The Bohr Model and Energy Levels玻尔模型与能级

Big Idea · Structure and Properties (SAP)AP Science Practice 1 · Models and RepresentationsAP link · Unit 1~35 min

Learning goals

  • State the main ideas of the Bohr model and distinguish the ground state from excited states
  • Use Eₙ = −2.18 × 10⁻¹⁸ J/n² to calculate the energy and wavelength of the photon absorbed or emitted when hydrogen's electron changes levels
  • Explain hydrogen's line spectrum with electron transitions, and describe the limits of the Bohr model

Before you start

  • Light and spectra (2.2.2)

Hydrogen’s four lines

Run electricity through a tube of hydrogen gas and it glows a pinkish purple. Spread that light with a prism, and the visible part is just four bright lines: red at 656 nm, blue-green at 486 nm, blue-violet at 434 nm, and violet at 410 nm. Why those four wavelengths and no others? In 1913, Niels Bohr found the answer.

Bohr’s model

  1. The electron in a hydrogen atom can move only in certain allowed orbits, each with a definite energy, called an energy level能级 and numbered n = 1, 2, 3, and so on. An electron in one of these orbits doesn’t radiate energy.
  2. The lowest-energy state (n = 1) is the ground state基态; any higher-energy state is an excited state激发态.
  3. When the electron jumps from one level to another, it absorbs or emits one photon whose energy equals the energy difference between the levels:
hν=∣ΔE∣=∣Efinal−Einitial∣h\nu = |\Delta E| = |E_{\text{final}} - E_{\text{initial}}|

Moving up to a higher level absorbs a photon; dropping to a lower level emits one.

Hydrogen’s energy levels

Bohr calculated the energy of hydrogen’s nth level:

En=−2.18×10−18 Jn2E_n = -\frac{2.18 \times 10^{-18}\ \text{J}}{n^2}

The energies are negative because zero is defined as an electron infinitely far from the nucleus (n = ∞, an ionized atom); an electron held by the nucleus has less energy than that. The larger n is, the higher the energy, and the closer together neighboring levels get.

For a jump from an initial to a final level, the energy change is:

ΔE=Efinal−Einitial=2.18×10−18 J×(1ninitial2−1nfinal2)\Delta E = E_{\text{final}} - E_{\text{initial}} = 2.18 \times 10^{-18}\ \text{J} \times \left(\frac{1}{n_{\text{initial}}^2} - \frac{1}{n_{\text{final}}^2}\right)

A positive ΔE means energy is absorbed; a negative ΔE means energy is released.

Explaining hydrogen’s spectrum

When hydrogen atoms are excited, their electrons drop back from higher levels to lower ones, emitting light of particular wavelengths. Drops that end at n = 2 (the Balmer series) happen to give visible light: those are the four lines. Drops to n = 1 (the Lyman series) release more energy and fall in the ultraviolet; drops to n = 3 (the Paschen series) fall in the infrared.

Because the levels are separate, the energy differences can take only certain values, so the spectrum is a set of separate lines instead of a continuous band. In reverse, when white light passes through hydrogen gas, those same wavelengths are absorbed, leaving dark lines in the continuous spectrum: an absorption spectrum.

Move the electron between levels and collect all four Balmer lines:

Try it: make hydrogen's electron jump between energy levels and see what light it absorbs or emits

Move the electron to:

The visible part of hydrogen's emission spectrum (lines appear here when the electron emits visible light):

Limits of the Bohr model

The Bohr model explains the spectra of hydrogen and other one-electron ions (such as He⁺) very well, but for atoms with more than one electron its predictions don’t match experiment. And electrons don’t really travel in fixed circular orbits. The Bohr model’s lasting legacy is the idea that energy is quantized, which lives on in the modern quantum mechanical model (lesson 2.2.4).

Example 1: The Hβ line

Hydrogen’s electron drops from n = 4 to n = 2. What is the energy of the emitted photon, and what is its wavelength?

Step-by-step solution

  1. Write the energy-level formula. In the Bohr model Eₙ = −2.18 × 10⁻¹⁸ J ÷ n², and ΔE = E(final) − E(initial):

    ΔE=2.18×10−18(142−122)\Delta E = 2.18 \times 10^{-18} \left(\frac{1}{4^2} - \frac{1}{2^2}\right)

  2. Calculate. The bracket is 0.0625 − 0.25 = −0.1875:

    ΔE=−4.09×10−19 J\Delta E = -4.09 \times 10^{-19}\ \text{J}

    The negative sign means the atom gives off energy; the photon’s energy is 4.09 × 10⁻¹⁹ J.

