Bohr Hydrogen Energy Levels

Equivalent forms

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

Integer quantum numbers determine atomic spectra.

Symbol	Name	SI unit	Dimension	Range
$E_n$	Energy of level noutput Bound-state energy of the nth orbit	J	ML^2T^-2	-2.18e-18 – 0
$n$	Principal quantum number Integer orbit index (1, 2, 3, ...)	dimensionless	1	1 – 20

Unit systems

Symbol	SI	CGS	Imperial
$E_n$	J	erg	ft*lbf
$n$	dimensionless	dimensionless	dimensionless

Where it holds

Valid for hydrogen and hydrogen-like ions (one electron). Fine structure, hyperfine splitting, and Lamb shift require relativistic/QED corrections.

Dimensional analysis

E_n \to [M*L^2*T^-2]

[M*L^2*T^-2] / (dimensionless

) \to [M*L^2*T^-2]

Both sides are energy.

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Discovery

Niels Bohr · 1913

Bohr combined Rutherford's nuclear model with Planck quantization to explain hydrogen's emission spectrum.

Try this

Why does hydrogen glow only in specific colors?

Electrons in hydrogen occupy quantized orbits with discrete energies. What is the energy of the n=2 level?

Research status: stable

Real-world applications

Astronomical spectroscopy (redshift measurements)
Hydrogen maser clocks
Plasma diagnostics
Derivation of Rydberg constant R_H

Common misconceptions

Electrons do NOT actually orbit in circles — this is a semiclassical approximation
The minus sign reflects that bound states are below the zero of free-electron energy
Bohr model fails for multi-electron atoms — full QM is required

Experimental verification

Hydrogen emission spectrum (Balmer, Lyman, Paschen series) matches to 4+ significant figures. Franck-Hertz experiment (1914) directly demonstrated quantized atomic energy levels.

Derivation

Bohr postulated quantized angular momentum

L = n

*hbar.

Combining Coulomb attraction

F = e^2/(4*pi*\varepsilon _0*r^2)

with circular motion

m_e*v^2/r = F

, and using

L = m_e*v*r = n

*hbar, solve for allowed radii

r_n = n^2

*a_0 (Bohr radius a_

0 \approx 5.29e-11\,\mathrm{m})

.

Total energy

E_n = KE + PE = -m_e*e^4/(8*\varepsilon _0^2*h^2*n^2) \approx -13.6\,\mathrm{eV} / n^2

.

Limiting cases

n = 1

⟶

E_1 = -13.6\,\mathrm{eV}

Ground state — the ionization energy of hydrogen.

n = 2

⟶

E_2 = -3.4\,\mathrm{eV}

First excited state — source of Balmer series transitions.

n \to \infty

⟶

E_n \to 0

Continuum limit — the electron is free.

What if…

What if

Z = 2

(He+)?

$E_n = -13.6*Z^2/n^2 = -54.4/n^2\,\mathrm{eV}$ — ionization energy quadruples.

What if

n \to \infty

?

$E_n \to 0$ , continuum states (ionized hydrogen).

What if the electron were heavier?

Muonic hydrogen has $m \approx 207*m_e$ — all energies scale by 207, Bohr radius shrinks by 207.

1

n=2 energy

Given ·

n:: 2

Find · E_2

Steps

$E_n = -13.6/n^2\,\mathrm{eV}$
$E_2 = -13.6/4 = -3.4\,\mathrm{eV}$
Convert: - $3.4 * 1.602176634e-19 \approx -5.45e-19\,\mathrm{J}$

Result ·

E_2 = -13.6/4 = -3.4\,\mathrm{eV} = -5.45e-19\,\mathrm{J}

2

Balmer alpha (n=3 → n=2)

Given ·

n i:: 3
n f:: 2

Find · lambda_photon

Steps

$\Delta _E = E_2 - E_3 = -3.4 - (-1.51) = -1.89\,\mathrm{eV}$ (emitted)
$E = hc/\lambda \to \lambda = hc/|\Delta _E$ |
$\lambda = 1240\,\mathrm{eV}*nm / 1.89\,\mathrm{eV} \approx 656\,\mathrm{nm}$

Result ·

\lambda \approx 656.3\,\mathrm{nm} (red H-\alpha

line)

Photon Energy (Planck Relation)→de Broglie Wavelength→Time-Dependent Schrödinger Equation→