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The hydrogen atom

Separating the Coulomb problem, energies $-13.6/n^2$ eV with $n^2$ states per level, radial probabilities, spectral series, and hydrogen-like ions.

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 find hydrogen's energies and degeneracies, predict its spectral lines, compute radial probabilities for its ground state, and scale the results to other one-electron systems.

2. What you already have

You know the Bohr model's energies $-13.6\ \text{eV}/n^2$ from modern physics, the angular momentum eigenvalues $l(l + 1)\hbar^2$ and $m\hbar$, and the spherical harmonics from the last lesson. You can compute probabilities from wave functions. This lesson solves the Schrödinger equation for the Coulomb potential and shows where Bohr's energies really come from and what the electron's state looks like.

3. Words for this lesson

4. Separating the Coulomb problem

The Coulomb potential $V = -e^2/(4\pi\varepsilon_0r)$ depends only on $r$, so the Hamiltonian commutes with $\hat{L}^2$ and $\hat{L}_z$, and the states separate as

$$\psi_{nlm}(r, \theta, \phi) = R_{nl}(r)\,Y_l^m(\theta, \phi).$$

Writing $u = rR$, the radial function obeys a one-dimensional Schrödinger equation with an effective potential

$$V_{\text{eff}}(r) = -\frac{e^2}{4\pi\varepsilon_0 r} + \frac{\hbar^2}{2m}\frac{l(l + 1)}{r^2},$$

the Coulomb attraction plus a centrifugal barrier that keeps states with $l > 0$ away from the nucleus. Solving it with the requirement that $u$ stay finite at infinity forces the series solution to terminate, and that happens only when

$$E_n = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2\hbar^2}\frac{1}{n^2} = -\frac{13.6\ \text{eV}}{n^2}, \qquad l = 0, 1, \ldots, n - 1.$$

The energy depends on $n$ alone, so each level holds $\sum(2l + 1) = n^2$ orbital states. The length scale is the Bohr radius, $a = 0.0529$ nm, and the ground state is

$$\psi_{100} = \frac{1}{\sqrt{\pi a^3}}e^{-r/a},$$

a spherical cloud, densest at the nucleus, whose radial probability density $4\pi r^2|\psi|^2$ peaks at $r = a$. Its average distance is $\tfrac{3}{2}a$, and states of higher $n$ are larger, with sizes that grow as $n^2a$.

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

1. Energies and lines. Use $E_n = -13.6\ \text{eV}\,Z^2/n^2$. A photon from $n_i$ to $n_f$ has energy $E_{n_i} - E_{n_f}$; for wavelengths use $1/\lambda = R_HZ^2(1/n_f^2 - 1/n_i^2)$ or $\lambda = 1239.8\ \text{eV nm}/E_\gamma$.
2. Quantum numbers. For each $n$, list $l$ from $0$ to $n - 1$ and $m$ from $-l$ to $l$. Spectroscopic letters: $s, p, d, f$ for $l = 0, 1, 2, 3$.
3. Probabilities. Square the wave function and multiply by the volume element $r^2\sin\theta\,dr\,d\theta\,d\phi$. For radial questions, integrate the angles out, leaving $r^2|R|^2\,dr$.
4. Scaling. For a hydrogen-like ion, energies scale as $Z^2$ and sizes as $1/Z$.

Checks. Bound energies are negative, approaching zero as $n \to \infty$. The Lyman series ($n_f = 1$) lies in the ultraviolet, the Balmer series ($n_f = 2$) in the visible, the Paschen series ($n_f = 3$) in the infrared. Every probability integral over all space must give one, and the ground state's most probable radius is $a$, not zero.

6. Why the energy depends only on $n$

For a general central potential, the energy depends on both $n$ and $l$, because the centrifugal term changes with $l$. For the $1/r$ potential alone, states with different $l$ but the same $n$ have exactly the same energy. The reason is a hidden symmetry: besides angular momentum, the Coulomb problem conserves the Laplace–Runge–Lenz vector, which in classical mechanics points along the ellipse's major axis and keeps Kepler orbits from precessing. Quantum mechanically, the extra conserved quantity links states of different $l$, forcing them to share an energy.

