Bohr Energy Levels (Hydrogen)

Equivalent forms

E_n = -\frac{m_e e^4}{8\epsilon_0^2 h^2 n^2}

\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

An integer n predicts every line of the hydrogen spectrum to spectacular accuracy.

Symbol	Name	SI unit	Dimension	Range
$E_n$	Energy of level noutput Bound-state energy of electron in level n	J	M·L²·T⁻²	-2.18e-18 – 0
$n$	Principal quantum number Energy level index	dimensionless	1	1 – 20

Unit systems

Symbol	SI	CGS	Imperial
$E_n$	J	erg	eV
$n$	dimensionless	dimensionless	dimensionless

Where it holds

Exact for one-electron hydrogenic atoms (H,

He^{+}

,

Li^{2+}

, ...) ignoring fine structure. Multi-electron atoms require quantum mechanics with electron-electron interactions; relativistic corrections are needed for heavy nuclei.

Dimensional analysis

$[E_n] = [m_e e^{4}]/[\varepsilon _0^{2} h^{2}]$ → energy $\checkmark$ (Rydberg constant has units of inverse length when expressed via $1/\lambda )$

Discovery

Niels Bohr · 1913

Bohr postulated quantized orbits to explain hydrogen's discrete spectrum, ushering in old quantum theory.

Try this

Why are atomic spectra made of sharp lines instead of a rainbow?

Find the energy of the n=2 level of hydrogen and the wavelength of the n=2 → n=1 transition.

Research status: stable

Real-world applications

Stellar spectroscopy — chemical composition and redshift of stars and galaxies.
Hydrogen masers and atomic clocks.
Lasers (e.g., HeNe, hydrogen plasma lines).
Astrophysical 21 cm hydrogen line — used to map galactic structure.

Common misconceptions

Electrons do not literally orbit the nucleus — quantum mechanics gives probability clouds, not classical paths.
The Bohr model fails for atoms with more than one electron (ignores electron-electron repulsion).
Negative energy means *bound*, not negative kinetic energy. Ionization sets $E = 0$ .

Experimental verification

Hydrogen Balmer series

(n\to 2)

measured

by \text{Å}

ngström and Balmer in the 1880s; fits

1/\lambda = R(1/n_f^{2} - 1/n_i^{2})

with

R = 1.097\times 10^{7} m^{-1} to 8

decimal places. Lyman, Paschen, Brackett, and Pfund series confirmed the formula. Spectral lines of stars are how we know the universe is made mostly of hydrogen.

Derivation

Bohr postulated quantized angular momentum

L = n\hbar for

circular orbits.

Combining centripetal force = Coulomb attraction

(m_e v^{2}/r = ke^{2}/r^{2})

with

L = m_e v r = n\hbar

yields

r_n = n^{2}a_0

(a_

0 \approx 0.529 \text{Å})

and

E_n = -ke^{2}/(2r_n) = -13.6\,\mathrm{eV/n}^{2}

.

The full Schrödinger solution in a Coulomb potential reproduces the same energies.

Limiting cases

n = 1

⟶

E_1 = -13.6\,\mathrm{eV}

Ground state — most tightly bound; ionization energy of hydrogen.

n \to \infty

⟶

E_n \to 0

Continuum threshold — electron is just barely free (ionization).

Z > 1 (hydrogenic ions)⟶

E_n = -13.6\,\mathrm{Z}^{2}/n^{2} eV

Higher nuclear charge binds the electron

Z^{2}

more tightly (e.g.,

He^{+}

:

-54.4\,\mathrm{eV}

ground state).

What if…

What if the nucleus

had Z = 2

(helium

ion He^{+})

?

$E_n = -13.6 \times Z^{2}/n^{2} = -54.4/n^{2} eV$ . Ground state is $4\times$ deeper, ionization energy 54.4 eV — observable in stellar coronas.

What if the electron were replaced by a muon

(200\times

heavier)?

Muonic hydrogen has E_n scaled by $m_\mu /m_e \approx 207$ → ground state $\approx -2.8\,\mathrm{keV}$ , and Bohr radius $\sim 200\times$ smaller. Used to probe proton charge radius.

What if n were not restricted to integers?

The classical orbits would form a continuum, atoms would be unstable, and the sharp spectral lines we observe would be impossible — quantization is essential.

1

Energy of n = 2 level

Given ·

n:: 2

Find · E_2

Steps

$E_n = -13.6\,\mathrm{eV} / n^{2}$
$n = 2 \to E_2 = -13.6/4 = -3.4\,\mathrm{eV}$
Less tightly bound than ground state by 10.2 eV.

Result ·

E_2 = -13.6/4 = -3.4\,\mathrm{eV}

2

n=2 → n=1 transition wavelength (Lyman α)

Given ·

n i:: 2
n f:: 1

Find ·

\lambda

Steps

$\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2\,\mathrm{eV}$
$\lambda (nm) = 1240/\Delta E(eV) = 1240/10.2 \approx 121.6\,\mathrm{nm}$
This is Lyman- $\alpha$ — the brightest UV line in the universe.

Result ·

\Delta E = 10.2\,\mathrm{eV} \to \lambda = 1240/10.2 \approx 121.6\,\mathrm{nm} (UV)

Photoelectric Effect→Schrödinger Equation (Time-Independent)→