Structure of Atom · Lesson 6 of 12

Bohr’s Model and Hydrogen Energy Levels

“Use stationary states to explain hydrogen spectra, excitation, and ionisation.”

Learning Objectives

• State Bohr’s postulates and explain why stationary states avoid classical radiative collapse within the model. • Calculate radii and energies for hydrogen and hydrogen-like ions. • Interpret negative energy, excitation, and ionisation using an energy-level diagram. • Derive transition frequencies from energy differences and connect them with spectral series. • Explain the model’s successes and limitations.

Allowed states replace arbitrary orbits

Rutherford’s nucleus explained scattering, but classical electron motion predicted collapse. Bohr introduced a new assumption: in a hydrogen atom, an electron may occupy only certain circular stationary orbits, each with a fixed radius and energy. While it remains in one allowed stationary state, it does not radiate energy. This is a postulate, not a conclusion obtained from classical electromagnetic theory.

Radiation is absorbed when an electron moves to a higher allowed state and emitted when it moves to a lower one. Its frequency is set by the energy difference, not by a continuous loss while the electron circles. The permitted states are labelled by the positive integer $n$ , the principal quantum number.

Bohr also restricted angular momentum. For a particle of mass $m_{e}$ moving tangentially at speed $v$ in a circle of radius $r$ , angular momentum has magnitude $m_{e} v r$ . This follows from moment of inertia $I = m_{e} r^{2}$ multiplied by angular speed $v / r$ . Only whole multiples of $h / (2 π)$ are allowed.

Bohr angular-momentum conditionLaTeX

m_{e} v r = \frac{nh}{2 π}, n = 1, 2, 3, \dots

The condition selects allowed stationary orbits in Bohr’s model. It is not the complete modern theory of atomic angular momentum.

Orbit size and energy

The model predicts that orbit radius grows with the square of $n$ . For hydrogen the first orbit has radius $a_{0} = 52.9 pm$ , called the Bohr radius. Doubling $n$ makes the radius four times as large, rather than twice as large.

The corresponding energy is negative relative to a free electron at rest infinitely far from the nucleus. We choose that separated state as zero energy. A bound electron has less energy, so energy must be supplied to free it. The most negative allowed value, at $n = 1$ , is the ground state; finite states with higher $n$ are excited states.

Hydrogen radii and energiesLaTeX

r_{n} = a_{0} n^{2}, E_{n} = - \frac{2.18 \times 1 0 ^{- 18}}{n ^{2}} J

a₀ = 52.9 pm. Eₙ is the energy of one hydrogen atom in state n with the stated zero of energy. The full dynamical derivation is outside this chapter’s scope.

Hydrogen energy levels— The energy gaps become smaller at larger n. The zero-energy level represents an ionised electron.

As $n$ approaches infinity, the electron’s energy approaches zero from below. The energy levels crowd together; they are not equally spaced. Ionisation from any bound level requires the positive magnitude of its binding energy. An electron already in an excited state needs less additional energy to escape than one in the ground state.

Example — A fifth orbit and its binding energy

Problem
Find the radius and energy of hydrogen in $n = 5$ .

1. $r_{5} = 52.9 \times 25 = 1322.5 pm = 1.3225 nm$ .
2. $E_{5} = - (2.18 \times 1 0^{- 18}) /25 = - 8.72 \times 1 0^{- 20} J$ .
3.Ionisation from this state needs $+ 8.72 \times 1 0^{- 20} J$ per atom. This is one twenty-fifth of the ground-state ionisation energy.

One-electron ions follow the same pattern

A hydrogen-like ion has one electron but a larger nuclear charge. Examples are helium with one electron, $He^{+}$ , and lithium with one electron, $Li^{2 +}$ . A neutral helium atom has two electrons and does not belong to this one-electron group.

Increasing nuclear charge strengthens attraction. For the same $n$ , the electron is more tightly bound and its predicted orbit is smaller. The energy magnitude increases as $Z^{2}$ , while radius decreases as $1/ Z$ . The qualitative orbital speed in Bohr’s model increases with $Z$ and decreases with $n$ .

Hydrogen-like ionsLaTeX

E_{n} = - 2.18 \times 1 0^{- 18} \frac{Z ^{2}}{n ^{2}} J, r_{n} = 52.9 \frac{n ^{2}}{Z} pm

Z is nuclear atomic number, not the positive charge written on the ion. These equations require exactly one electron.

Example — The first orbit of a helium ion

Problem
Calculate the energy and radius of the first Bohr orbit of $He^{+}$ .

1.Helium has $Z = 2$ and the first orbit has $n = 1$ .
2. $E_{1} = - 2.18 \times 1 0^{- 18} (2^{2}) = - 8.72 \times 1 0^{- 18} J$ .
3. $r_{1} = 52.9/2 = 26.45 pm$ . Relative to hydrogen at the same n, the binding-energy magnitude is four times larger and the radius is half as large.

Deriving the spectral relationship

A transition changes the atom’s energy by final energy minus initial energy. Substituting the allowed energy expressions gives a difference between reciprocal squares of integers. This is the origin of the pattern Rydberg found from measured wavelengths.

