6.2 Electronic Transitions (Absorption and Emission)
PRE-HEALTH PREP
Bohr Model Electronic Transitions
Absorption vs. Emission
General Chemistry 6.2
The Bohr model describes electrons occupying fixed orbits around the nucleus, each corresponding to a specific quantized energy level calculated using E = −2.18 × 10−18 J / n². Electronic transitions between these orbits involve the absorption or emission of photons whose energies exactly match the difference between energy levels, with positive ΔE indicating absorption and negative ΔE indicating emission. The resulting spectral patterns—including the ultraviolet Lyman series, the visible Balmer series, and the infrared Paschen and Brackett series—serve as unique fingerprints for identifying elements.
The Bohr Model of the Atom
What the Bohr Model Got Right (and Wrong)
The Bohr model of the atom holds a fascinating place in the history of chemistry: it worked beautifully for the hydrogen atom yet failed for virtually every other element. For hydrogen, the model explained both emission spectra and absorption spectra with remarkable precision. However, it broke down completely for any multi-electron system. Hydrogen is the only neutral atom with a single electron, which is precisely why the Bohr model cannot handle any other neutral atom.
Despite these limitations, Bohr's contributions were revolutionary enough that his model remains a cornerstone of introductory general chemistry. So what exactly did he propose?
Electrons in Fixed, Quantized Orbits
Bohr agreed with Rutherford that every atom contains a central nucleus. His key addition was the idea that electrons orbit this nucleus at specific, fixed distances — set orbits. Because each orbit corresponds to a particular distance from the nucleus, each orbit also corresponds to a specific quantized energy level.
To understand why, consider an analogy from physics. An object held above the ground possesses gravitational potential energy — it has the potential to fall. Gravity attracts any two objects with mass toward each other. When you release the object, it falls, losing potential energy and gaining kinetic energy. The key principle: whenever two objects attract each other, the closer they get, the lower their potential energy.
Example: The Gravity Analogy
Hold a marker above the ground — it has gravitational potential energy. Release it, and as it falls toward the Earth, it loses potential energy and gains kinetic energy. The same principle governs electrons and nuclei: as a negatively charged electron moves closer to a positively charged nucleus, its potential energy decreases.
The same logic applies to an electron (negatively charged) attracted to the nucleus (positively charged, containing protons). An electron in the first orbit — the one closest to the nucleus — has the lowest potential energy. The farther an electron sits from the nucleus, the higher its energy.
The Energy Equation for Electron Orbits
Bohr's Energy Formula
Bohr derived an equation for the energy of an electron in any orbit:
Energy of an Electron in Orbit n:
E = −2.18 × 10−18 J ÷ n²
where n = 1, 2, 3, … (the orbit number, also called the principal quantum number)
The closest orbit corresponds to n = 1, the next to n = 2, then n = 3, and so on through an infinite number of possible orbits. However, the spacing between orbits decreases as n increases. The gap between n = 1 and n = 2 is much larger than the gap between n = 2 and n = 3, which in turn is larger than the gap between n = 3 and n = 4.
At the theoretical limit of n = ∞, the energy reaches zero: dividing by infinity squared yields zero. This represents an electron so far from the nucleus that it is essentially free — no longer bound to the atom. Since zero is the maximum possible energy, all bound electrons have negative energies, growing more negative as they approach the nucleus.
Calculating Energies for Each Orbit
Plugging in values of n makes this concrete:
| Orbit (n) | Calculation | Energy (J) |
|---|---|---|
| 1 | −2.18 × 10−18 ÷ 1² | −2.18 × 10−18 |
| 2 | −2.18 × 10−18 ÷ 2² | −5.45 × 10−19 |
| 3 | −2.18 × 10−18 ÷ 3² | −2.42 × 10−19 |
| ∞ | −2.18 × 10−18 ÷ ∞² | 0 |
Tip: Notice that −5.45 × 10−19 is less negative than −2.18 × 10−18. Less negative means closer to zero, which means higher energy. The pattern holds: as n increases, the energy becomes less negative (higher).
For any orbit, you simply plug the appropriate value of n into the equation — n = 4, n = 5, n = 6, or any positive integer — to find the energy of an electron in that orbit according to Bohr's model.
Electronic Transitions: Absorption vs. Emission
How Electrons Absorb Photons to Move to Higher Energy Orbits
The real power of the Bohr model lies in explaining why atoms absorb or emit photons during electronic transitions. Consider an electron sitting in the first orbit (n = 1). It can jump to the second orbit (n = 2) or the third orbit (n = 3) — but doing so requires energy, because outer orbits are higher in energy. Where does this energy come from? The electron absorbs a photon whose energy exactly matches the difference between the two orbits.
Quantized Energy: Why Only Specific Photons Are Absorbed or Emitted
Here is what made Bohr's proposal revolutionary: because electrons can only exist in specific orbits with no intermediate positions allowed, only specific energies are possible. There is no orbit between n = 1 and n = 2, and therefore no energy level between the two. This means the electron's energy in an atom is quantized — it exists only at discrete, specific values.
