Photoelectron Spectroscopy (PES) practice questions

Question 1

A student analyzes a purified sample of an element using a mass spectrometer and obtains the mass spectrum shown. Based on the mass spectrum, what is the average atomic mass of the element?

Accepted Answer

$10.8 \text{ amu}$, because the isotope with a mass of $11 \text{ amu}$ is four times as abundant as the isotope with a mass of $10 \text{ amu}$.

Answer

$10.2 \text{ amu}$, because the peak at $m/z = 10$ represents the isotope with the greater relative abundance.

Answer

$10.5 \text{ amu}$, because the average atomic mass is the simple unweighted mean of the masses of the two isotopes.

Answer

$11.0 \text{ amu}$, because the average atomic mass equals the mass of the peak with the greatest relative abundance.

Question 2

A student obtains a mass spectrum of a natural sample of chlorine after the atoms are ionized to form singly charged ions. The resulting spectrum is shown.

Based on the relative heights of the peaks, what is the estimated relative atomic mass of chlorine?

Accepted Answer

$35.5$

Answer

$35.0$

Answer

$36.0$

Answer

$36.5$

Question 3

A mass spectrometer is used to analyze an unknown elemental vapor composed of diatomic molecules, $\text{X}_2$. All detected ions have a charge of $+1$. The peaks near $m/z=80$ result from $\text{X}^+$ fragments, whereas the peaks near $m/z=160$ result from intact $\text{X}_2^+$ molecular ions.

Which particulate representation is consistent with the relative abundances of the three molecular ions shown in the spectrum?

Accepted Answer

Diagram B

Answer

Diagram A

Answer

Diagram C

Answer

Diagram D

Question 4

The complete photoelectron spectra of two neutral, second-period elements, X and Y, are shown below.

Which of the following correctly identifies Element Y and explains why its $\text{1s}$ peak is located at a higher binding energy than the $\text{1s}$ peak of Element X?

Accepted Answer

Element Y is $\text{O}$; its nucleus contains a greater number of protons, exerting a stronger coulombic attraction on the core electrons.

Answer

Element Y is $\text{C}$; its $\text{1s}$ electrons experience less shielding from valence electrons, causing them to be held more tightly by the nucleus.

Answer

Element Y is $\text{O}$; its $\text{1s}$ electrons experience greater electron-electron repulsion within the core subshell, which increases their binding energy.

Answer

Element Y is $\text{N}$; its $\text{2p}$ subshell is exactly half-filled, which provides extra stability that shifts all binding energy peaks to higher values.

Question 5

A student hypothesizes that inner-shell (core) electrons shield outer-shell (valence) electrons from the full attractive force of the nucleus. Which of the following observations from photoelectron spectroscopy (PES) data provides the best evidence to support this hypothesis?

Accepted Answer

The observation that the binding energy of the valence electron drops sharply from neon ($Z = 10$) to sodium ($Z = 11$), despite the increase in nuclear charge.

Answer

The observation that the binding energy of $1s$ electrons increases continuously across Period 2 elements from lithium ($Z = 3$) to neon ($Z = 10$).

Answer

The observation that the relative peak height for the $2p$ subshell in neon is three times greater than the relative peak height for its $2s$ subshell.

Answer

The observation that the binding energy of $2s$ electrons is slightly higher than the binding energy of $2p$ electrons within the same shell of a neon atom.

Question 6

A student compares the photoelectron spectra of gaseous phosphorus and sulfur atoms. The displayed portion of each spectrum contains the peak produced by ejecting a $3s$ electron, and the peak intensities have been normalized.

Which statement best interprets the relative positions of the peaks?

Accepted Answer

The $3s$ electrons in $\text{S}$ experience a greater effective nuclear charge because $\text{S}$ has one more proton, while its additional same-shell electron provides only partial additional shielding.

Answer

The $3s$ electrons in $\text{S}$ experience a greater effective nuclear charge because $\text{S}$ has more neutrons, which increase the attraction between the nucleus and the electrons.

Answer

The $3s$ electrons in $\text{P}$ experience a greater effective nuclear charge because the lower binding-energy peak indicates that more energy is required to remove an electron.

Answer

The $3s$ electrons in $\text{P}$ experience a greater effective nuclear charge because the additional $3p$ electron in $\text{S}$ completely shields the charge of its additional proton.

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