For excitation, the final level is higher and the atom’s energy change is positive. For emission, it is negative. Photon energy, however, is always positive: it equals the magnitude of the atomic energy change. Keeping these two signs separate avoids a negative photon frequency.

Atomic energy changeLaTeX

Δ E_{atom} = E_{f} - E_{i} = 2.18 \times 1 0^{- 18} Z^{2} (\frac{1}{n _{i}^{2}} - \frac{1}{n _{f}^{2}}) J

nᵢ and n𝒻 denote initial and final state. For emission nᵢ > n𝒻, giving a negative atomic energy change.

Photon frequency and wavenumberLaTeX

ν = \frac{∣Δ E _{atom} ∣}{h}, \tilde{ν} = R_{H} Z^{2} \frac{1}{n _{i}^{2}} - \frac{1}{n _{f}^{2}}

For hydrogen, the energy constant divided by h is approximately 3.29 × 10¹⁵ s⁻¹. The energy constant in joules and the Rydberg wavenumber constant have different units.

Transitions ending at $n = 1$ form the Lyman emission series; those ending at 2 form Balmer; those ending at 3 form Paschen. Higher starting levels approach a series limit. In a collection of excited atoms, many different transitions can occur. Line intensity depends partly on how many photons of that wavelength are emitted, not simply on the energy of each photon.

Example — A transition from the fifth to the second state

Problem
Find the frequency and wavelength emitted by hydrogen during $n = 5 \to 2$ .

1. $Δ E_{atom} = 2.18 \times 1 0^{- 18} (1/25 - 1/4) = - 4.578 \times 1 0^{- 19} J$ .
2.The emitted photon carries $4.578 \times 1 0^{- 19} J$ . Therefore $ν = E_{γ} / h \approx 6.91 \times 1 0^{14} Hz$ .
3. $λ = c / ν \approx 4.34 \times 1 0^{- 7} m = 434 nm$ . The final state is 2, so this is a Balmer line.

Example — Exciting and then ionising hydrogen

Problem
How much energy excites ground-state hydrogen to $n = 2$ , and how much more is needed to ionise it from there?

1. $E_{1} = - 2.18 \times 1 0^{- 18} J$ and $E_{2} = - 5.45 \times 1 0^{- 19} J$ .
2.Excitation requires $E_{2} - E_{1} = 1.635 \times 1 0^{- 18} J$ .
3.Ionising from state 2 requires $0 - E_{2} = 5.45 \times 1 0^{- 19} J$ . The two inputs add to $2.18 \times 1 0^{- 18} J$ , the direct ground-state ionisation energy.

What Bohr’s model can and cannot explain

Bohr’s model explains the main hydrogen and hydrogen-like-ion energy pattern and the corresponding line frequencies. It gives a clear account of ground states, excitation, emission, and ionisation. Its success is specific: it does not justify applying the same single-electron formula to atoms with several electrons.

It cannot fully explain fine spectral structure, the spectra of multi-electron atoms, splitting in magnetic fields (the Zeeman effect) or electric fields (the Stark effect), or chemical bonding. The assumed definite circular path also ignores the electron’s wave character and conflicts with the uncertainty principle. The next lesson develops those deeper reasons for replacing the orbit picture.

Common mistake

A negative bound-state energy does not mean negative kinetic energy. It is the total energy relative to a chosen zero. Also, use the absolute energy difference for photon frequency, and use the nuclear atomic number Z rather than the ion’s charge in hydrogen-like formulas.

Quick check

Which species can be treated by the one-electron Bohr energy formula?

Quick check

What does a negative energy for bound hydrogen mean?

Quick check

For hydrogen, increasing n from 1 to 2 changes the predicted radius by a factor of:

Quick check

During emission from a higher to a lower level:

Quick check

Which statement about high-n hydrogen levels is correct?

Practice Problems

Find the radius and energy of the third orbit of hydrogen.
Calculate the wavelength for $n = 4 \to 2$ emission in hydrogen and identify its series.
Find the ground-state ionisation energy of $Li^{2 +}$ .
Compare hydrogen ionisation energies from $n = 1$ , $n = 2$ , and $n = 5$ .
Show algebraically how subtracting two Bohr energies produces a reciprocal-square spectral expression.
For atoms initially excited to $n = 6$ , determine the maximum number of different downward transition lines possible across an ensemble. Explain the counting.
Explain two experimental limitations of Bohr’s model and one conceptual limitation.

Key Takeaways

Bohr postulated non-radiating stationary states and quantised angular momentum. For one-electron species, energy scales as −Z²/n² and orbit radius as n²/Z. Excitation raises atomic energy; emission lowers it; ionisation reaches the separated zero-energy reference. Photon energy is the positive magnitude of a transition’s energy difference. Bohr’s model explains the main one-electron spectral pattern but is not a general model for all atoms.

Previous · Lesson 5

Atomic Spectra and the Hydrogen Pattern

Next · Lesson 7

Matter Waves and the Limits of Orbits

Allowed states replace arbitrary orbits

Orbit size and energy

One-electron ions follow the same pattern

Deriving the spectral relationship

What Bohr’s model can and cannot explain

Quiz

Practice Problems

Key Takeaways