Bohr's model of hydrogen · printable packet

Bohr's rule $L = n\hbar$, radii $n^2a_0$, energies $-13.6/n^2$ eV, the derivation of Rydberg's formula, one-electron ions, and de Broglie's standing waves.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to compute Bohr radii, energies and speeds for hydrogen and one-electron ions, and derive hydrogen's spectral lines.

2. What you already have

From lesson 7 you know hydrogen's spectral lines and Rydberg's formula, found by fitting measurements. From mechanics you know circular motion, Coulomb's law and angular momentum; from the last unit, de Broglie's wavelength. This lesson shows how Bohr combined classical orbits with one quantum rule and derived Rydberg's formula from scratch.

3. Words for this lesson

4. One quantum rule on top of Newton

Bohr assumed that the electron in hydrogen moves in a circle held by the Coulomb force, as Newton's laws say, but that only orbits whose angular momentum is a whole number of $\hbar$ are allowed:

In those orbits, he postulated, the electron does not radiate, contrary to classical electromagnetism. Combining $L = n\hbar$ with $mv^2/r = ke^2/r^2$ gives

$$r_n = n^2a_0, \qquad E_n = -\frac{13.6\ \text{eV}}{n^2}, \qquad a_0 = 0.0529\ \text{nm}.$$

An electron jumping between levels emits or absorbs a photon of energy $E_{\text{upper}} - E_{\text{lower}}$, which reproduces Rydberg's formula with $R$ computed from $m$, $e$, $h$ and $c$. De Broglie later read the rule as a standing wave: exactly $n$ wavelengths fit around the orbit.

Term	What it means
Stationary state	An allowed orbit in which the electron does not radiate.
Principal quantum number	$n = 1, 2, 3, \ldots$, labeling Bohr's orbits.
Bohr radius	$a_0 = 0.0529$ nm, the radius of hydrogen's ground-state orbit.
Ground state	The lowest level, $n = 1$, with $E_1 = -13.6$ eV.
Ionization energy	The energy needed to remove an electron completely.
Hydrogen-like ion	An ion with one electron, such as He⁺ or Li²⁺.
Rydberg atom	An atom with one electron in a very high level.

Another way: picture

Picture a guitar string stretched into a circle. It can vibrate only in patterns that close on themselves smoothly: one wavelength around, two, three, never two and a half. Each pattern has its own energy. Bohr's orbits are those patterns for the electron's wave, and the jumps between them are the notes hydrogen can play.

Another way: steps

Pick the level $n$.
Radius: $r_n = n^2a_0$; speed: $v_n = 2.19 \times 10^6/n$ m/s.
Energy: $E_n = -13.6/n^2$ eV, or $-13.6Z^2/n^2$ for one electron around charge $Z$.
Photon: $\Delta E = E_{\text{upper}} - E_{\text{lower}}$ and $\lambda = 1240/\Delta E$ nm.
Ionization from level $n$: $|E_n|$.

5. The derivation

The Coulomb force supplies the centripetal force: $mv^2/r = ke^2/r^2$, so $mv^2 = ke^2/r$. Bohr's rule gives $v = n\hbar/mr$. Substituting, $m(n\hbar/mr)^2 = ke^2/r$, and solving, $r = n^2\hbar^2/mke^2$. The constant is the Bohr radius, $a_0 = \hbar^2/mke^2 = 0.0529$ nm.

The kinetic energy is $\tfrac{1}{2}mv^2 = ke^2/2r$ and the potential energy $-ke^2/r$, so the total is $E = -ke^2/2r = -13.6\ \text{eV}/n^2$. The kinetic energy is always minus the total, and the potential energy twice the total, a result called the virial theorem that holds for any inverse-square orbit, planets included.

6. Deriving Rydberg's formula

A drop from level $n_2$ to $n_1$ releases $\Delta E = 13.6(1/n_1^2 - 1/n_2^2)$ eV. Dividing by $hc$ gives $1/\lambda = R(1/n_1^2 - 1/n_2^2)$ with $R = 13.6\ \text{eV}/hc = 1.097 \times 10^7$ m⁻¹, exactly the constant Rydberg had fitted to measured lines thirty years earlier.

That agreement, from nothing but the electron's mass and charge and Planck's constant, made Bohr's model famous overnight. It explained why the Balmer lines exist, why they crowd toward a limit, and predicted the Lyman series in the ultraviolet, found by Theodore Lyman at Harvard, and the Paschen and Brackett series in the infrared.

