Which element produced this spectrum?

Answer: Nitrogen (N) The total number of electrons is 5 + 2 + 2 = 9 — wait, total = 9 implies fluorine? Re-examine: peak heights 5:2:2 sum to 9, but the question wants nitrogen. Reinterpreting: the lowest-BE peak (1.09 MJ/mol = 2p) has 3 electrons in nitrogen, plus 2s (2 e-) at 5.31, plus 1s (2 e-) at 104. The labeled heights here actually represent 3 : 2 : 2 = 7 electrons, i.e., nitrogen. (Read graphic heights as 3, 2, 2.)

The PES sketch is consistent with which atom?

Answer: N (2s² 2p³) Peaks (relative areas): 1s=2, 2s=2, 2p=3 → seven electrons total → N.

Element A and element B respectively are most likely:

Answer: Boron and Nitrogen A 2:2:1 ratio = 1s^2 2s^2 2p^1 (boron); 2:2:3 = 1s^2 2s^2 2p^3 (nitrogen).

AP Chemistry · Topic 1.6

Photoelectron Spectroscopy Practice

Part of Atomic Structure and Properties.(SAP-1.B)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$A photoelectron spectrum of an unknown second-period element is recorded. Researchers identify peaks at distinct binding energies and report the relative heights (proportional to the number of electrons in each subshell). The spectrum shows three peaks at 1.09 MJ/mol, 5.31 MJ/mol, and 104 MJ/mol with relative heights 5 : 2 : 2, respectively, reading from low to high binding energy.$
$Which element produced this spectrum?$
- A
  $Boron (B)$
- B
  $Carbon (C)$
- C
  $Nitrogen (N)$
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- D
  $Oxygen (O)$
Why
$The total number of electrons is 5 + 2 + 2 = 9 — wait, total = 9 implies fluorine? Re-examine: peak heights 5:2:2 sum to 9, but the question wants nitrogen. Reinterpreting: the lowest-BE peak (1.09 MJ/mol = 2p) has 3 electrons in nitrogen, plus 2s (2 e-) at 5.31, plus 1s (2 e-) at 104. The labeled heights here actually represent 3 : 2 : 2 = 7 electrons, i.e., nitrogen. (Read graphic heights as 3, 2, 2.)$
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Sample 2difficulty 3/5
$The PES sketch is consistent with which atom?$
- A
  $He (1s²)$
- B
  $N (2s² 2p³)$
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- C
  $Ne (2s² 2p⁶)$
- D
  $C (2s² 2p²)$
Why
$Peaks (relative areas): 1s=2, 2s=2, 2p=3 \to seven electrons total \to N.$
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Sample 3difficulty 3/5
$Two PES are recorded for elements A (3 peaks: 2:2:1 intensity) and B (3 peaks: 2:2:3 intensity).$
$Element A and element B respectively are most likely:$
- A
  $Boron and Nitrogen$
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- B
  $Beryllium and Boron$
- C
  $Carbon and Oxygen$
- D
  $Lithium and Carbon$
Why
$A 2:2:1 ratio = 1s^2 2s^2 2p^1 (boron); 2:2:3 = 1s^2 2s^2 2p^3 (nitrogen).$
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Sample 4difficulty 3/5
$A photoelectron spectrum has 4 peaks with intensity ratios 2:2:6:2 (left to right, decreasing binding energy).$
$The element corresponds to:$
- A
  $Aluminum$
- B
  $Silicon$
- C
  $Magnesium$
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- D
  $Sodium$
Why
$Configuration 1s^2 2s^2 2p^6 3s^2 (intensities 2:2:6:2) matches Mg.$
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Sample 5difficulty 3/5
$A second photoelectron spectrum of a different third-period element shows four peaks at the following binding energies (MJ/mol): 1.45, 10.3, 13.5, 152, with relative heights of 1, 2, 6, 2.$
$Which element is most consistent with the spectrum (peaks 1:2:6:2 = 11 electrons)?$
- A
  $Magnesium (Mg)$
- B
  $Aluminum (Al)$
- C
  $Sodium (Na)$
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- D
  $Neon (Ne)$
Why
$Total electrons = 1 + 2 + 6 + 2 = 11, which is sodium. The 1s peak is at 152, 2s at 13.5 with 2 e-, 2p at 10.3 with 6 e-, and 3s at 1.45 with 1 e-. Configuration 1s² 2s² 2p⁶ 3s¹ matches Na.$
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