As a direct consequence, only photons with exactly the right energy can be absorbed or emitted. A photon that is slightly too energetic or slightly too weak will not cause a transition. This selectivity is precisely what produces the characteristic absorption and emission line spectra we observe in hydrogen and other elements.
Key Concept: Quantized energy means that electron energy in an atom is restricted to specific, discrete values — no in-between values are allowed. This is fundamentally different from classical physics, where energy can take any value along a continuous range.
Calculating ΔE for an Electronic Transition
To find the energy change for an electronic transition, calculate the difference between the final and initial energy levels. For an electron jumping from n = 1 to n = 3:
Example: ΔE for the n = 1 → n = 3 Transition
ΔE = E3 − E1
ΔE = (−2.42 × 10−19 J) − (−2.18 × 10−18 J)
ΔE = −2.42 × 10−19 + 2.18 × 10−18
ΔE = 1.938 × 10−18 J
This positive value indicates the electron must absorb this much energy to make the transition.
The Direct ΔE Formula Using 2 Quantum Numbers
Rather than calculating each energy level separately and subtracting, you can compute ΔE directly using both n values in a single expression:
Direct ΔE Formula:
ΔE = −2.18 × 10−18 J × (1/nf² − 1/ni²)
where ni is the initial orbit and nf is the final orbit. This ΔE equals the energy of the absorbed or emitted photon.
Positive vs. Negative ΔE: How to Tell Absorption from Emission
The sign of ΔE tells you the direction of the electronic transition:
- Positive ΔE — the electron moves to a higher orbit and a photon is absorbed
- Negative ΔE — the electron drops to a lower orbit and a photon is emitted
This is analogous to enthalpy changes in thermochemistry: just as a negative ΔH indicates an exothermic process (energy released), a negative ΔE indicates emission (energy released as a photon).
Important: The energy of a photon itself is always positive — there is no such thing as a photon with negative energy. When ΔE comes out negative, simply take the absolute value to find the photon's energy. The negative sign merely indicates the direction of the transition (emission rather than absorption).
Calculating Wavelength from ΔE
Converting Photon Energy to Wavelength
Once you know the energy of the photon involved in an electronic transition, you can determine its frequency or wavelength. Recall from the previous lesson that the energy of a photon relates to its wavelength through:
Photon Energy Relationships:
E = hν = hc / λ
Rearranging for wavelength: λ = hc / E
Applying this to the n = 1 → n = 3 transition:
Example: Wavelength of the n = 1 → n = 3 Photon
λ = hc / E
λ = (6.626 × 10−34 J·s)(3.0 × 108 m/s) / (1.938 × 10−18 J)
λ = 1.03 × 10−7 m = 103 nm
At 103 nm, this photon falls well below the visible range (400–700 nm), placing it squarely in the ultraviolet region of the electromagnetic spectrum.
Constant Reminders: Planck's constant (h = 6.626 × 10−34 J·s) and the speed of light (c = 3.0 × 108 m/s) are typically provided on exams. The Bohr energy constant (−2.18 × 10−18 J) is also usually given.
The Rydberg Equation: A Direct Route to Wavelength
Historically, scientists measured the wavelengths of emitted and absorbed photons before energy calculations became standard. Rydberg developed an equation that goes directly to wavelength without an intermediate energy step:
The Rydberg Equation:
1/λ = RH × (1/n1² − 1/n2²)
where RH is the Rydberg constant (1.097 × 107 m−1), n1 is the smaller n value, and n2 is the larger n value.
Tip: In the Rydberg equation, n1 must always be the smaller value and n2 the larger value. This convention ensures a positive result, since wavelength cannot be negative.
Emission Line Spectra
Absorption Spectrum: Dark Lines as an Elemental Fingerprint
White light — the kind produced by the Sun — contains a continuous spectrum of all visible wavelengths. Pass it through a prism, and you see the familiar rainbow from red through violet. However, if you examine sunlight carefully, you find that certain narrow bands of color are missing, appearing as dark lines in the otherwise continuous spectrum.
These dark lines correspond to wavelengths that hydrogen gas in the Sun's outer layers has absorbed. The white light generated by nuclear reactions in the Sun's core must pass through vast amounts of hydrogen gas before reaching us, and that gas absorbs photons at very specific wavelengths — exactly those corresponding to electronic transitions in hydrogen.
Key Concept: An absorption spectrum shows dark lines at specific wavelengths against a continuous background of color. Each element produces a unique pattern of absorption lines, serving as a chemical fingerprint.
Using Absorption Spectra to Identify Elements in Stars
This is how scientists determined the composition of the Sun — and indeed of any star — without ever collecting a physical sample. The Sun turns out to be roughly three-quarters hydrogen and one-quarter helium, with trace amounts of other elements. We know this because the dark lines in the solar absorption spectrum match precisely the absorption patterns of hydrogen and helium measured in laboratories here on Earth.