  3. Wavelength. From E = hc/λ, λ = hc/E:

    λ=(6.626×10−34)(3.00×108)4.09×10−19=4.86×10−7 m=486 nm\begin{aligned} \lambda &= \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{4.09 \times 10^{-19}} \\ &= 4.86 \times 10^{-7}\ \text{m} \\ &= \boxed{486\ \text{nm}} \end{aligned}

  4. Check against experiment. 486 nm is exactly the blue-green line in hydrogen’s spectrum.

  5. A common mistake. Carrying the negative sign of ΔE into the wavelength formula and getting a negative wavelength. Use the size of the photon’s energy.

Example 2: How much energy ionizes hydrogen?

How much energy does it take to remove the electron completely from a ground-state hydrogen atom (n = 1 → n = ∞)? How much for a mole of atoms? (A mole is 6.022 × 10²³ atoms; see lesson 2.2.2.)

Step-by-step solution

  1. Per atom. At n = ∞ the electron has left the atom and its energy is zero; the ground state is E₁ = −2.18 × 10⁻¹⁸ J:

    ΔE=0−E1=2.18×10−18 J\begin{aligned} \Delta E &= 0 - E_1 \\ &= \boxed{2.18 \times 10^{-18}\ \text{J}} \end{aligned}

  2. Per mole. Multiply by Avogadro’s number and convert to kJ:

    Em=(2.18×10−18)(6.022×1023)=1.31×106 J/mol=1.31×103 kJ/mol\begin{aligned} E_\text{m} &= (2.18 \times 10^{-18})(6.022 \times 10^{23}) \\ &= 1.31 \times 10^{6}\ \text{J/mol} \\ &= \boxed{1.31 \times 10^{3}\ \text{kJ/mol}} \end{aligned}

  3. Check against experiment. Hydrogen’s measured first ionization energy is 1312 kJ/mol, matching the calculation. Lesson 2.3.4 covers ionization energy in depth.

Practice: does the hydrogen atom absorb or emit a photon?
  1. The electron moves from n = 1 to n = 3
  2. The electron drops from n = 4 to n = 2
  3. The electron moves from n = 2 to n = 5
  4. The electron drops from n = 3 to n = 1
  5. The atom goes from the ground state to an excited state
  6. The atom returns from an excited state to the ground state
  7. A hydrogen tube gives off red Hα light
  8. Dark lines appear after white light passes through hydrogen gas
  9. A freed electron is recaptured into the n = 2 level

Common mistakes

Mistake 1: “The emitted photon’s energy equals the energy of the level the electron started from.” The photon’s energy equals the difference between the two levels.

Mistake 2: “The energy levels are evenly spaced.” Eₙ is proportional to 1/n², so the levels crowd closer together as n increases. That’s why the Balmer lines bunch up toward the violet end.

Mistake 3: “The Bohr model explains every atom’s spectrum.” It works only for hydrogen and one-electron ions. Atoms with more electrons need the quantum mechanical model.

Quick check

Which hydrogen transition emits the highest-energy photon?

When hydrogen's electron drops from n = 3 to n = 2, the light it emits is:

Summary

  • Bohr model: the electron occupies only certain energy levels; the ground state is lowest; a jump absorbs or emits a photon with hν = |ΔE|.
  • Hydrogen’s levels: Eₙ = −2.18 × 10⁻¹⁸ J/n², with zero at n = ∞ (ionization). Drops to n = 2 give visible light (Balmer), to n = 1 ultraviolet, and to n = 3 infrared.
  • Separate levels give a line spectrum. The Bohr model works only for one-electron species, but its idea of quantized energy carries on in the quantum mechanical model.
On to AP (for later)

AP doesn’t ask you to memorize Bohr’s energy formula, but topic 3.11 expects you to explain absorption and emission with energy-level diagrams and ΔE = hν. For atoms with many electrons, AP uses ionization energy and photoelectron spectroscopy (PES, topic 1.6). The 1.31 × 10³ kJ/mol from Example 2 (written 1.31 MJ/mol on PES plots) is exactly where hydrogen’s peak appears in its PES spectrum.

Practice this chapter: 2.2 · Atomic Structure · Basic, Medium, and Hard levels

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