We will now construct two possibilities for the photoelectron spectrum of \(\mathrm{K}\). a) First, consider the first three shells \((18\) electrons) of \(\mathrm{K}\). For these 18 electrons, estimate the IEs [Hint: compare to Ar.] and indicate their relative intensities. b) If the \(19^{\text {th }}\) electron of \(\mathrm{K}\) is found in the \(n=4\) shell, would the ionization energy be closest to \(0.42,1.4\), or \(2.0 \mathrm{MJ} / \mathrm{mole}\) ? Explain. [Hint: compare to Na and Li.] Show a predicted photoelectron spectrum based on this assumption. c) If the \(19^{\text {th }}\) electron of \(\mathrm{K}\) is found in the third subshell of \(n=3\), would the ionization energy be closest to \(0.42,1.4\), or \(2.0 \mathrm{MJ} / \mathrm{mole}\) ? Explain. [Hint: compare to other cases in which a new subshell appears.] Show a predicted photoelectron spectrum based on this assumption.

Step by step solution

01

Analyze the first three shells of K

For the first 18 electrons, the structure matches up with argon (Ar). The IE values should be very similar to those of Ar, with a 1s2 peak with the highest energy (and lowest intensity), followed by the 2s2, 2p6, 3s2 and 3p6 peaks.

02

Predict the Ionization energy in the 4th shell

Comparing with similar elements, the 19th electron in the 4th shell should have an IE closer to sodium and lithium's outer shell electrons, due to similar shell architecture. Based on this, it would likely be closest to 0.42 MJ/mole, which is representative of an electron further away from the nucleus, hence easier to remove.

03

Projecting the Photoelectron spectrum

Based on the assumptions, the photoelectron spectrum should reflect a new peak at much lower ionization energy, representing the 4s1 electron.

04

Predict the Ionization energy in third subshell of n=3

If the 19th electron is located in third subshell of the 3rd shell, the IE would likely be closer to the rest, due to increase in effective nuclear charge in the same shell. Therefore the IE would likely be closer to 2.0 MJ/mole.

05

Projecting the Photoelectron spectrum

In this case, the photoelectron spectrum will reflect a new peak for the 3d1 electron, but at a much higher energy compared to the 4s scenario.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization Energy

Ionization energy (IE) refers to the amount of energy required to remove an electron from an atom or ion in its gaseous state. In the context of the photoelectron spectrum of potassium, understanding its ionization energy is crucial for analyzing the energy required to detach electrons from different shells.

As the exercise suggests, comparing potassium to argon (which has a similar electron configuration for the first 18 electrons) offers insight into the IE of potassium's electrons. Generally, the IE increases as electrons are removed from closer to the nucleus, due to the increased electromagnetic attraction. For potassium, the outermost electron (19th) will have the lowest IE, as it experiences the least nuclear attraction and is the most easily removed.

Electron Configuration

Electron configuration describes the distribution of electrons in the atomic orbitals of an atom. Potassium, with an atomic number of 19, has the electron configuration [Ar] 4s1, meaning it has the same configuration as argon plus one additional electron in the 4s orbital.

Argon's electron configuration is 1s2 2s2 2p6 3s2 3p6, fully filling the first three shells. For potassium, those inner 18 electrons would have similar energies to those in argon, which helps in predicting the photoelectron spectrum. The photoelectron spectrum reflects the energy levels of these electrons, showing peaks correlating to electrons' ionization energies by shell. The 19th electron is in a higher energy level (the 4s shell), which affects its ionization energy compared to inner-shell electrons.

Photoelectron Spectroscopy

Photoelectron spectroscopy is a technique used to determine the energy levels of electrons within an atom or molecule. In essence, a photon beam (often from an X-ray or UV source) is directed towards a sample, and electrons are ejected as a result of the photoelectric effect.

The energy distribution of these ejected electrons is measured, creating a spectrum that represents the different ionization energies of electrons in their respective shells or orbitals. For potassium, the spectrum would show a peak for the inner 18 electrons similar to argon's, as well as an additional peak representing the 19th electron. The position of this peak offers insight into the ionization energy of the outermost electron.

Energy Levels in Atoms

The concept of energy levels in atoms is foundational for understanding atomic structure and the emission or absorption of light. Electrons occupy discrete energy levels or shells around the nucleus, and the energy required to move an electron from one level to another is quantized.

In potassium, the first three energy levels (1s, 2s, 2p, 3s, 3p) are filled similarly to argon, which informs us about the energies associated with those shells. The presence of the 19th electron in the 4s orbital introduces a new energy level, which affects the overall photoelectron spectrum. The analysis of the energy required for ionizing this 19th electron helps in understanding the furthermost energy level's characteristics relative to the others within the atom.