Question

A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A gaseous sample is irradiated with UV light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write $$ h v=I E+\frac{1}{2} m u^{2} $$ where \(v\) is the frequency of the UV light, and \(m\) and \(u\) are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be \(5.34 \times 10^{-19} \mathrm{~J}\) using a UV source of wavelength \(162 \mathrm{nm}\). Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (i.e., the most loosely held electron)?

Short answer

The ionization energy of potassium is about \(6.97 \times 10^{-19}\, \text{J}\), corresponding to the valence electron because it is the most loosely held.

Step-by-step solution

Convert Wavelength to Frequency

First, convert the wavelength of the UV light to frequency using the formula \( v = \frac{c}{\lambda} \), where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light and \( \lambda = 162 \, \text{nm} = 162 \times 10^{-9} \, \text{m} \). Thus, \( v = \frac{3 \times 10^8}{162 \times 10^{-9}} \).

Calculate the Frequency

Calculate the frequency \( v \) using the values from Step 1. This gives \( v \approx 1.85 \times 10^{15} \, \text{Hz} \).

Use Planck's Equation to Find Photon Energy

Use Planck's equation \( E = h v \) to find the energy of the UV photon, where \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \). So, \( E = 6.626 \times 10^{-34} \times 1.85 \times 10^{15} \approx 1.23 \times 10^{-18} \, \text{J} \).

Write the Photoelectron Spectroscopy Equation

The given equation for photoelectron spectroscopy is \( hv = IE + \frac{1}{2} mu^2 \). Rearrange to solve for ionization energy (IE): \( IE = hv - \frac{1}{2} mu^2 \).

Insert Known Values to Find Ionization Energy

Substitute the known values: \( hv = 1.23 \times 10^{-18} \, \text{J} \) (photon energy from Step 3) and \( \frac{1}{2} mu^2 = 5.34 \times 10^{-19} \, \text{J} \) (given kinetic energy of the electron), to find \( IE \). Thus, \( IE = 1.23 \times 10^{-18} - 5.34 \times 10^{-19} = 6.97 \times 10^{-19} \, \text{J} \).

Verify Correlation to Valence Electron

The valence electron in potassium is the most loosely held, which corresponds to its ionization energy. The calculated ionization energy represents this because it's the energy required to remove the outermost electron, confirming that this corresponds to the valence shell electron.

Key concepts

Ionization Energy

Ionization energy refers to the amount of energy required to remove an electron from an atom or molecule in its gaseous state. This process is important because it reveals how strongly an atom's electrons are held. In photoelectron spectroscopy, ionization energy is determined by calculating the difference between the energy of incident UV photons and the kinetic energy of ejected electrons.

• It gauges how difficult it is to strip an electron from an atom.
• In the case of potassium, the calculation showed an ionization energy of approximately \( 6.97 \times 10^{-19} \) Joules.
• This energy correlates with removing an electron from the valence shell, which is the outermost electron shell.

This value is essential for understanding the behavior and reactivity of elements. Low ionization energy implies that an electron can be easily removed, making the element more reactive.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. When electrons are ejected from an atom by absorbing UV photons, they exhibit kinetic energy, which can be measured.
The kinetic energy of the ejected electrons is crucial for calculating ionization energy using the formula:
\[ h v = IE + \frac{1}{2} m u^{2} \]

• Here, \( \frac{1}{2} m u^{2} \) represents the kinetic energy of the electron.
• The value given for potassium's ejected electron was \( 5.34 \times 10^{-19} \) Joules.
• This data helps bridge the gap between photon energy and ionization energy.

The measurement of kinetic energy in this context assists in deciphering the initial energy that the electron had before it was ejected.

Valence Electron

Valence electrons are the electrons found in the outermost shell of an atom. They play a vital role in chemical bonding and reactivity. In the context of photoelectron spectroscopy, these are typically the electrons that are ejected first, as they are the most loosely held.

• Valence electrons determine how an element reacts with others.
• For potassium, the valence electron is the one removed during the photoelectron spectroscopy experiment.
• This removal provides data on ionization energy, as it's related to how easily the valence electron can be detached.

The information obtained from such experiments gives insight into the behavior of elements during chemical reactions and helps understand their periodic table trends.

UV Light

Ultra Violet (UV) light is a part of the electromagnetic spectrum with a shorter wavelength than visible light, making it ideal for initiating electronic excitations in atoms. In photoelectron spectroscopy, UV light is used to eject electrons from atoms, starting with those in the valence shell.

• Has a high enough energy to influence electrons significantly.
• Used in the experiment because its photons can provide sufficient energy to overcome the atom’s ionization energy.
• The wavelength employed in the experiment was \( 162 \text{nm} \), a typical value in the UV range.

This use of UV light highlights its importance in studying atomic properties and conducting experiments requiring precise energy measurements.

Planck's Equation

Planck's equation relates the energy of a photon to its frequency through the constant \( h \) (Planck’s constant). The equation is expressed as \( E = h v \), where \( E \) is the energy and \( v \) is the frequency. In photoelectron spectroscopy, this equation helps determine the energy of UV photons used to eject electrons.

• Planck’s constant: \( 6.626 \times 10^{-34} \) Joules-second.
• Using this, energy of a photon can be calculated from its frequency, as demonstrated in the potassium experiment.
• Helps bridge the theoretical and experimental aspects of photon interaction with matter.

Understanding Planck’s equation is crucial for translating electromagnetic wave properties into practical energy values in scientific studies.