The degeneracy is broken by anything that spoils the pure $1/r$ form. In atoms with more electrons, the inner electrons screen the nucleus, and states of low $l$, which penetrate the screening, are pulled lower: that is why $4s$ fills before $3d$ in the periodic table. In hydrogen itself, relativity and spin split levels of the same $n$ by about $10^{-4}$ eV, the fine structure.

7. The radial probability and where the electron is

The ground-state density $|\psi_{100}|^2 = e^{-2r/a}/\pi a^3$ is largest at the nucleus. But the probability of finding the electron at a distance between $r$ and $r + dr$ includes the volume of a thin shell, $4\pi r^2dr$, which vanishes at the center. The product, $P(r) = (4r^2/a^3)e^{-2r/a}$, is zero at the nucleus, rises, peaks where $dP/dr = 0$ at $r = a$, and decays.

The average distance is larger than the peak, $\langle r\rangle = \tfrac{3}{2}a$, because the tail stretches out. Integrating, the chance of finding the electron inside the Bohr radius is only $32$ percent, and beyond $2a$ it is still $24$ percent. For the $2p$ state, $P(r) \propto r^4e^{-r/a}$ peaks at $4a$, the Bohr-model radius for $n = 2$; in general the state with $l = n - 1$ peaks at $n^2a$, which is why Bohr's radii are right for those states.

8. Spectral series

Transitions end on a given level form a series. The Lyman series, ending on $n = 1$, runs from $121.6$ nm to $91.2$ nm, in the ultraviolet; the Lyman-alpha line at $121.6$ nm is the brightest line of hydrogen in space and is seen in the spectra of distant quasars, stretched by the expansion of the universe into the visible. The Balmer series, ending on $n = 2$, gives the four visible lines at $656$, $486$, $434$ and $410$ nm that color hydrogen discharge tubes and the Orion Nebula red.

The Rydberg constant used here, $1.0968 \times 10^{7}$ m⁻¹, is slightly smaller than the infinite-mass value $1.0974 \times 10^{7}$ m⁻¹ because the proton moves too: the electron's mass is replaced by the reduced mass. Deuterium's heavier nucleus shifts its lines by about $0.18$ nm, the difference by which Harold Urey discovered deuterium in 1931.

9. Hydrogen-like ions

Any nucleus with a single electron is a hydrogen atom scaled by its charge $Z$. The potential is $Z$ times stronger, so the Bohr radius shrinks to $a/Z$ and the energies grow to $-13.6Z^2/n^2$ eV. Singly ionized helium, $Z = 2$, has a ground energy of $-54.4$ eV; lithium's last electron, $Z = 3$, is bound by $122.4$ eV.

For heavy nuclei these energies reach the x-ray range. Uranium stripped to one electron, $Z = 92$, would have $|E_1| = 13.6 \times 92^2 = 115$ keV by this formula; the true value, about $132$ keV, is larger because the electron moves at over half the speed of light and relativity matters. Such ions are made in heavy-ion storage rings and in the hottest plasmas, where their lines measure temperatures of tens of millions of kelvins in solar flares and fusion experiments.

10. The electron is not in an orbit

The Bohr model pictured the electron circling at radius $n^2a$ with angular momentum $n\hbar$. The Schrödinger solution corrects both pictures. The ground state has $l = 0$: zero orbital angular momentum, not $\hbar$. There is no path, only a probability cloud, and that cloud is spherically symmetric, with no preferred plane of motion.

Bohr's energies survived because they depend only on $n$, and his model happens to get $n$ right. His radii and angular momenta do not. Evidence for the quantum picture is direct: a nonzero density at the nucleus lets the electron interact with the proton's magnetic moment in $s$ states, producing the $21$ cm hyperfine line; and electron capture, where a nucleus absorbs one of its own $1s$ electrons, is possible only because $s$ electrons are sometimes found inside the nucleus.

11. In the world: neutral-atom quantum computers

Companies such as QuEra, Atom Computing and Pasqal, and university groups at Harvard and Caltech, build quantum computers from hundreds of neutral atoms held in arrays of laser tweezers a few micrometers apart. To make two atoms interact, a laser excites them to a Rydberg state with $n$ around $50$ to $100$. The hydrogen scaling tells why this works. Such an atom has a radius of order $n^2a$: for $n = 70$, $4900 \times 0.0529 = 260$ nm, thousands of times larger than an ordinary atom, and its huge electric dipole makes it interact with a neighbor several micrometers away.

