The question is incomplete. Here is the complete question.
The photoelectron spectroscopy is shwon below.
(a) Based on the photoelectron spectrum, identify the unknown element and write its electron configuration.
(b) Consider the element in the periodic table that is directly to the right of the element identified in part (a). Would the 1s peak of this element appear to the left of, the right of, or in the same position as the 1s peak of the element in part (a)? Explain your reasoning.
Answer and Explanation: Photoelectron Spectroscopy is a method of determinining the relative energy of electrons in atoms and molecules.
It is based on the photoelectric effect: when a radiation energy incides on a substance, an electron is ejected from it. If we know the kinetic energy of the ejected electron, known as photoelectrons, and the energy of the incident radiation, it is possible to find the energy of the electron in the substance.
The energy needed to eject an electron from the sample is called Binding Energy and in an atom, depends on which shell the electron is: valence eletrons (outermost shell), binding energy is lower; core eletrons (innermost shell), binding energy is highest.
In the graph, vertical axis shows 5 peaks for different energies. The peak closer to the origin, the leftmost peak, correspond to the 1s subshell, since their are closest to the nucleus, and so, has the highest binding energy.
Following from left to the right, we noticed:
(a) Then, we can conclude the eletron configuration of the element is
[tex]1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{3}[/tex]
which is Phosphorus with atomic number of 15.
(b) The element to the right of element P is Sulfur (S). The peak 1s of sulfur will appear in the same position as the 1s peak of Phosphorus, because the elements in the Periodic Table are grouped according to certain properties. Elements in the same horizontal line are elements in the same period, which one of the characteristics is they have the same total number of electron shells.
