Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. \(\infty \infty\) (Section 6.2) In PES, monochromatic light is directed onifference between the energy of the photons and the kinetic energy of the electrons corresponds to the to a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The denergy 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 \(\mathrm{eV}\) ? (b) Write an equation that shows the process corresponding to the first ionization energy of Hg. (c) The kinetic energy of the emitted electrons is measured to be \(10.75 \mathrm{eV}\). What is the first ionization energy of \(\mathrm{Hg}\), in \(\mathrm{kJ} / \mathrm{mol}\) ? (d) Using Figure 7.10, determine which of the halogen elements has a first ionization energy closest to that of mercury.

Short answer

(a) The energy of a photon with a wavelength of \(58.4 \mathrm{nm}\) is \(E = 21.22 \mathrm{eV}\). (b) The first ionization energy of Hg is represented by the equation: Hg(g) → Hg+(g) + e^- (c) The first ionization energy of Hg is \(10.47 \mathrm{kJ/mol}\). (d) Among the halogen elements, iodine (I) has a first ionization energy closest to that of mercury.

Step-by-step solution

(a) Calculate the energy of a photon

: To find the energy of the photon, we will use the following equation: \(E = h\frac{c}{\lambda}\) Here, \(E\) is the energy of the photon, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), \(c\) is the speed of light (\(2.998 \times 10^8 \mathrm{m/s}\)), and \(\lambda\) is the wavelength of the photon (in our case, \(58.4 \mathrm{nm}\), which is \(58.4 \times 10^{-9} \mathrm{m}\)). Now, we will calculate the energy of the photon: \(E = (6.626 \times 10^{-34} \mathrm{Js})\frac{2.998 \times 10^8 \mathrm{m/s}}{58.4 \times 10^{-9} \mathrm{m}}\) After calculating the energy in Joules, convert it to electron volts by dividing by the charge of an electron (\(1.602 \times 10^{-19}\mathrm{C}\)): \(E [\mathrm{eV}] = \frac{E [\mathrm{J}]}{1.602 \times 10^{-19} \mathrm{C}}\)

(b) Write the ionization equation for Hg

: The first ionization energy refers to the energy required to remove an electron from the outermost shell of an atom. For mercury (Hg), the equation will look like: Hg(g) → Hg+(g) + e^-

(c) Calculate the first ionization energy of Hg

: We are given that the kinetic energy of the emitted electrons is \(10.75 \mathrm{eV}\). According to the photoelectric effect, the difference between the photon's energy and the kinetic energy of the electrons is equal to the ionization energy: Ionization energy (Hg) = Photon energy - Kinetic energy of emitted electrons Now, convert the ionization energy from electron volts to kJ/mol. To do this, first, convert electron volts to Joules by multiplying by the charge of an electron (\(1.602 \times 10^{-19}\)C), and then convert this value to kJ/mol by using Avogadro's number (\(6.022 \times 10^{23}\) mol^-1). Ionization energy (Hg) [kJ/mol] = Ionization energy (Hg) [J] * \(\frac{1}{1.602 \times 10^{-19}\mathrm{C}}\) * \(\frac{6.022 \times 10^{23}\mathrm{mol^{-1}}}{10^3\mathrm{kJ/J}}\)

(d) Determine the halogen with a similar ionization energy to Hg

: Once the first ionization energy of Hg is calculated, refer to a table of ionization energies or Figure 7.10 (as mentioned in the exercise) to compare the values for the halogen elements (F, Cl, Br, I, At). Determine which halogen element has a first ionization energy closest to that of mercury.

Key concepts

Ionization Energy

Ionization energy is a fundamental concept in chemistry that describes the minimum amount of energy required to remove an electron from a neutral atom in its gaseous state. In the context of photoelectron spectroscopy (PES), it is particularly important because PES measures this energy. It helps us understand the strength of the electron binding within atoms, which in turn can offer insights into reactivity, bonding, and electronic structure.

• The higher the ionization energy, the more difficult it is to remove an electron.
• Ionization energies generally increase across a period and decrease down a group in the periodic table.
• Knowledge of ionization energies helps predict molecular behavior during chemical reactions.

To enhance understanding, it's useful to discuss the process of ionization in PES experiments like the one described, emphasizing the direct relationship between the energy of light used and the kinetic energy of emitted electrons.

Photoelectric Effect

The photoelectric effect is an essential principle in both physics and chemistry, which involves the emission of electrons from a material when light hits its surface. Discovered by Heinrich Hertz and later explained by Albert Einstein, who received the Nobel Prize for this work, the photoelectric effect underpins the operation of photoelectron spectroscopy.

• Only light with sufficient photon energy can eject electrons, with the threshold energy being specific to the material.
• The emitted electrons' kinetic energy depends on the photon's energy minus the work function of the material.

When a PES experiment is conducted, it is this principle that allows us to observe and measure the ionization energies of elements such as mercury, by shedding light with enough energy on its atoms.

Electron Volts

An electron volt (eV) is a unit of energy that's particularly convenient when dealing with atomic and subatomic systems. It's defined as the amount of kinetic energy gained or lost by a single electron moving across an electric potential difference of one volt. In photoelectron spectroscopy, the energy of photons and electrons is often expressed in electron volts because of the small energy scales involved.

• An eV is a small unit of energy equivalent to approximately \(1.602 \times 10^{-19}\) joules.
• It is a practical unit in the quantum realm, helping bridge the gap between everyday energy scales and those at the atomic level.

In the example problem, it's noteworthy that converting the photon's energy from joules to eV, and vice versa, is necessary to calculate the first ionization energy of mercury in a familiar unit.

Planck's Constant

Planck's constant is a crucial quantity in quantum mechanics, symbolized by \(h\). It relates the energy of a photon to its frequency and was first introduced by Max Planck. This fundamental constant signifies the quantized nature of energy in the field of quantum physics, and it plays a pivotal role in calculations involving the photoelectric effect and photoelectron spectroscopy.

• Planck's constant has a value of approximately \(6.626 \times 10^{-34}\) Joule-seconds (Js).
• It allows us to calculate the energy of photons, as seen in the PES exercise with the equation \(E = h \frac{c}{\lambda}\), linking energy (E), speed of light (c), and wavelength (\lambda).

Understanding Planck's constant is crucial for students as it facilitates comprehension of how light behaves both as a wave and a particle (wave-particle duality), which is a cornerstone of modern physics and chemistry.