The interaction is so strong that, within a certain distance, two atoms cannot both be excited — the Rydberg blockade — and that conditional behavior is a two-qubit logic gate. The binding energy, $13.6/n^2$ eV, is only about $2.8$ meV at $n = 70$, so the lasers must be tuned very precisely, and the states are fragile: stray electric fields and blackbody radiation can ionize them, limiting their lifetimes to around $100$ microseconds.

12. In the world: the redshift of quasars

Hydrogen's Lyman-alpha line, the $n = 2 \to 1$ transition at $121.6$ nm, lies deep in the ultraviolet, blocked by Earth's atmosphere. Yet ground-based telescopes see it in the spectra of distant quasars, because the expansion of the universe has stretched it into the visible. A quasar whose Lyman-alpha line appears at $486.4$ nm has a redshift $z = 486.4/121.6 - 1 = 3.0$: its light left when the universe was a quarter of its present size.

Between the quasar and us, every cloud of neutral hydrogen absorbs at its own redshifted Lyman-alpha wavelength, printing a thicket of absorption lines called the Lyman-alpha forest. The Dark Energy Spectroscopic Instrument, on the Mayall Telescope at Kitt Peak in Arizona, has measured this forest toward hundreds of thousands of quasars to map the distribution of matter billions of years ago and measure the expansion history of the universe. All of it rests on the energy difference $\tfrac{3}{4} \times 13.6$ eV between hydrogen's two lowest levels.

13. The electron does not orbit at the Bohr radius

The Bohr model's picture of an electron circling at $0.0529$ nm is still widespread, but the ground state has no orbit: it is a spherical probability cloud with zero angular momentum, densest at the nucleus. The Bohr radius is only the most probable distance, the peak of $4\pi r^2|\psi|^2$. The electron is found inside that radius about $32$ percent of the time and outside it about $68$ percent, and there is a small but real probability of finding it inside the proton.

A second misconception is that the energy depends on the shape of the orbital. In hydrogen, $2s$ and $2p$, or $3s$, $3p$ and $3d$, have identical energies. Only in atoms with more than one electron does screening split them, which is where the familiar filling order of chemistry comes from.

14. The Balmer-alpha line

1. Find the energy of level $n = 3$.

   $E_3 = -\dfrac{13.6}{9} = -1.511\ \text{eV}$

   $E_n = -13.6/n^2$ eV.

2. Find the energy of level $n = 2$.

   $E_2 = -\dfrac{13.6}{4} = -3.400\ \text{eV}$

   The lower level of the Balmer series.

3. Find the photon energy.

   $E_\gamma = E_3 - E_2 = -1.511 + 3.400 = 1.889\ \text{eV}$

   The energy released in the drop.

4. Convert to a wavelength.

   $\lambda = \dfrac{1239.8\ \text{eV nm}}{1.889\ \text{eV}} = 656.3\ \text{nm}$

   $\lambda = hc/E$.

5. Identify the color.

   $656\ \text{nm} \approx \text{red}$

   The H-alpha line that makes emission nebulae glow red.

15. The ground state's most probable and average radius

1. Write the radial probability density.

   $P(r) = 4\pi r^2|\psi_{100}|^2 = \dfrac{4r^2}{a^3}e^{-2r/a}$

   Density times shell area.

2. Differentiate to find the peak.

   $\dfrac{dP}{dr} = \dfrac{4}{a^3}e^{-2r/a}\left(2r - \dfrac{2r^2}{a}\right)$

   Apply the product rule.

3. Set the derivative to zero.

   $2r - \dfrac{2r^2}{a} = 0 \quad\Rightarrow\quad r = a$

   The most probable radius is the Bohr radius.

4. Set up the average radius.

   $\langle r\rangle = \displaystyle\int_0^\infty r\,P(r)\,dr = \dfrac{4}{a^3}\int_0^\infty r^3e^{-2r/a}\,dr$

   Weight each radius by its probability.

5. Use the standard integral.

   $\displaystyle\int_0^\infty r^3e^{-2r/a}\,dr = \dfrac{3!}{(2/a)^4} = \dfrac{6a^4}{16}$

   $\int_0^\infty x^ne^{-bx}\,dx = n!/b^{n+1}$.

