Question

identify the element that is most likely to produce the pes graph displayed below.

Explanation:

Step1: Calculate total number of electrons

Sum the relative number of electrons: \(528 + 61.6+51.3 + 7.43+4.53+1.16 + 0.71\approx655\). Since the relative number of electrons is proportional to the actual number of electrons in each sub - shell, and we assume the ratios are based on whole numbers (after considering significant figures and possible rounding in the data), we can also use the fact that the electron configuration pattern.
The binding energy levels correspond to sub - shells. The highest binding energy is for the \(1s\) sub - shell. The relative number of electrons in the \(1s\) sub - shell is approximately \(2\) (if we consider the ratio of intensities in a proper PES interpretation, and using the rule that \(n = 1\) shell (\(1s\)) has \(2\) electrons, \(n=2\) shell: \(2s\) and \(2p\) (sum of relative intensities for \(n = 2\) gives \(2 + 6\)), \(n = 3\) shell: \(3s\), \(3p\) (sum gives \(2+6\)), \(n=4\) shell: \(4s\), \(3d\), \(4p\) (sum gives \(2 + 10+6\)).
Another way is to use the electron configuration rules. The peaks (from high to low binding energy) correspond to sub - shells. If we assume the first (highest binding energy) peak is \(1s\) (2 electrons), then we check the electron configuration.
The electron configuration of \(Br\) (bromine) is \([Ar]3d^{10}4s^{2}4p^{5}\). The sum of electrons: \(2 + 2+6+2 + 6+10+2+5=35\).
Let's count the electrons from the relative intensities (assuming proper scaling). If we consider the ratio of intensities to actual electron numbers (using the Aufbau principle). The first (highest binding energy) peak (let's assume it's \(1s\)): if we scale the intensities. For example, if we consider the lowest - energy (highest binding energy) peak (let's assume it's \(1s\)) and we know that \(1s\) has \(2\) electrons. If we take the ratio of the first peak (say intensity \(I_1\)) to the actual \(1s\) electrons (\(n_1 = 2\)). Then for each peak \(i\), \(n_i=\frac{2}{I_1}\times I_i\).
If we assume the first peak (highest binding energy) has an intensity that scales to \(2\) electrons (since \(1s\) has \(2\) electrons). The electron configuration of \(Br\):

• \(1s^{2}\): binding energy is highest.
• \(2s^{2}2p^{6}\)
• \(3s^{2}3p^{6}\)
• \(4s^{2}3d^{10}4p^{5}\)

The sum \(2+2 + 6+2+6+10+2+5 = 35\)

Answer:

Bromine (\(Br\))