Problem 52 - Chemistry An Atoms First Approach

Question

Ionization energy is the energy required to remove an electron from an atom in the gas phase. The ionization energy of gold is \(890.1 \mathrm{kJ} / \mathrm{mol} .\) Is light with a wavelength of \(225 \mathrm{nm}\) capable of ionizing a gold atom (removing an electron) in the gas phase? ( 1 mol gold \(=6.022 \times 10^{23}\) atoms gold.)

Step-by-Step Solution

Answer

The ionization energy per atom of gold is \(1.478\times 10^{-15}\, \mathrm{J/atom}\) and the energy per photon of light with a wavelength of \(225\, \mathrm{nm}\) is \(8.884 \times 10^{-19}\, \mathrm{J/photon}\). Since the energy per photon of light is less than the ionization energy per atom for gold, the light with a wavelength of 225 nm is not capable of ionizing a gold atom in the gas phase.

Step 1: Write down the given information

Ionization energy of gold: \(890.1\, \mathrm{kJ/mol}\) Wavelength of light: \(\lambda = 225\, \mathrm{nm}\) Speed of light: \(c = 2.998 \times 10^8\, \mathrm{m/s}\) Planck's constant: \(h = 6.626 \times 10^{-34}\, \mathrm{Js}\) Number of atoms in one mol: \(N = 6.022 \times 10^{23} \,\mathrm{atoms/mol}\)

Step 2: Convert ionization energy to energy per atom

To compare the ionization energy of gold with the energy of light, we need to convert the ionization energy from kJ/mol to kJ/atom. Divide the ionization energy by the number of atoms in one mol: Ionization energy per atom = \( \dfrac{890.1\, \mathrm{kJ/mol}}{6.022\times 10^{23}\, \mathrm{atoms/mol}}\) Ionization energy per atom = \(1.478\times 10^{-18}\, \mathrm{kJ/atom}\)

Step 3: Convert ionization energy to joules per atom

To make the units consistent, we need to convert the ionization energy per atom from kJ/atom to J/atom. Ionization energy per atom in Joules = \(1.478\times 10^{-18}\, \mathrm{kJ/atom} \times 1000\, \mathrm{J/kJ}\) Ionization energy per atom in Joules = \(1.478\times 10^{-15}\, \mathrm{J/atom}\)

Step 4: Calculate the energy of light per photon

Calculate the energy of light per photon using the formula: Energy per photon = \(h \ c / \lambda\) First, convert the wavelength of light given in nanometers to meters. \(\lambda\) = 225 nm = \(225\times 10^{-9}\, \mathrm{m}\) Now, plug in the values of Planck's constant, speed of light, and wavelength into the formula: Energy per photon = \(\dfrac{6.626 \times 10^{-34}\, \mathrm{Js} \times 2.998 \times 10^8\, \mathrm{m/s}}{225\times 10^{-9}\, \mathrm{m}}\) Energy per photon = \(8.884 \times 10^{-19}\, \mathrm{J/photon}\)

Step 5: Compare the ionization energy and the energy of light

Compare the ionization energy per atom of gold with the energy of light per photon. Ionization energy per atom = \(1.478\times 10^{-15}\, \mathrm{J/atom}\) Energy per photon = \(8.884 \times 10^{-19}\, \mathrm{J/photon}\) Since the energy per photon of light is less than the ionization energy per atom for gold, we can conclude that the light with a wavelength of 225 nm is not capable of ionizing a gold atom in the gas phase.

Key Concepts

Photon energy

To understand photon energy, we first need to grasp that light is made up of particles called photons. Each photon carries a quantum of energy, which is determined by its frequency or, equivalently, its wavelength. The energy of a photon is calculated using the formula:\[ E = \frac{h \cdot c}{\lambda} \]where:

• \(E\) is the energy of the photon.
• \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34} \, \mathrm{Js} \).
• \(c\) is the speed of light, approximately \(2.998 \times 10^8 \, \mathrm{m/s} \).
• \(\lambda\) is the wavelength of the light, expressed in meters.

This energy relationship tells us how much energy is carried by a single photon. Shorter wavelengths mean higher frequencies and thus, higher energy photons. For instance, ultraviolet light has a higher photon energy than infrared light due to its shorter wavelength.
In the exercise, we're dealing with photons having a wavelength of \(225\, \mathrm{nm} \). Using the formula, we convert this wavelength to meters and calculate the resulting energy per photon. This calculated energy will help determine if the photons have enough energy to ionize a gold atom.

Wavelength conversion

The conversion of wavelength from nanometers to meters is crucial in calculations involving photon energy. This conversion ensures consistency in units, which is necessary for accurate computation. Wavelength is usually measured in nanometers (nm) when dealing with light, but in the formula for photon energy, it's essential to convert it to meters.
In mathematical terms, this conversion is straightforward:

• Since \(1\, \mathrm{nm} = 1 \times 10^{-9} \, \mathrm{m} \), the conversion from nanometers to meters involves multiplying the wavelength in nanometers by \(10^{-9} \).

For example, the exercise states a wavelength of \(225\, \mathrm{nm}\), which would be converted to:\[ 225 \times 10^{-9}\, \mathrm{m} \]
By converting the wavelength to meters, we can directly input it into the photon energy formula, aiding our calculation of whether the light can ionize a gold atom. Such conversions are standard in scientific calculations, keeping all mathematical expressions unit-compatible.

Energy calculations

Energy calculations are crucial in determining whether a photon can ionize an atom. In our exercise, we compare two energies: the ionization energy of a gold atom and the photon energy. This comparison tells us if the photon has enough energy to remove an electron from the atom, effectively ionizing it.
First, let's understand the ionization energy. It is the amount of energy required to remove an electron from an atom in its gaseous state. For gold, this energy is provided as \(890.1 \mathrm{kJ/mol}\). To compare the photon and ionization energies, it is vital to convert ionization energy from per mole to per atom:
- Divide the ionization energy by Avogadro's number (\(6.022 \times 10^{23} \)) to find the energy required for one atom.

• This gives us the ionization energy per atom in kilojoules, then convert it to joules by multiplying by 1000.

Next, calculate the energy of one photon using its wavelength, as described in the photon section.
Comparing these energies is the final step. If the photon energy exceeds the ionization energy per atom, the photon can ionize the atom. However, in our case, the photon energy is insufficient, implying ionization of gold by this light isn't feasible.

Gold atom ionization

Ionization of a gold atom involves removing an electron, which requires a specific amount of energy, known as the ionization energy. For gold, this value is quite high, which means photons need considerable energy to achieve ionization.
Ionization energies can reflect an atom's hold on its electrons. Gold, being a metal, typically has lower ionization energies than non-metals, yet it's still significantly high due to its electron's arrangement and atomic structure. The high value for gold signifies that interaction with low-energy light like the example given (225 nm) is insufficient for ionization.
To understand this: when light strikes an atom, the photons will only ionize an electron if their energy exceeds the necessary ionization energy. When the exercise asks if 225 nm light can ionize a gold atom, it beckons you to explore the relationship between photon energy and expected ionization needs.

• Unfortunately, the calculated photon energy from 225 nm light is not adequate compared to the single-atom ionization energy, leading to the conclusion that a photon of this wavelength doesn't suffice to ionize a gold atom.

This concept underlines the significance of energy calculations when discussing interactions between photons and matter.