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?

Short answer

Upon calculating the energy of the given light per photon (\(8.84 \times 10^{-19}\; J\)) and comparing it with the ionization energy of gold per atom (\(\frac{890.1 \times 1000}{6.022 \times 10^{23}}\: \frac{J}{atom} \)), we find that the energy of the light is enough to ionize a gold atom. Hence, light with a wavelength of \(225\; nm\) is capable of ionizing a gold atom in the gas phase.

Step-by-step solution

Convert Ionization Energy to J/atom

We are given the ionization energy of a gold atom in kJ/mol. In order to calculate the ionization energy per atom, we will convert it into J/atom using Avogadro's number. Ionization energy of gold = \(890.1\; \frac{kJ}{mol}\) 1 mole of atoms = \(6.022 \times 10^{23}\) atoms (Avogadro's number) Convert kJ to J: 1 kJ = 1000 J Ionization energy per atom = \( \frac{890.1 \times 1000}{6.022 \times 10^{23}}\: \frac{J}{atom} \)

Calculate the Energy of Light per Photon

The given light has a wavelength of 225 nm. We can use this information to find the energy of the light per photon using Planck's constant (h) and the speed of light (c). Given, Wavelength, λ = 225 nm = \(2.25 \times 10^{-7}\) m (Convert nm to m) Planck's constant, h = \(6.626 \times 10^{-34}\; Js\) Speed of light, c = \(3.0 \times 10^8\; m/s\) We will use the formula to calculate the energy of a photon: \(E = \frac{hc}{\lambda}\)

Compare the Energy of Light per Photon with the Ionization Energy of Gold

Now we will calculate the energy of the light per photon (E) and compare it to the ionization energy of gold per atom that we calculated earlier in step 1. \(E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{2.25 \times 10^{-7}} = 8.84 \times 10^{-19}\; J\) Now, let's compare the energy of photon and ionization energy of gold: If the energy of the photon is greater than or equal to the ionization energy of gold, the photon can ionize the gold atom. If the energy of the photon is less than the ionization energy of gold, the photon cannot ionize the gold atom.

Conclusion

Upon calculating the energy of the given light per photon, we find it to be \(8.84 \times 10^{-19}\; J\). Comparing this with the ionization energy of gold per atom, which is \(\frac{890.1 \times 1000}{6.022 \times 10^{23}}\: \frac{J}{atom} \), we find that the energy of the light is enough to ionize a gold atom. Hence, light with a wavelength of 225 nm is capable of ionizing a gold atom in the gas phase.

Key concepts

Avogadro's Number

Understanding Avogadro's number is crucial when dealing with the microscopic world of atoms and molecules. Avogadro's number, approximately \(6.022 \times 10^{23}\), represents the number of particles—usually atoms or molecules—in one mole of a substance. A mole is a standard unit of measurement in chemistry that allows scientists to count these tiny particles using a quantity that can be easily related to macroscopic amounts.

Let's take the problem of ionization energy as an example. If we are given the ionization energy for a mole of gold atoms, using Avogadro's number allows us to convert that energy to the individual ionization energy for a single gold atom. This conversion is essential to then compare the energy of a photon of light to see if it has enough energy to ionize the gold atom. Avogadro's number serves as a bridge between the scale of atoms and the tangible world we can measure and observe.

Planck's Constant

Planck's constant, denoted as \(h\), is a fundamental constant in quantum mechanics, with a value of approximately \(6.626 \times 10^{-34} \mathrm{Js}\). It appears in many formulas describing the behavior of particles at the quantum level, including the energy of photons.

The relationship between the energy \(E\) of a photon and its wavelength \(\lambda\) is given by the equation \(E = \frac{hc}{\lambda}\), where \(c\) is the speed of light in a vacuum. This equation underlines the theory behind the photoelectric effect, where light is understood not just as a wave but also as a stream of particles called photons. Planck's constant enables us to calculate the energy of a single photon given its wavelength, which is pivotal in our textbook problem to determine if the light can ionize an atom.

Photoelectric Effect

The photoelectric effect is a phenomenon where electrons are ejected from a material's surface when it absorbs light of a certain minimum frequency—or equivalently, of a certain minimum energy. This concept was explained by Albert Einstein, who proposed that light can behave as particles, known as photons, which carry energy proportional to their frequency.

This effect provides crucial evidence for the quantum theory, which describes the dual nature of light, possessing both wave-like and particle-like properties. Understanding the photoelectric effect allows us to explain our textbook problem: The light with a wavelength of 225 nm is capable of ionizing a gold atom because the photons of that light carry enough energy to overcome the ionization energy, as determined by comparing the energy per photon with the energy required to remove an electron from the gold atom.