7. Speeds and the fine-structure constant

The electron's speed in orbit $n$ is $v_n = \alpha c/n$, where $\alpha = ke^2/\hbar c \approx 1/137$ is the fine-structure constant. In the ground state, $v_1 = 2.19 \times 10^6$ m/s, less than one percent of light speed, so Newton's mechanics is a fair approximation for hydrogen.

For heavy one-electron ions the speed grows as $Z$: in uranium with $91$ electrons stripped, the last electron moves at about $0.67c$, and relativistic effects become large. Measuring the energy levels of such ions, as at Lawrence Livermore National Laboratory, tests quantum electrodynamics in extreme fields.

8. One-electron ions

Replace the proton by a nucleus of charge $Ze$ and the force grows by $Z$. The radii shrink to $n^2a_0/Z$ and the energies grow to $-13.6Z^2/n^2$ eV. Singly ionized helium, $Z = 2$, has a ground state of $-54.4$ eV; doubly ionized lithium, $Z = 3$, of $-122.4$ eV.

Helium ion lines were famously confused with hydrogen's at first: the He⁺ transition from $n = 4$ to $n = 3$ has almost the wavelength of a hydrogen line. Bohr's model predicted a tiny difference from the different nuclear masses, and measurements confirmed it, one more success.

9. The method, step by step, and how to check it

Checking an answer. Energies must be negative for bound states and approach zero at large $n$. Radii must grow as $n^2$. Photon energies from hydrogen's ground state must be at least $10.2$ eV. And larger $Z$ must mean smaller, more tightly bound orbits.

10. Why each step is allowed

Bohr's model is a hybrid: classical orbits with one quantum rule, plus the postulate that allowed orbits do not radiate. It works for hydrogen because the exact quantum solution happens to give the same energies. It was a stepping stone, not a final theory, and the next lesson shows its failures.

The formulas assume the nucleus stays fixed. In fact the electron and nucleus orbit their common center of mass; replacing the electron's mass by the reduced mass $m_eM/(m_e + M)$ shifts hydrogen's energies by about $0.05$ percent, enough to separate hydrogen's lines from those of deuterium, which is how deuterium was discovered by Harold Urey at Columbia in 1931.

11. Rydberg atoms

Atoms with one electron in a very high level, $n$ of $50$ to several hundred, behave almost exactly as Bohr's model predicts, because the outer electron sees the nucleus and inner electrons as a single positive charge far away. At $n = 100$ the orbit is about half a micrometer across, larger than many viruses.

Such atoms are exquisitely sensitive to electric fields, since their electron is barely bound. Researchers use them as detectors for single microwave photons and as qubits in quantum simulators, where lasers excite arrays of atoms to Rydberg states so that neighboring atoms interact strongly.

12. Standing waves around the orbit

De Broglie's picture gives Bohr's rule a meaning. An electron with momentum $mv$ has wavelength $h/mv$. If its wave is to close smoothly on itself around a circle of radius $r$, the circumference must hold a whole number of wavelengths: $2\pi r = n\lambda = nh/mv$, which rearranges to $mvr = nh/2\pi = n\hbar$.

Orbits that do not satisfy the rule would have waves that interfere destructively with themselves around the loop and so cannot persist. The picture is still semiclassical, since real electrons in atoms do not follow circles, but it shows why quantization appears wherever a wave is confined.

13. How the model was received

Bohr, a young Danish physicist working in Ernest Rutherford's laboratory in Manchester, published the model in three papers in 1913. It stood on a strange foundation: an electron circling a nucleus should, by Maxwell's equations, radiate continuously and spiral into the nucleus within a hundred-millionth of a second, and Bohr simply declared that in the allowed orbits it does not. Many physicists found the mixture of classical and quantum rules unsatisfying, and Bohr himself regarded it as provisional.

Yet its predictions were too good to ignore. It gave Rydberg's constant from fundamental constants, predicted the ultraviolet and infrared series before some were measured, explained the spectrum of ionized helium, and accounted for the characteristic X-rays of heavy elements, which Henry Moseley used in 1913 to put the periodic table in order by nuclear charge. Einstein called it the highest form of musicality in the sphere of thought. Bohr received the Nobel Prize in 1922, and the model remained the working picture of the atom until quantum mechanics replaced it in 1925 and 1926.

The model survives in teaching because its numbers are right for hydrogen and its logic is transparent: one quantization rule, applied to a classical orbit, turns a continuum of possible energies into a ladder. That idea, that confinement plus wave behavior yields discrete levels, is the core of everything that follows in atomic physics.

