An atom has a core and outside the core an electron occupies an orbital for...

Explanation

The question is asking which atom is likely to have an electron in an orbital with the quantum numbers \(n = 4, L = 0, m = 0\) and \(s = +\frac{1}{2}\) or \(-\frac{1}{2}\). Let's break down what these quantum numbers mean. \(n\) is the principal quantum number and represents the energy level of the electron. In this case, the electron is in the fourth energy level. \(L\) is the angular momentum quantum number and determines the shape of the orbital. When \(L = 0\), the orbital is an s-orbital, which has a spherical shape. Finally, \(m\) is the magnetic quantum number and represents the orientation of the orbital in space. When \(m = 0\), the orbital is oriented along the z-axis. Now, let's look at the options. Option A is boron, which has an atomic number of 5. Sodium, option B, has an atomic number of 11. Potassium, option C, has an atomic number of 19. Fluorine, option D, has an atomic number of 9. Since we know the electron is in an s-orbital with \(n = 4, L = 0, m = 0\) and \(s = +\frac{1}{2}\) or \(-\frac{1}{2}\), we can eliminate boron and sodium because they both have electrons in lower energy levels and do not have the correct quantum numbers. Fluorine has an electron in the \(2s\) orbital, which also does not match the given quantum numbers. Therefore, the correct answer is option C, potassium, which has an electron in the \(4s\) orbital with the correct quantum numbers. Understanding quantum numbers and the energy levels of electrons is important in understanding the behavior of atoms and their interactions with other atoms and molecules. To learn more about this, please read the relevant sections of the recommended textbooks.