Emission Spectrum vs. Absorption Spectrum: 2 Key Differences
An emission spectrum is essentially the inverse of an absorption spectrum. Instead of looking for missing colors in white light, you start with no light and observe what colors appear.
The process works like this: apply a voltage to a sample of gas, exciting its electrons to higher energy orbits. As those electrons fall back to lower orbits, they emit photons at specific wavelengths. The result is a spectrum showing only certain bright lines of color against a dark background — the exact wavelengths that were missing in the absorption spectrum. The 2 key differences between these spectra are: (1) an absorption spectrum shows dark lines on a continuous color background, while (2) an emission spectrum shows bright, colored lines on a dark background.
Example: Hydrogen Emission Spectrum
When hydrogen gas is excited by an electrical discharge, its electrons jump to high-energy orbits. As they cascade back down, they emit photons at specific wavelengths. The visible lines appear at 656 nm (red), 486 nm (blue-green), 434 nm (blue-violet), and 410 nm (violet) — the characteristic Balmer series of the hydrogen emission spectrum.
One important pattern to notice: the energy levels get closer together as n increases. The gap between n = 5 and n = 6 is smaller than the gap between n = 4 and n = 5, and so on. The largest energy gap in the hydrogen atom is between n = 1 and n = 2.
The 4 Hydrogen Spectral Series
When excited electrons in hydrogen fall back to lower orbits, the electronic transitions group naturally into 4 spectral series based on the final orbit the electron drops to. Each series produces photons in a distinct region of the electromagnetic spectrum.
| Series | Final Orbit (nf) | Example Transitions | Spectral Region |
|---|---|---|---|
| Lyman | 1 | 2→1, 3→1, 4→1, 5→1, 6→1 | Ultraviolet |
| Balmer | 2 | 3→2, 4→2, 5→2, 6→2 | Visible |
| Paschen | 3 | 4→3, 5→3, 6→3 | Infrared |
| Brackett | 4 | 5→4, 6→4, 7→4, 8→4 | Infrared |
The Lyman Series (Transitions to n = 1): Ultraviolet
The Lyman series includes all transitions where electrons drop to the n = 1 orbit. Because these transitions span the largest energy gaps in the hydrogen atom, the emitted photons carry high energy and short wavelengths, placing them in the ultraviolet region — invisible to the human eye.
The Balmer Series (Transitions to n = 2): Visible Light
The Balmer series encompasses transitions to the n = 2 orbit (3→2, 4→2, 5→2, and so on). This series holds special importance because many of its lines fall in the visible portion of the electromagnetic spectrum — these are the colored lines you can actually see when hydrogen gas glows. The Balmer series is the most commonly tested spectral series in general chemistry.
Exam Priority: Of all 4 spectral series, the Balmer series is by far the most frequently tested. Remember that it corresponds to transitions ending at n = 2 and produces visible light.
The Paschen and Brackett Series: Infrared
The Paschen series (transitions to n = 3) and Brackett series (transitions to n = 4) involve progressively smaller energy differences. Their photons fall in the infrared region, with wavelengths longer and energies lower than visible light.
Pattern to Remember: As the final orbit increases (nf = 1 → 2 → 3 → 4), the energy differences get smaller, so the emitted photons shift from ultraviolet → visible → infrared. Higher nf means lower energy photons and longer wavelengths.
Key Takeaways
- The Bohr model places electrons in fixed, quantized orbits around the nucleus, each with a specific energy. It works perfectly for hydrogen but fails for multi-electron atoms.
- Electron energy is calculated using E = −2.18 × 10−18 J / n², where n is the principal quantum number. Energy becomes less negative (higher) as n increases.
- Electron energy in atoms is quantized — only specific energy values are allowed, with no in-between states. This is why only specific photon energies can trigger electronic transitions.
- A positive ΔE means the electron absorbs a photon and jumps to a higher orbit (absorption); a negative ΔE means it emits a photon and drops to a lower orbit (emission).
- The energy of a transition equals the photon energy: ΔE = hc/λ. Rearranging gives the wavelength of the absorbed or emitted photon.
- The Rydberg equation (1/λ = RH(1/n1² − 1/n2²)) provides a shortcut to calculate wavelength directly from orbit numbers, bypassing the intermediate ΔE step.
- Absorption spectra show dark lines (missing wavelengths) in continuous light; emission spectra show bright lines (specific wavelengths) against darkness. Both serve as elemental fingerprints.
- The hydrogen emission spectrum is organized into 4 spectral series: the Lyman series (ultraviolet, nf = 1), the Balmer series (visible, nf = 2), the Paschen series (infrared, nf = 3), and the Brackett series (infrared, nf = 4). The Balmer series is the most important for exams.