Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. exo (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 joules? (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 \(1.72 \times 10^{-18} \mathrm{~J}\). 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

To find the energy of a photon of light, we use the energy formula: \(E = h \times v\), where h is Planck's constant, \(6.626\times 10^{-34}\, \mathrm{J\cdot s}\), and the frequency (v) is found using the formula: \(v = \frac{c}{\lambda}\), with c being the speed of light and \(\lambda\) the wavelength. For the ionization of Hg, the equation is: \(\mathrm{Hg} \xrightarrow{\mathrm{UV\,light}} \mathrm{Hg}^+ + e^-\). The first ionization energy of Hg is calculated using the formula: \(E_{\mathrm{photon}} - E_{\mathrm{kinetic}} = E_{\mathrm{ionization}}\). To convert the ionization energy to \(\mathrm{kJ/mol}\), we use the formula: \(E_{\mathrm{ionization\, kJ/mol}} = E_{\mathrm{ionization}} \times \frac{1000\, \mathrm{J}}{1\, \mathrm{kJ}} \times N_A\), with N_A being Avogadro's number. Finally, by referring to the ionization energy trends in Figure 7.10, we can find the closest halogen element with a first ionization energy similar to that of mercury.

Step-by-step solution

Find the energy of a photon of ultraviolet light

We are given the wavelength of the ultraviolet (UV) light: \(58.4\,\mathrm{nm}\). Let's first convert the wavelength to meters: s \[ 58.4\,\mathrm{nm}\times\frac{1\,\mathrm{m}}{10^9\,\mathrm{nm}}=5.84\times 10^{-8}\,\mathrm{m}. \] We will now find the frequency (v) of the light using the formula: speed of light (c) = frequency (v) * wavelength (λ) \[ v = \frac{c}{\lambda} \] where c is the speed of light, \(3.00\times 10^{8}\, \mathrm{m/s}\), and \(\lambda\) is the wavelength. Now, we will find the energy (E) of a photon of this light using the formula: \[ E = h \times v \] where h is Planck's constant, \(6.626\times 10^{-34}\, \mathrm{J\cdot s}\).

Write the ionization equation for Hg

Write an equation representing the first ionization energy of mercury (Hg): \[ \mathrm{Hg} \xrightarrow{\mathrm{UV\,light}} \mathrm{Hg}^+ + e^- \]

Find the first ionization energy of Hg

We are given the kinetic energy of the emitted electrons as \(1.72\times 10^{-18} \, \mathrm{J}\). Then, we will find the energy of the emitted photon using the measured kinetic energy. \[ E_{\mathrm{photon}} - E_{\mathrm{kinetic}} = E_{\mathrm{ionization}} \]

Convert the ionization energy to kJ/mol

To have the ionization energy in \(\mathrm{kJ/mol}\), we will first multiply the energy by \(\frac{1000\,\mathrm{J}}{1\,\mathrm{kJ}}\) and the Avogadro's number, \(N_A=6.022\times 10^{23}\, \mathrm{mol^{-1}}\): \[ E_{\mathrm{ionization\, kJ/mol}} = E_{\mathrm{ionization}} \times \frac{1000\, \mathrm{J}}{1\, \mathrm{kJ}} \times N_A \]

Determine the closest halogen element

Using the ionization energy trends for halogens given in Figure 7.10, we can determine which of the halogen elements has a first ionization energy closest to that of mercury.

Key concepts

Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an atom or molecule. It’s a fundamental property that indicates how strongly an atom holds onto its electrons. Each element has a unique ionization energy, and it plays an important role in understanding chemical reactivity. There are a few key points to remember about ionization energy:

• It generally increases across a period on the periodic table, as atoms hold onto their electrons more tightly.
• It often decreases as you move down a group due to the increase in atomic size, making it easier to remove an outer electron.

In an ultraviolet photoelectron spectroscopy (PES) experiment, ionization energy is determined by the difference between the energy of incoming photons and the kinetic energy of emitted electrons. For mercury, if we know the energy of the photons and observe the kinetic energy, we can calculate the ionization energy using the formula:\[E_{\text{ionization}} = E_{\text{photon}} - E_{\text{kinetic}}\]Remember to convert this energy to \( \mathrm{kJ/mol} \) when necessary for standard scientific reporting.

The Photoelectric Effect

The photoelectric effect is a phenomenon where electrons are ejected from a material after absorbing light. This effect is crucial in our understanding of quantum mechanics. Albert Einstein explained the photoelectric effect by proposing that light consists of packets of energy called photons.
The main concepts to grasp about the photoelectric effect include:

• Photons need sufficient energy to overcome the work function of the material, which is the energy required to release electrons.
• Once the energy threshold is met, the kinetic energy of the ejected electrons depends on the energy of the incident photons.

In photon-based experiments, like ultraviolet photoelectron spectroscopy, scientists utilize the photoelectric effect to determine ionization energies by bombarding atoms with energetic light and measuring the ejected electrons' kinetic energies.

Planck's Constant

Planck's constant is fundamental in quantum mechanics and is crucial for understanding energy interactions in the atomic realm. It is denoted by \( h \) and has a value of \( 6.626 \times 10^{-34} \, \mathrm{J \cdot s} \). This constant is key to calculations involving the energy of photons, using the equation:\[E = h \times v\]where \( E \) is the energy of the photon and \( v \) is its frequency. Planck's constant signifies the smallest possible unit of energy, reflecting the idea that energy levels are quantized. In our ultraviolet photoelectron spectroscopy example, this constant helps calculate the energy of ultraviolet photons based on their frequency, further allowing us to explore the ionization energies of various elements, such as mercury.

Halogen Elements

Halogens are a group of elements in Group 17 of the periodic table, including fluorine, chlorine, bromine, iodine, and astatine. They are highly reactive, especially with alkali metals and alkaline earth metals, forming a wide range of salts. Halogens have several distinctive features:

• They all have seven valence electrons, making them one electron short of a full outer shell, which accounts for their high reactivity.
• Their reactivity decreases as you move down the group; for instance, fluorine is more reactive than iodine.

When considering ionization energies, halogens exhibit relatively high values due to their desire to gain an electron to achieve a stable octet configuration. In an experimental context, like in an ultraviolet photoelectron spectroscopy experiment, comparing mercury’s ionization energy with that of halogens helps scientists draw conclusions about the relative stability and reactivity of the elements.