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Explain why the 1s peak in the PES spectrum of oxygen appears at a higher binding energy than the 1s peak in the PES spectrum of carbon?
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Oxygen has a greater nuclear charge (8+) than carbon (6+). The 1s electrons in oxygen experience a stronger electrostatic attraction to the nucleus, so more energy is required to remove them, resulting in a higher binding energy.
Identify the element that produces a PES spectrum with four peaks having relative heights of 2, 2, 6, and 2 (listed from highest to lowest binding energy).
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The total number of electrons is 2 + 2 + 6 + 2 = 12, which corresponds to magnesium (Z = 12). The four peaks correspond to the subshells 1s², 2s², 2p⁶, and 3s², matching the ground-state electron configuration of Mg.
Explain why the 2s electrons in a PES spectrum of fluorine appear at a higher binding energy than the 2p electrons in the same spectrum?
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The 2s subshell is lower in energy (closer to the nucleus on average) than the 2p subshell. Because 2s electrons penetrate closer to the nucleus and experience less shielding from other electrons, they are held more tightly. Removing them therefore requires more energy, so the 2s peak appears at a higher binding energy than the 2p peak.
Predict how the PES spectrum of a Na⁺ ion would differ from that of a neutral Ne atom, given that both species have 10 electrons?
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Both Na⁺ and Ne have 10 electrons with the same electron configuration (1s² 2s² 2p⁶), so both spectra show three peaks with relative heights of 2, 2, and 6. However, Na⁺ has a nuclear charge of 11+ compared to Ne’s 10+. The stronger nuclear charge in Na⁺ causes every peak to shift to a higher binding energy because all electrons are held more tightly by the larger nucleus.