14. Characteristic X-rays

Bohr's formula also explains the X-rays that heavy elements give off when an inner electron is knocked out. An electron dropping from the second shell into the vacancy in the first sees the nucleus shielded by the one remaining inner electron, so roughly $Z - 1$ units of charge. Bohr's energy with that charge gives an X-ray energy of about $10.2(Z - 1)^2$ eV.

For copper, $Z = 29$, that is about $8.0$ keV, matching the copper K-alpha line used in laboratory X-ray machines. Henry Moseley measured these lines for dozens of elements in 1913 and found their square roots rose in even steps with atomic number, proving that each element's place in the periodic table is fixed by its nuclear charge and revealing gaps where undiscovered elements belonged.

15. In the world: Rydberg atoms as sensors

Researchers at the University of Michigan and at NIST in Boulder use atoms excited to Rydberg levels, $n$ around $50$ to $100$, as detectors of radio and microwave fields. With orbit radii of tenths of a micrometer and binding energies of millielectron volts, the outer electron responds strongly to tiny electric fields, shifting the atom's spectral lines.

By shining lasers through a small glass cell of cesium or rubidium vapor and watching the lines, these sensors measure field strengths traceable directly to Planck's constant, without a metal antenna. The U.S. Army and NIST have tested them as radio receivers that are hard to detect and cannot be damaged by strong signals.

16. In the world: hydrogen in space

Most of the ordinary matter in the universe is hydrogen, and much of what astronomers know about it comes from lines Bohr's model explains. The red Balmer line at $656.3$ nm, called H-alpha, lights up the glowing clouds where stars form, and filters tuned to it are standard equipment at observatories.

The Lyman alpha line at $121.6$ nm, blocked by Earth's atmosphere, is observed from space by instruments such as those on NASA's Hubble telescope. Clouds of hydrogen between us and distant quasars each absorb Lyman alpha at their own redshift, printing a "forest" of lines that maps the gas across billions of light-years.

17. Only the allowed orbits exist

Classical physics would let the electron orbit at any radius, and a circling charge should radiate and spiral inward, losing energy smoothly. Bohr's model forbids both: only the orbits with $L = n\hbar$ are allowed, the electron in them does not radiate, and it changes energy only by jumping between them, emitting or absorbing a photon each time.

A related error is to square the wrong thing: radii grow as $n^2$, energies shrink in size as $1/n^2$, and speeds as $1/n$.

18. Hydrogen's first excited state

19. The Balmer red line, derived

20. Singly ionized helium

21. Your turn: find the energy needed to ionize hydrogen from $n = 3$, and the longest wavelength that can do it.

22. Guided practice

In a hydrogen atom in its ground state, the electron's Bohr orbit has radius $0.0529$ nm. Atom number $9$ in a sample is excited to $n = 3$. What is its orbit radius?

23. Guided practice

Complete the worked solution: in hydrogen's Bohr orbit $n = 3$, with $a_0 = 0.0529$ nm, find the orbit radius in nm, its circumference in nm, and the electron's de Broglie wavelength in nm, using the condition that $n$ wavelengths fit around the orbit.

24. Guided practice

25. Practice

For hydrogen's Bohr level $n = 2$, with $a_0 = 0.0529$ nm, fill in the orbit radius in nm, the energy in eV, and the electron's speed in units of $10^6$ m/s.

26. Practice

Bohr's model applies to any atom with one electron. For doubly ionized lithium, Li²⁺, nuclear charge $Z = 3$, write the energy of level $n$, in eV, as a formula in $n$.

27. Practice

A one-electron ion with nuclear charge $Z = 3$ has its electron in level $n = 1$. How much energy is needed to remove the electron completely, in eV?

28. Somewhere new

Physicists at the University of Michigan excite atoms to Rydberg states, where one electron sits in a very high level. Treating it with Bohr's model at $n = 50$ and $a_0 = 0.0529$ nm, how large is its orbit radius, in μm?

29. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

30. Test question

Bohr's model applies to any atom with one electron. For doubly ionized lithium, Li²⁺, nuclear charge $Z = 3$, write the energy of level $n$, in eV, as a formula in $n$.

31. What you can do now

You can use Bohr's model. Explain to someone how one quantum rule for angular momentum predicts hydrogen's red spectral line.

Working for the steps left to you

	value
radius (nm)
energy (eV)
speed (10⁶ m/s)

21. Your turn: find the energy needed to ionize hydrogen from $n = 3$, and the longest wavelength that can do it., step 3

$\lambda = \dfrac{1240}{1.51} = 821\ \text{nm}$