Atomic Spectra: Photons, Quantized Energy Levels & PES

🌈 Light, Photons & Energy

Part 1 of 7 — The Electromagnetic Spectrum and Photon Energy

Topics in This Part

🔑 Why this matters: Atomic spectra are the fingerprint of an element. Before we can understand why hydrogen produces a specific pattern of lines, we need the language of light — wavelength, frequency, and the photon.

What You'll Master in Part 1

• Relating wavelength ($\lambda$), frequency ($\nu$), and the speed of light
• Calculating the energy of a single photon
• Predicting whether radiation lies in the UV, visible, or IR region

💡 What Is Light?

Light is electromagnetic radiation — oscillating electric and magnetic fields that travel through space at a fixed speed:

$\boxed{c = 3.00 \times 10^{8} \; \text{m/s}}$

It has both wave-like and particle-like behavior. The particle of light is called a photon — a discrete bundle of energy.

The Wave Equation

For any wave, the speed equals frequency times wavelength:

$\boxed{c = \lambda \nu}$

💡 Key relationship: Wavelength and frequency are inversely proportional. Short wavelength ⇒ high frequency ⇒ high energy.

📌 The Electromagnetic Spectrum

🌈 Visible memory aid (low → high energy): Red, Orange, Yellow, Green, Blue, Indigo, Violet — "ROY G. BIV". Red is longest wavelength (~700 nm); violet is shortest (~400 nm).

⚡ Photon Energy: $E = h\nu$

Each photon carries a discrete amount of energy:

$\boxed{E_{\text{photon}} = h\nu = \frac{hc}{\lambda}}$

where Planck's constant is:

$h = 6.626 \times 10^{-34} \; \text{J}\cdot\text{s}$

Worked Example — Energy of a Green Photon

Problem: A green laser pointer emits light at $\lambda = 532$ nm. What is the energy of one photon?

Step 1. Convert wavelength to meters: $532 \; \text{nm} = 5.32 \times 10^{-7} \; \text{m}$.

Step 2. Apply $E = hc/\lambda$:

$E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{5.32 \times 10^{-7}} = 3.74 \times 10^{-19} \; \text{J}$

That is the energy delivered by one photon. A typical 1-mW laser emits roughly $10^{15}$ photons per second.

Light & Photon Concepts 🎯

Photon Energy Drill 🧮

1) A radio station broadcasts at $f = 1.00 \times 10^{8}$ Hz. What is the wavelength in meters? (Enter as a decimal.)

2) What is the energy (in J) of a single 656-nm photon (hydrogen's red line)? Express as $a \times 10^{-19}$ J — enter only $a$ to 3 sig figs.

3) A photon has energy $4.42 \times 10^{-19}$ J. What is its wavelength in nm? (To the nearest whole nm.)

Spectrum Match-Up 🔽

Exit Quiz — Light & Photons ✅

🔥 Atomic Emission & Absorption Spectra

Part 2 of 7 — How Atoms Make Light

Topics in This Part

🔑 Big idea: A glowing solid produces a continuous rainbow. A glowing gas of atoms produces only specific colors — sharp lines. Each element's pattern of lines is unique.

🌈 Continuous vs Line Spectra

💡 Hot vs cold gases: A hot, low-pressure gas emits specific wavelengths. A cool gas in front of a continuous source absorbs those same wavelengths — leaving dark gaps in the rainbow.

🔥 Emission Spectra

When an atom is supplied with energy (heat, electricity, light), an electron jumps to a higher-energy state. When it falls back down, the energy difference is released as a photon:

$\boxed{\Delta E_{\text{atom}} = E_{\text{photon}} = h\nu}$

The photon's frequency depends on the energy gap between levels. Because atomic energy levels are discrete, only certain photon energies — and therefore certain colors — are possible.

Example — The Sodium D-Line

When you sprinkle table salt (NaCl) into a Bunsen-burner flame, the flame turns bright yellow-orange at 589 nm. This single bright line is the signature of the sodium atom relaxing from its $3p$ state back to $3s$.

This is the basis of flame tests in introductory chemistry.

🌫️ Absorption Spectra

If white light shines through a cool gas, atoms in the gas can absorb photons whose energies match an allowed transition. Those wavelengths are missing from the transmitted beam — appearing as dark lines.

