Given a photoelectron spectrum, either #"F"# or #"C"# is peak #"X"# and the other is peak #"Y"#; both are the first ionization. #"X"# is further to the left and the x-axis reports increasing electron energy. Which atom corresponds to which peak and why?

The ionization source is #"21.2 eV"#.

#A)# Carbon corresponds to peak #"Y"# because the ionization energy generally increases from the right to the left of the periodic table.
#B)# Carbon corresponds to peak #"Y"# because the binding energy of #"F"# is larger, leading to a lower electron energy as it is ejected from the atom.
#C)# Fluorine corresponds to peak #"Y"# because peak #"X"# has a lower electron energy, and the electron was harder to remove from that atom.
#D)# Fluorine corresponds to peak #"Y"# because peak #"Y"# has a higher electron energy, and the electron was easier to remove from that atom.

Answer 1

Fluorine atom is more electronegative than carbon atom since its effective nuclear charge is higher. Another perspective is that its atomic radius is smaller, so it holds onto its valence electrons more strongly than carbon with its valence electrons.

Therefore, it is more difficult to ionize #"F"# (remove one valence electron), and the binding energy is larger. Larger binding energy is further left on the photoelectron spectrum.

A is incorrect because the ionization energy trend is stated backwards.

C is incorrect because it chooses the wrong peak (peak Y) but gives a correct physical reason that would logically lead to choosing peak X.

D is incorrect because it chooses the wrong peak (peak Y) but gives a correct physical reason that would logically lead to choosing peak X.

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Answer 2

Peak "X" corresponds to the fluorine atom, and peak "Y" corresponds to the carbon atom. This is because fluorine has a higher electron affinity than carbon, leading to a greater energy release when an electron is removed, which is reflected in the spectrum where peak "X" is further to the left, indicating a higher energy level.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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