no orbital angular momentum at all, where Bohr gave it $\hslash$.

The electron's spin has two possible orientations, which doubles the count to $2n^2$: 2, 8 and 18 states for $n=1$, 2 and 3. Together with the exclusion principle, which allows at most one electron in each state, these counts shape the periodic table. In atoms with many electrons, states with different $l$ no longer share one energy; once they separate, the rows of the table hold 2, 8, 8, 18, 18, 32 and 32 elements.

The same results hold for any ion with a single electron and nuclear charge $Ze$. Replacing $k{e}^2$ by $Zk{e}^2$ in the formulas above gives $E_n=-13.6\mathrm{eV}{Z}^2/n^2$ and $r_n=n^2{a}_0/Z$. For ${\mathrm{He}}^{+}$, with $Z=2$, the ground state lies at $-54.4\mathrm{eV}$.

One more correction: the nucleus is not infinitely heavy. Electron and nucleus both circle their common centre of mass, and the formulas stay the same if the electron mass $m$ is replaced by the reduced mass $\mu =mM/(m+M)$, where $M$ is the mass of the nucleus. For hydrogen this lowers the binding energies by 0.054 %; for deuterium, whose nucleus is twice as heavy, by only 0.027 %. Deuterium's lines therefore carry slightly more energy, and its Hα line lies 0.18 nm shorter than hydrogen's: the small shift that revealed deuterium in the early 1930s.

Worked problem: from the red line to radio waves

Bohr required his quantum rules to agree with classical physics wherever quantum effects become small: the correspondence principle. Classically, an orbiting charge radiates at its orbital frequency $f_{\mathrm{orb}}=v/(2\pi r)$, the number of trips round per second. This problem compares that frequency with the photon's, first low on the ladder and then high up.

Step 1, the orbital frequency of level $n$. With $v_n=\alpha c/n$ and $r_n=n^2{a}_0$,
$f_{\mathrm{orb}}=\frac{v_n}{2\pi r_n}=\frac{\alpha c}{2\pi a_0{n}^3}=\frac{6.58\times 10^{15}\mathrm{Hz}}{n^3}$

Step 2, the red line. The electron circles $2.44\times 10^{14}$ times a second on $n=3$ and $8.22\times 10^{14}$ times on $n=2$. The Hα photon of the jump $3\to 2$ has $f=c/\lambda =4.57\times 10^{14}\mathrm{Hz}$: between the two orbital frequencies, and equal to neither. Low on the ladder, a quantum jump has no classical counterpart.

Step 3, a jump high up, from $n=110$ to $n=109$. The photon energy is
$hf=13.6\mathrm{eV}\left(\frac{1}{109^2}-\frac{1}{110^2}\right)=2.07\times 10^{-5}\mathrm{eV}$
Dividing by $h=4.14\times 10^{-15}\mathrm{eV}\mathrm{s}$ gives $f=5.01\times 10^9\mathrm{Hz}$, and $\lambda =c/f\approx 6.0\mathrm{cm}$: a radio wave. The orbital frequencies are now 5.08 GHz on $n=109$ and 4.94 GHz on $n=110$, both within 1.5 % of the photon's frequency.

Step 4, why they merge. For a step $\Delta n=1$ at large $n$, the energy difference is close to the slope of $E_n$ with respect to $n$:
$hf\approx \frac{d{E}_n}{dn}=\frac{2\cdot 13.6\mathrm{eV}}{n^3}=\frac{27.2\mathrm{eV}}{n^3}$
And $h{f}_{\mathrm{orb}}=\hslash \alpha c/(a_0{n}^3)$ is exactly the same, because $\hslash /a_0=\alpha mc$, so $\hslash \alpha c/a_0={\alpha }^{2}m{c}^2=27.2\mathrm{eV}$. At large $n$ the quantum and classical frequencies merge, as the correspondence principle demands. Jumps with $\Delta n=2$ or 3 likewise give $2f_{\mathrm{orb}}$ and $3f_{\mathrm{orb}}$, the overtones that a classical elliptical orbit radiates.

Such giant atoms are real. The $n=110$ orbit has a radius of $110^2{a}_0\approx 0.64\mu \mathrm{m}$, the size of a small bacterium, so the atom survives only in very thin gas, where collisions are rare. In the ionised clouds around hot young stars, protons recapture electrons, which then cascade down the ladder step by step. Radio telescopes observe the resulting 5.01 GHz line, named H109α, from clouds that also glow red in Hα.

Limits of validity

Bohr's model is only a first approximation even for hydrogen, and for atoms with more than one electron it fails. Knowing where it stops tells you when the full theory is needed.

For hydrogen, relativity and the electron's spin shift the levels by relative amounts of order ${\alpha }^2\approx 5\times 10^{-5}$; this is expected, since the electron moves at $v=\alpha c$ and relativistic corrections grow as $(v/c{)}^2$. This fine structure splits the $n=2$ level by $4.5\times 10^{-5}\mathrm{eV}$ and turns Hα into components about 0.016 nm apart. Smaller still are the Lamb shift ($4.4\times 10^{-6}\mathrm{eV}$, from quantum electrodynamics) and the hyperfine splitting of the ground state ($5.9\times 10^{-6}\mathrm{eV}$, from the magnetic interaction of the electron and proton spins), whose 21 cm line maps hydrogen across the Galaxy.

In heavy one-electron ions the electron's speed, $Z\alpha c$ on the lowest orbit, approaches $c$, and relativity dominates. In ${\mathrm{U}}^{91+}$, $Z\alpha \approx 0.67$, and the ground state is bound by about 132 keV, against Bohr's $13.6\mathrm{eV}\cdot 92^2\approx 115\mathrm{keV}$.

With two or more electrons the model fails outright, as the note in the animation warns, because each electron also repels the others. For helium, ignoring that repulsion gives $2\cdot (-54.4)=-108.8\mathrm{eV}$ for the total energy, while the measured total is $-79.0\mathrm{eV}$.

Bohr's rules also say nothing about how bright a line is or how long a level lasts before it emits (hydrogen's 2p state, with $n=2$ and $l=1$, decays in 1.6 ns). And the orbits cannot be real paths: placing the electron on a circle of radius $a_0$ means knowing its position to within about $a_0$, which leaves its momentum uncertain by about $\hslash /a_0=m{v}_1$, its entire momentum. One effect the model can safely ignore is recoil: an atom emitting Hα recoils at only 0.60 m/s, which takes just $2\times 10^{-9}\mathrm{eV}$ from the photon.

Where this leads

The Bohr atom is where several lessons meet. Its photon rule applies to atoms the light quanta of Photoelectric Effect and of Blackbody Radiation and the Quantum. Its force balance is Coulomb's Law inside Circular Motion. Its quantised angular momentum brings Angular Momentum and Its Conservation down to the atomic scale. Wave–Particle Duality explains the allowed orbits as standing electron waves, and Schrödinger's Equation and Tunnelling replaces orbits by wave functions while keeping $E_n$.

In Semiconductors, Diodes and LEDs the levels of a crystal's atoms merge into bands, and an LED's colour is set by a band gap just as the colour of Hα is set by $E_3-E_2$. Radioactivity and Half-Life moves to the nucleus, whose levels lie thousands to millions of times further apart. In Particles and Forces: the Standard Model, $\alpha \approx 1/137$ returns as the strength of the electromagnetic interaction itself.