🔬 Astronomy connection: The dark lines in the Sun's spectrum (Fraunhofer lines) come from atoms in the cooler outer layers absorbing specific wavelengths from the hotter interior. They tell us what the Sun is made of.

Conservation of Photon Energies

For a given element, the emission lines and the absorption lines appear at exactly the same wavelengths — because both correspond to the same energy gaps in the atom. Only the direction of the transition is reversed.

Spectrum Quick Check 🎯

Atom-to-Photon Drill 🧮

1) An excited mercury atom drops from a state at $-3.71 \times 10^{-19}$ J to one at $-7.26 \times 10^{-19}$ J. What is $\Delta E$ released as a photon, in $\times 10^{-19}$ J? (Enter just the magnitude, to 2 decimals.)

2) What wavelength (nm) does that photon have? (Round to nearest whole nm.)

3) A sodium atom emits a photon of energy $3.37 \times 10^{-19}$ J. What wavelength (nm)? (Round to nearest whole nm.)

Match Each Source to Its Spectrum 🔽

Exit Quiz ✅

🪐 The Bohr Model & Quantized Energy Levels

Part 3 of 7 — Why Spectra Have Lines

Topics in This Part

🔑 Big picture: In 1913, Niels Bohr proposed that electrons orbit the nucleus in only certain allowed energy levels — quantized states. This explains why hydrogen's emission spectrum has a few sharp lines instead of a continuous rainbow.

🪐 Bohr's Big Idea

The Bohr model has three key postulates:

1. Electrons can only occupy specific orbits (energy levels), labeled $n = 1, 2, 3, \ldots$
2. While in an allowed orbit, the electron does not radiate.
3. Energy is emitted or absorbed only when an electron jumps from one orbit to another, in discrete amounts:

$\boxed{\Delta E = E_{\text{higher}} - E_{\text{lower}} = h\nu}$

💡 Why "$n = 1$" matters: $n = 1$ is the ground state — the lowest, most stable level. All higher levels ($n = 2, 3, \ldots$) are excited states.

⚡ Hydrogen's Energy Levels

Experiments on hydrogen give the following allowed energies. On the AP exam, energy-level values like these are always given to you; you use them, you do not derive them.

(Note the negative signs — a bound electron has lower energy than a free electron, which has $E = 0$.)

⚠️ Watch the signs: Higher $n$ ⇒ less-negative (more positive) energy. The electron is more loosely bound in higher levels. Notice also that the gaps between levels shrink as $n$ increases.

Energy of a Photon Emitted in a Transition

When an electron drops from a higher level to a lower level, the photon carries away exactly the energy difference:

$\boxed{E_{\text{photon}} = E_{\text{upper}} - E_{\text{lower}}}$

Example: $n = 3 \to n = 2$: $E_{\text{photon}} = (-2.42 \times 10^{-19}) - (-5.45 \times 10^{-19}) = 3.03 \times 10^{-19}$ J, which corresponds to $\lambda = hc/E = 656$ nm (red light).

The magnitude $|\Delta E|$ equals the energy of the photon emitted.

🔑 Excitation vs Ionization

Worked Example — Ground-State Ionization Energy of H

Problem: What is the minimum energy needed to ionize a hydrogen atom from its ground state?

Solution. From $n = 1$ to $n = \infty$:

$\Delta E = E_\infty - E_1 = 0 - (-2.18 \times 10^{-18}) = 2.18 \times 10^{-18} \; \text{J}$

Per mole: $\;(2.18 \times 10^{-18})(6.022 \times 10^{23}) = 1.31 \times 10^{6}$ J/mol = 1310 kJ/mol.

This matches the measured first ionization energy of hydrogen, about 1312 kJ/mol.

Energy Levels Quick Check 🎯

Energy-Level Calculations 🧮

Use the hydrogen energy levels given in the table above.

1) Calculate the photon energy emitted when an H electron falls from $n=4$ to $n=2$. Express as $\times 10^{-19}$ J — enter just the magnitude (3 sig figs).

2) What wavelength (nm) does that correspond to? (Nearest whole nm.)

3) What energy (in J) is needed to ionize an H atom from $n = 2$? Express as $\times 10^{-19}$ J — enter just the magnitude (3 sig figs).

