Photoelectron Spectroscopy (PES) is an experimental technique that measures the energy required to remove electrons from an atom or molecule.

Process:

1. High-energy photons strike a sample
2. Photons transfer energy to electrons
3. Electrons are ejected from various orbitals
4. Instrument measures kinetic energy of ejected electrons
5. Data reveals ionization energies for each electron

Key Equation:

$E_{\text{photon}} = IE + KE$

Where:

• $E_{\text{photon}}$ = energy of incoming photon (constant)
• $IE$ = ionization energy (binding energy of electron)
• $KE$ = kinetic energy of ejected electron

Rearranged:

$IE = E_{\text{photon}} - KE$

PES Spectrum Features

X-axis: Binding Energy (Ionization Energy)

• Units: MJ/mol or eV
• Direction: Usually increases left to right
• Higher binding energy = electron held more tightly
• Represents energy needed to remove electron

Y-axis: Relative Number of Electrons

• Also called: Intensity, Signal, or Counts
• Height of peak indicates number of electrons with that binding energy
• Area under peak is proportional to number of electrons

Peaks

Each peak represents electrons from a specific orbital/sublevel:

• Position (x): Binding energy (how tightly held)
• Height (y): Number of electrons in that orbital

Example: A peak at 1.4 MJ/mol with height of 2 means 2 electrons have binding energy of 1.4 MJ/mol

Reading PES Data

General Trends

Binding energy increases with:

1. Closer to nucleus (lower n value)
2. Higher nuclear charge (more protons)
3. Less shielding (inner vs. outer electrons)

Typical order (increasing binding energy):

$\text{Valence orbitals} < \text{Inner orbitals} < \text{Core orbitals}$

For multi-electron atoms:

$ns < np < (n-1)d < (n-1)p < (n-1)s < ... < 1s$

Interpreting Peaks

Example: Nitrogen (N)

Electron configuration: $1s^2 2s^2 2p^3$

Expected PES spectrum:

1. Peak at ~1.4 MJ/mol, height = 3: Three 2p electrons (valence)
2. Peak at ~1.8 MJ/mol, height = 2: Two 2s electrons (valence)
3. Peak at ~10 MJ/mol, height = 2: Two 1s electrons (core)

Three peaks total (three different orbitals)

Information from PES

1. Number of Occupied Orbitals

Number of peaks = Number of occupied sublevels

• 3 peaks → 3 occupied sublevels (e.g., 1s, 2s, 2p)
• 5 peaks → 5 occupied sublevels (e.g., 1s, 2s, 2p, 3s, 3p)

2. Number of Electrons in Each Orbital

Height/area of peak = Number of electrons

• Peak height of 2 → 2 electrons
• Peak height of 6 → 6 electrons

3. Relative Energies

Peak position reveals how tightly electrons are held:

• Rightmost peak (highest binding energy) = 1s (closest to nucleus)
• Leftmost peak (lowest binding energy) = valence electrons (farthest from nucleus)

4. Identifying Elements

Combine information to write electron configuration, then identify element:

Example:

• 4 peaks with heights 2, 2, 6, 2 (left to right)
• Configuration: $1s^2 2s^2 2p^6 3s^2$
• Element: Magnesium (Mg)

Comparing PES Spectra

Same Period (e.g., C, N, O)

Trend: Binding energies increase across period

• More protons → stronger nuclear charge
• Same shielding (same inner electrons)
• Electrons held more tightly

Example:

• 1s electrons: $BE_C < BE_N < BE_O$
• More protons in O pull electrons tighter

Same Group (e.g., Li, Na, K)

Trend: Binding energies decrease down group

• More electron shells → more shielding
• Valence electrons farther from nucleus
• Weaker attraction despite more protons

Example:

• 2s in Li > 3s in Na > 4s in K

Core vs. Valence

Core electrons (inner shells):

• Much higher binding energy
• Closer to nucleus
• Less shielding

Valence electrons (outer shell):

• Lower binding energy
• Farther from nucleus
• More shielding

Large gap between core and valence peaks

Common PES Questions

Type 1: Write electron configuration from spectrum

Given: Peak positions and heights Find: Electron configuration and element

Steps:

1. Count peaks (number of sublevels)
2. Note heights (electrons per sublevel)
3. Assign orbitals from right to left (high to low BE)
4. Write configuration and identify element

Type 2: Predict spectrum from element

Given: Element or electron configuration Find: Number of peaks, relative heights, relative positions

Steps:

1. Write electron configuration
2. Count occupied sublevels (number of peaks)
3. Count electrons per sublevel (peak heights)
4. Order by binding energy (1s highest, valence lowest)

Type 3: Compare spectra

Given: Two elements Find: Differences in binding energies

Steps:

1. Consider nuclear charge (more protons → higher BE)
2. Consider shielding (more shells → lower BE)
3. Compare same orbitals between elements

Example: Sodium (Na)

Electron configuration: $1s^2 2s^2 2p^6 3s^1$

PES spectrum:

• Peak 1: 3s¹ (height = 1, lowest BE ~0.5 MJ/mol) - valence
• Peak 2: 2p⁶ (height = 6, medium BE ~3 MJ/mol) - inner
• Peak 3: 2s² (height = 2, high BE ~6 MJ/mol) - inner
• Peak 4: 1s² (height = 2, highest BE ~100 MJ/mol) - core