6. Evaluate the average.

   $\langle r\rangle = \dfrac{4}{a^3} \cdot \dfrac{6a^4}{16} = \dfrac{3}{2}a = 0.0794\ \text{nm}$

   Half again the most probable radius, pulled out by the tail.

16. Counting and labeling the $n = 3$ states

1. List the allowed values of $l$.

   $l = 0, 1, 2 \quad (3s, 3p, 3d)$

   $l$ runs up to $n - 1 = 2$.

2. Count the $m$ values for $3s$.

   $l = 0: \quad m = 0 \quad (1 \text{ state})$

   $2l + 1 = 1$.

3. Count the $m$ values for $3p$.

   $l = 1: \quad m = -1, 0, 1 \quad (3 \text{ states})$

   $2l + 1 = 3$.

4. Count the $m$ values for $3d$.

   $l = 2: \quad m = -2, \ldots, 2 \quad (5 \text{ states})$

   $2l + 1 = 5$.

5. Add the orbital states.

   $1 + 3 + 5 = 9 = 3^2$

   The $n^2$ rule.

6. Include the spin states.

   $2 \times 9 = 18$

   Two spin states each: the third shell holds up to $18$ electrons.

7. Find the common energy.

   $E_3 = -\dfrac{13.6}{9} = -1.51\ \text{eV}$

   All eighteen share it, in hydrogen.

8. Find the size of these states.

   $r \sim n^2a = 9 \times 0.0529 = 0.48\ \text{nm}$

   Nine times the ground state's.

17. Your turn: how much energy does it take to ionize a hydrogen atom from its $n = 2$ level?

1. Write the energy of the level.

   $E_2 = -\dfrac{13.6}{2^2}$

   Ionization raises the electron to $E = 0$.

2. Evaluate the energy.

   $E_2 = -3.40\ \text{eV}$

   A quarter of the ground-state binding.

3. Your turn: work this step out. Its working is at the end of the packet.

   State the ionization energy.

18. Guided practice

Ignoring spin, how many orbital states $|n, l, m\rangle$ of hydrogen have the energy $E_{4}$?

• $16$
• $9$
• $4$
• $1$, since each energy has a single state

19. Guided practice

Complete the worked solution: a hydrogen-like ion with $Z = 5$ drops from $n = 2$ to $n = 1$. Find the magnitudes of the two energies and the photon energy.

1. Find the magnitude of the ground energy.

   $|E_1| = 13.6 \times 5^2 =$ _a eV

   The ionization energy of the ion.

2. Find the magnitude of the first excited energy.

   $|E_2| = \dfrac{13.6 \times 5^2}{4} =$ _b eV

   A quarter of the ground energy.

3. Subtract to find the photon energy.

   $E_\gamma = |E_1| - |E_2| =$ _c eV

   Three quarters of the ionization energy.

20. Guided practice

Match each property of hydrogen to its value.

21. Practice

A hydrogen-like ion has nuclear charge $Z = 3$ and one electron. Fill in the energy, in eV, and the number of orbital states for $n = 1$, $2$ and $4$.

22. Practice

What is the wavelength, in nm, of the photon a hydrogen atom emits when its electron drops from $n = 6$ to $n = 2$? Use $R_H = 1.0968 \times 10^{7}$ m⁻¹.

Answer: nm

23. Practice

A hydrogen atom is in its ground state, $\psi_{100} = e^{-r/a}/\sqrt{\pi a^3}$. What is the probability that the electron is found within a distance $b$ of the nucleus, where $b/a = 1$?

Answer:

24. Somewhere new

Neutral-atom quantum computers entangle atoms by exciting them to Rydberg states of very high $n$. Treating the excited electron like hydrogen's, with orbital size $n^2a$ and $a = 0.0529$ nm, how large is an atom excited to $n = 44$, in nm?

Answer: nm

25. Lesson test

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

26. Test question

A hydrogen-like ion has nuclear charge $Z = 5$ and one electron. Fill in the energy, in eV, and the number of orbital states for $n = 1$, $2$ and $4$.

27. What you can do now

You can solve problems about the hydrogen atom. Explain to someone why the Bohr radius is not the size of an orbit.

Working for the steps left to you

17. Your turn: how much energy does it take to ionize a hydrogen atom from its $n = 2$ level?, step 3