Bohr Model Concepts 🔽

Exit Quiz — Bohr Model ✅

🌈 Energy-Level Diagrams & Spectral Regions

Part 4 of 7 — Energy-Level Diagrams and Spectral Regions

Topics in This Part

🔑 Big idea: An energy-level diagram tells you every photon an atom can emit or absorb. Different regions of the spectrum probe different kinds of motion: microwaves change how molecules rotate, infrared changes how bonds vibrate, and ultraviolet/visible light moves electrons between levels.

📊 Reading an Energy-Level Diagram

Here are the four lowest energy levels of a hypothetical atom X:

Emission Lines

An excited atom can drop from any higher level to any lower one. From level 4, the possible drops are $4 \to 3$, $4 \to 2$, $4 \to 1$, $3 \to 2$, $3 \to 1$, and $2 \to 1$, so six emission lines are possible.

Absorption Lines

In a cool gas, nearly every atom is in the ground state, so absorption lines come from transitions that start at level 1: $1 \to 2$, $1 \to 3$, and $1 \to 4$. The atom absorbs only photons whose energy exactly matches one of these gaps.

💡 Key Rule: Bigger gap → higher photon energy → higher frequency → shorter wavelength.

🔑 Matching Spectral Regions to Transitions

Molecules have several kinds of quantized energy, and each kind is probed by a different region of the electromagnetic spectrum:

Why the Order Makes Sense

• Rotational levels are very closely spaced, so tiny energy jumps (microwaves) are enough.
• Vibrational levels are farther apart and need infrared photons.
• Electronic levels are the most widely spaced, so electron jumps need visible or ultraviolet photons.

⚠️ AP Tip: Infrared light does not move electrons to higher levels, and microwaves do not make bonds vibrate. Always match the region to the correct type of transition.

Regions and Transitions Check 🎯

🧪 Worked Examples

Example 1 — From Diagram to Wavelength

Problem: For atom X, find the wavelength of the photon emitted in the $4 \to 2$ transition.

$E = E_4 - E_2 = (-2.00 \times 10^{-19}) - (-5.00 \times 10^{-19}) = 3.00 \times 10^{-19} \; \text{J}$

$\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{3.00 \times 10^{-19}} = 6.63 \times 10^{-7} \; \text{m} = \boxed{663 \; \text{nm}}$

That is red visible light, so this is an electronic transition.

Example 2 — Identifying the Type of Transition

Problem: A gas absorbs radiation with a wavelength of $4.3 \times 10^{-6}$ m. What kind of transition is this?

The wavelength is about 4300 nm, longer than visible light (700 nm) but much shorter than microwaves (about 1 mm and longer). It is infrared, so the absorption changes the vibrational energy of the molecules.

$E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{4.3 \times 10^{-6}} = 4.6 \times 10^{-20} \; \text{J}$

This energy is smaller than every electronic gap in atom X, consistent with vibrational levels being more closely spaced than electronic levels.

Energy-Level Diagram Drill 🧮

Use the energy levels of atom X and $h = 6.626 \times 10^{-34}$ J·s, $c = 3.00 \times 10^{8}$ m/s.

1) Suppose atom X could also be excited to a fifth level. How many different emission lines would then be possible from level 5 and below?

2) What is the wavelength (in nm, nearest whole) of the photon emitted in the $2 \to 1$ transition?

3) What photon energy is needed to excite a ground-state atom of X to level 4? Express as $\times 10^{-19}$ J (3 sig figs).

Match the Region 🔽

Exit Quiz — Diagrams & Regions ✅

🌌 Beyond Hydrogen — Multi-Electron Atoms & PES

Part 5 of 7 — Why Real Spectra Are More Complex

Topics in This Part

🔑 Big idea: The Bohr model handles hydrogen perfectly but breaks down for atoms with more than one electron. The energies depend not only on $n$ but also on the subshell ($s, p, d, f$). Photoelectron spectroscopy (PES) measures these energies directly.

⚛️ Multi-Electron Atoms

For atoms with more than one electron, three new effects break Bohr's simple picture:

1. Electron–electron repulsion raises the energy of every level.
2. Shielding — inner electrons partially screen outer electrons from the full nuclear charge.
3. Penetration — $s$ orbitals dip closer to the nucleus than $p$ or $d$ orbitals at the same $n$, so $s$ feels a larger effective charge.

The result: within a given shell $n$, the subshells split:

$E_{ns} < E_{np} < E_{nd} < E_{nf}$

That is why the emission spectrum of, say, sodium has many more lines than hydrogen, and why hydrogen's energy-level values cannot be reused for any other element.