Total: 4 peaks

Note: Large gap between 3s and 2p (valence vs. core)

Key Relationships

Binding Energy and Atomic Number

Same orbital, different elements:

Higher atomic number (Z) → Higher binding energy

Example: 1s electrons

• Li (Z=3): ~6 MJ/mol
• C (Z=6): ~11 MJ/mol
• F (Z=9): ~80 MJ/mol

Sublevels in Same Shell

Energy order (increasing BE):

$ns < np < (n-1)d$

Example for n=3: $3s < 3p < 2d$

However, 3d is usually empty or partially filled, and 2d doesn't exist.

More common: $3s < 3p < 2p < 2s < 1s$

Applications

1. Elemental analysis: Identify unknown elements
2. Electron configuration confirmation: Verify predicted configurations
3. Bonding studies: Understand how electrons participate in bonds
4. Material characterization: Analyze surface composition
5. Support for quantum theory: Experimental evidence for orbital model

Tips for AP Exam

1. Right to left: Highest BE (rightmost) is always 1s
2. Count electrons: Sum of peak heights = total electrons = atomic number
3. Valence electrons: Leftmost peak(s) with lowest BE
4. Core electrons: Large gap separates from valence
5. Period = shells: Number of principal energy levels (n values)
6. Peaks ≠ Electrons: Number of peaks is number of sublevels, not total electrons

Nitrogen (N): Z = 7 $1s^2 2s^2 2p^3$

Step 2: Compare nuclear charge

• C: 6 protons
• N: 7 protons (one more)

Step 3: Consider shielding for 1s electrons

For 1s electrons:

• Same principal energy level (n = 1)
• No inner electrons to provide shielding
• Experience nearly full nuclear charge

Step 4: Apply trend

Higher nuclear charge → Stronger attraction → Higher binding energy

Since N has more protons (7 vs. 6), its 1s electrons experience:

• Greater nuclear charge (+7 vs. +6)
• Stronger electrostatic attraction
• Higher energy needed to remove

Answer: Nitrogen (N) will have higher binding energy for 1s electrons

Quantitative comparison:

• C: 1s BE ≈ 11 MJ/mol
• N: 1s BE ≈ 15 MJ/mol

Explanation:

Both elements have the same electron configuration for 1s (1s²), but nitrogen has one additional proton in the nucleus. This extra positive charge pulls all electrons, including the 1s electrons, more tightly toward the nucleus, requiring more energy to remove them.

General principle: For elements in the same period, binding energy increases from left to right due to increasing nuclear charge with minimal change in shielding for the same orbital.

Note: This trend applies to ALL orbitals (1s, 2s, 2p), not just 1s.

Step 2: Calculate total electrons

Total = 1 + 3 + 1 = 5 electrons

Wait! Let's reconsider. The heights represent relative numbers.

Actually, looking at typical PES data and the pattern of BE values:

Re-analysis:

• BE values are relatively close (1.09, 1.31) - both are valence
• 6.84 is much higher - core electrons
• The 3 binding energies suggest period 2 element

Step 3: Assign orbitals correctly

For a period 2 element with these patterns:

• Highest BE (6.84 MJ/mol): 1s electrons (peak height should be 2, but given as 1 - likely relative scale)
• Medium BE (1.31 MJ/mol): 2s electrons (peak height 3 - likely means this is 2p with 3 electrons)
• Lowest BE (1.09 MJ/mol): Could be 3s with 1 electron

Wait, let me reconsider with actual chemistry knowledge:

Better interpretation:

If we have 3 peaks and the middle one has height 3:

• Right peak (6.84 MJ/mol, height 1): Needs to be 1s, but 1s should have 2 electrons

Actually, let's solve this properly:

Given the binding energies and the fact that the second peak (1.31 MJ/mol) has height = 3:

This suggests:

• Peak at 6.84 MJ/mol (height = 1): This seems too low for 1s with only 1 electron

Let me reconsider: Heights might mean something different, or there's a scaling issue.

Most likely scenario: Element is Nitrogen (N)

Electron configuration: $1s^2 2s^2 2p^3$

Corrected assignment:

• 6.84 MJ/mol: 1s² (should show 2 electrons, not 1)
• 1.31 MJ/mol: 2p³ (3 electrons) ✓
• 1.09 MJ/mol: 2s² (should show 2 electrons, not 1)

Answer: Element is Nitrogen (N), configuration: $1s^2 2s^2 2p^3$

Why does the 2p peak (1.31 MJ/mol) have the greatest height?

The 2p orbital contains 3 electrons (2p³), which is more than:

• 1s orbital (2 electrons)
• 2s orbital (2 electrons)

The height/intensity of a PES peak is proportional to the number of electrons in that sublevel. Since nitrogen has three 2p electrons (following Hund's rule with one electron in each 2p orbital), this peak shows the greatest signal.

Note: The problem statement may have simplified the relative heights. Typically, 1s² and 2s² would both show heights of 2, but the 2p³ peak would indeed be the tallest at 3.