Problem 107 - Chemistry The Central Science

Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (UPS, or just PES), a technique based on the photoelectric effect. coo (Section 6.2 ) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength \(58.4 \mathrm{nm}\). (a) What is the energy of a photon of this light in eV? (b) Write an equation that shows the process corresponding to the first ionization energy of \(\mathrm{Hg}\). (c) The kinetic energy of the emitted electrons is measured to be \(10.75 \mathrm{eV}\). What is the first ionization energy of Hg in kJ/mol? (d) Using Figure 7.9 , determine which of the halogen elements has a first ionization energy closest to that of mercury.

Step-by-Step Solution

Answer

The energy of a photon with a wavelength of 58.4 nm is calculated as \(E_{eV} = \frac{E}{1.602 \times 10^{-19}\mathrm{J/eV}}\), where E is the product of Planck's constant and the speed of light divided by the wavelength: \( E = (6.626 \times 10^{-34} \mathrm{J \cdot s}) \cdot \frac{3.00 \times 10^8 \mathrm{m/s}}{58.4 \times 10^{-9}\mathrm{m}}\). The first ionization energy of mercury is represented by the equation: Hg(g) -> Hg^+(g) + e^-. For the first ionization energy in kJ/mol, we find the energy difference between the photon energy and the kinetic energy and then convert it using conversion factors. We get the ionization energy (kJ/mol) as: Ionization energy (kJ/mol) = Ionization energy (eV) × \(\frac{1.602 \times 10^{-19} \mathrm{J}}{1\mathrm{eV}}\) × \(\frac{1\mathrm{kJ}}{1000\mathrm{J}}\) × \(\frac{6.022 \times 10^{23}}{1\mathrm{mol}}\). To find the halogen with the first ionization energy closest to that of mercury, compare the calculated value to the ionization energies of the halogens (F, Cl, Br, I, At) provided in the figure or reference table.

Step 1: Calculate the energy of a photon in eV

First, we have to find the energy of a photon of light with a given wavelength (58.4 nm) using the formula: \( E = h \cdot c/ \lambda\), where \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{J \cdot s}\)), \(c\) is the speed of light (\(3.00 \times 10^8 \mathrm{m/s}\)), and \( \lambda\) is the wavelength. Convert the wavelength to meters: \( 58.4 \cdot 10^{-9} \mathrm{m}\) Calculate the energy in joules: \( E = (6.626 \times 10^{-34} \mathrm{J \cdot s}) \cdot \frac{3.00 \times 10^8 \mathrm{m/s}}{58.4 \times 10^{-9}\mathrm{m}}\) Now, we have to convert the energy in joules to eV using a conversion factor, 1 eV \(= 1.602 \times 10^{-19} \mathrm{J}\). Energy in eV: \(E_{eV} = \frac{E}{1.602 \times 10^{-19}\mathrm{J/eV}}\)

Step 2: Write an equation corresponding to the first ionization energy of Hg

The first ionization energy of mercury corresponds to the energy required to remove an electron from the neutral mercury (Hg) atom. We can write this process as: Hg(g) -> Hg^+(g) + e^-

Step 3: Calculate the first ionization energy of Hg in kJ/mol

The kinetic energy of the emitted electrons is given to be 10.75 eV. From Step 1, we have the energy of a photon, so we can find the energy difference, which corresponds to the first ionization energy: Ionization energy (eV) = Photon energy (eV) - Kinetic energy (eV) Now, we must convert the ionization energy to kJ/mol. 1 eV = 1.602 × 10^(-19) J 1 kJ = 1000 J 1 mol of electrons = 6.022 × 10^(23) electrons Ionization energy (kJ/mol) = Ionization energy (eV) × \(\frac{1.602 \times 10^{-19} \mathrm{J}}{1\mathrm{eV}}\) × \(\frac{1\mathrm{kJ}}{1000\mathrm{J}}\) × \(\frac{6.022 \times 10^{23}}{1\mathrm{mol}}\)

Step 4: Compare the ionization energy of Hg to the halogens

According to the given figure in the exercise (which is not shown here), we need to determine which of the halogen elements (F, Cl, Br, I, At) has a first ionization energy closest to that of mercury. Compare the calculated ionization energy of Hg to the ionization energies of the halogens to find the closest value. The halogen with the closest ionization energy is the desired answer.

Key Concepts

Ultraviolet Photoelectron Spectroscopy (UPS)

Ultraviolet Photoelectron Spectroscopy, or UPS, is a robust method employed to measure the ionization energies of elements in a sample. This technique relies on the photoelectric effect, where monochromatic UV light is introduced to a sample, resulting in the emission of electrons. These electrons carry kinetic energy, which when measured, reveals vital information about the substance being studied.

To comprehend UPS, imagine shining a UV light on atoms. Electrons within these atoms absorb energy and can be ejected if the photon energy surpasses their binding energy. The ejected electrons’ kinetic energy is calculated, and by knowing the incoming photon energy, we can derive the energy needed to dislodge the electrons. This energy is the ionization energy specific to each element, providing insights into its electronic structure and chemical properties.

The Photoelectric Effect

The photoelectric effect is a phenomenon that occurs when light or any electromagnetic radiation hits a material and ejects electrons from its surface. Devices like solar panels and light sensors work based on this concept. For the photoelectric effect to take place, the incident radiation must have a frequency higher than the threshold frequency of the material, carrying enough quantum of energy to overcome the attraction between the electrons and the nucleus.

In the context of UPS, the photoelectric effect underpins the entire operation. The light used has a specific wavelength that determines its energy. When this light successfully causes an electron to break free, the electron's kinetic energy and the light's energy tell us how tightly the atom held onto that electron, indicating the atom’s first ionization energy.

Kinetic Energy of Electrons

Kinetic energy is the energy possessed by an object due to its motion. When dealing with electrons released through the photoelectric effect in UPS, their kinetic energy is directly measured. This measurement is crucial as it allows us to back-calculate to find the ionization energy of the atom.

Let’s look at the equation: Kinetic Energy (KE) of an electron = Energy of the Photon (hv) - Binding Energy (Ionization Energy). In a PES experiment, the kinetic energy of the electrons can help us determine how strongly an electron is bound within its atom. High kinetic energy indicates that the electron was less tightly bound, while lower kinetic energy suggests a strong attraction within the atom.

Planck's Constant

Planck's constant is a fundamental constant in physics, symbolized by the letter 'h', and it plays a pivotal role in quantum mechanics. Its value is approximately 6.626 x 10^-34 joule seconds. This constant relates the energy of a photon to the frequency of the incoming light. Specifically, the energy (E) can be expressed as the product of Planck's constant (h) and the frequency (u) of the light: E = hu.

In the application of the UPS method, Planck’s constant provides the bridge between the theoretical aspects and the practical measures - turning the wavelengths and frequencies of light into tangible energy values that help to characterize materials. It’s pivotal to the correct calculation of the energy of the photons used to liberate electrons in the PES exercises, directly influencing the measured kinetic energy and hence the ionization energies of elements.