📌 Effective Nuclear Charge ($Z_{\text{eff}}$)

Each electron experiences not the full nuclear charge $Z$, but a smaller effective charge:

$\boxed{Z_{\text{eff}} = Z - S}$

where $S$ is the shielding constant from inner electrons.

💡 Higher $Z_{\text{eff}}$ ⇒ deeper (more negative) orbital energy ⇒ photon released when an electron falls into that orbital has higher energy.

🔬 Connecting Spectra to PES

Photoelectron spectroscopy (PES) is the modern technique that directly measures the binding energy of every occupied orbital in an atom.

A high-energy UV or X-ray photon (energy $h\nu$) ejects an electron with kinetic energy $KE$. Conservation of energy gives the binding energy (BE):

$\boxed{BE = h\nu - KE}$

Each orbital contributes a peak in the PES spectrum. The peak's position = orbital binding energy. The peak's height = number of electrons in that orbital.

Together, emission spectra and PES give a complete picture of the electronic structure of the atom.

⚠️ What a One-Electron Picture Can't Explain

A model built for hydrogen (one electron, one proton) works well for hydrogen but cannot account for:

The modern model replaces fixed orbits with orbitals (regions where an electron is likely to be found) and explains subshell energies, which PES confirms experimentally.

Beyond Bohr — Quick Check 🎯

PES & Binding Energy 🧮

1) A PES experiment uses photons of energy $1.00 \times 10^{-15}$ J. An ejected electron has kinetic energy $7.50 \times 10^{-16}$ J. What is the binding energy of that electron, in $\times 10^{-16}$ J? (To 2 decimals.)

2) In hydrogen, the binding energy of the 1s electron is $2.18 \times 10^{-18}$ J. What minimum photon wavelength (in nm, nearest whole) is needed to ionize hydrogen from the ground state?

3) Convert the 1s binding energy of hydrogen, $2.18 \times 10^{-18}$ J per atom, to kJ/mol. (Nearest whole number.)

Multi-Electron Concepts 🔽

Exit Quiz — Beyond Bohr ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Putting It All Together

Workshop Goals

📋 Hydrogen energy levels (J): $E_1 = -2.18 \times 10^{-18}$, $E_2 = -5.45 \times 10^{-19}$, $E_3 = -2.42 \times 10^{-19}$, $E_4 = -1.36 \times 10^{-19}$, $E_5 = -8.72 \times 10^{-20}$, $E_6 = -6.06 \times 10^{-20}$

🛠️ Reference equations:
$c = \lambda \nu, \quad E = h\nu = hc/\lambda$
$E_{\text{photon}} = E_{\text{upper}} - E_{\text{lower}}$ (levels given)
$BE = h\nu - KE$
Constants: $h = 6.626 \times 10^{-34}$ J·s, $c = 3.00 \times 10^{8}$ m/s, $N_A = 6.022 \times 10^{23}$

🧪 Worked Walkthrough 1 — Identifying a Transition

Problem: A spectral line at 434 nm is observed in hydrogen's emission spectrum. Using the given energy levels, identify the transition.

Step 1. Find the photon energy:

$E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{4.34 \times 10^{-7}} = 4.58 \times 10^{-19} ; \text{J}$

Step 2. Look for the pair of levels whose difference matches. Visible photons in hydrogen come from drops to $n = 2$, so try $E_{\text{upper}} = E_2 + 4.58 \times 10^{-19}$:

$E_{\text{upper}} = (-5.45 \times 10^{-19}) + (4.58 \times 10^{-19}) = -8.7 \times 10^{-20} ; \text{J}$

Step 3. That matches $E_5 = -8.72 \times 10^{-20}$ J.

Answer: The transition is $n=5 \to n=2$. ✅

🧪 Worked Walkthrough 2 — Energy per Mole

Problem: Light of wavelength 350 nm is absorbed by a mole of atoms, each absorbing one photon. How much energy (kJ) is absorbed in total?

Step 1. Energy per photon:

$E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{3.50 \times 10^{-7}} = 5.68 \times 10^{-19} \; \text{J}$

Step 2. Multiply by Avogadro's number:

$E_{\text{mole}} = (5.68 \times 10^{-19})(6.022 \times 10^{23}) = 3.42 \times 10^{5} \; \text{J/mol} = \boxed{342 \; \text{kJ/mol}}$

💡 This is comparable in magnitude to a typical chemical bond energy — UV light can break bonds!

🧪 Worked Walkthrough 3 — A PES Peak

Problem: A PES experiment uses 21.2 eV photons. The kinetic energy of ejected electrons from a hydrogen atom is 7.6 eV. Verify that the binding energy matches the ground-state ionization energy of H (13.6 eV).

Step 1. Apply $BE = h\nu - KE$:

$BE = 21.2 \; \text{eV} - 7.6 \; \text{eV} = 13.6 \; \text{eV} \;\;\checkmark$

This matches the binding energy of hydrogen's 1s electron, 13.6 eV (that is, $2.18 \times 10^{-18}$ J).

Workshop Multiple Choice 🎯

Workshop Calculations 🧮

1) A photon ionizes a ground-state hydrogen atom and leaves the ejected electron with KE = $5.45 \times 10^{-19}$ J. What was the photon's energy in $\times 10^{-18}$ J? (To 3 sig figs.)

2) What was the photon's wavelength in nm? (Nearest whole.)

3) A ground-state hydrogen atom absorbs a photon of energy $2.09 \times 10^{-18}$ J. Using the given levels, which level does the electron reach?

4) How much energy, in kJ/mol, is needed to move one mole of hydrogen atoms from $n = 1$ to $n = 2$? (Nearest whole number.)

Exit Quiz — Workshop Wrap-Up ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Tying Atomic Spectra to the AP Curriculum

What You'll Master

🎯 AP CED alignment: This topic draws on AP Chemistry Topics 1.5 (Atomic Structure & Electron Configuration), 1.6 (Photoelectron Spectroscopy), and 3.11 (Spectroscopy & the Electromagnetic Spectrum). Expect FRQs that ask you to: • Calculate photon energy or wavelength from a transition • Sketch or interpret an emission/PES spectrum • Justify the magnitude of $Z_{\text{eff}}$ for a labeled peak • Compare line spectra of H to multi-electron atoms

🌟 Big-Picture Summary

Cross-Topic Connections

• Electron configuration (Topic 1.5): Order of subshell filling reflects the same energy ordering seen in PES peaks.
• Periodic trends (Topic 1.7): $Z_{\text{eff}}$ explains both trends and PES binding-energy shifts.
• Properties of photons (Topic 3.12): $E = h\nu$ and $c = \lambda\nu$ connect every line to a photon energy.
• Spectroscopy regions (Topic 3.11): Molecules absorb microwave, infrared, and UV/visible light for different reasons.

⚠️ Common Pitfalls to Avoid

1. Sign errors in $\Delta E$. $E_n$ is negative. The energy of the emitted photon is the positive magnitude $|E_{n_f} - E_{n_i}|$.
2. Forgetting unit conversions. Always convert nm → m before plugging into $E = hc/\lambda$.
3. Mixing up "shorter wavelength" with "less energy." Shorter $\lambda$ ⇒ HIGHER energy.
4. Mismatching region and transition. Microwaves change rotation, infrared changes vibration, and only UV/visible light moves electrons between levels.
5. Confusing emission and absorption. They occur at the same wavelengths, but the direction of the transition is opposite.
6. Reading PES wrong. Higher binding energy = closer to nucleus = lower $n$ orbital.

AP-Style Multiple Choice 🎯

Final Mastery Drill 🧮

Hydrogen levels (J): $E_1 = -2.18 \times 10^{-18}$, $E_2 = -5.45 \times 10^{-19}$, $E_3 = -2.42 \times 10^{-19}$, $E_4 = -1.36 \times 10^{-19}$.

1) What energy (in J, $\times 10^{-19}$, 3 sig figs) does it take to ionize an H atom that is in $n = 3$?

2) A photon of energy $2.04 \times 10^{-18}$ J is absorbed by a ground-state H atom. What level does the electron end up in?

3) A PES experiment uses photons with $h\nu = 200$ eV. An ejected electron has KE = 50 eV. Express the binding energy in eV.

4) Convert 91 nm into the corresponding photon energy in $\times 10^{-18}$ J. (3 sig figs.)

AP-Style Conceptual Connections 🔽

Final Exit Quiz — You've Mastered Atomic Spectra! 🎓