Problem 108 - Chemistry Matter and Change

Question

Calculate It takes \(8.17 \times 10^{-19} \mathrm{J}\) of energy to remove one electron from a gold surface. What is the maximum wavelength of light capable of causing this effect?

Step-by-Step Solution

Answer

The maximum wavelength capable of removing an electron from the gold surface is 243 nm.

Step 1: Understand the Problem

The energy given is the work function, which is the energy required to remove an electron from a surface. We need to find the maximum wavelength of light that can provide this energy to an electron.

Step 2: Use the Energy-Wavelength Relationship

The energy of a photon is related to its wavelength by the formula: \[ E = \frac{hc}{\lambda} \] where \( E \) is the energy, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \ \mathrm{Js} \), \( c \) is the speed of light \( 3.00 \times 10^8 \ \mathrm{m/s} \), and \( \lambda \) is the wavelength.

Step 3: Rearrange the Formula to Solve for Wavelength

We need to rearrange the formula to solve for the wavelength \( \lambda \): \[ \lambda = \frac{hc}{E} \] This equation will allow us to calculate the maximum wavelength for the given energy.

Step 4: Substitute Values and Calculate

Substitute \( h = 6.626 \times 10^{-34} \ \mathrm{Js} \), \( c = 3.00 \times 10^8 \ \mathrm{m/s} \), and \( E = 8.17 \times 10^{-19} \ \mathrm{J} \) into the equation: \[ \lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{8.17 \times 10^{-19}} \] Calculate the result to find \( \lambda \).

Step 5: Calculate the Value

Perform the calculation: \[ \lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{8.17 \times 10^{-19}} \approx 2.43 \times 10^{-7} \ \mathrm{m} \] Convert meters to more relevant units if needed. In this case, convert to nanometers by noting that 1 m = \(10^9\) nm.

Step 6: Convert the Wavelength to Nanometers

Convert the wavelength to nanometers: \[ 2.43 \times 10^{-7} \ \mathrm{m} = 243 \ \mathrm{nm} \] Thus, the maximum wavelength of light capable of removing an electron from the gold surface is 243 nm.

Key Concepts

Photon Energy

Photon energy is the energy carried by a single photon, the basic unit of light. This energy is crucial in phenomena like the photoelectric effect, where light shone onto certain materials can eject electrons.
Photons are both wave-like and particle-like, making their energy calculation interesting.To find the energy of a photon, use the formula: \[ E = \frac{hc}{\lambda} \]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 \(3.00 \times 10^8 \ \mathrm{m/s}\).
• \(\lambda\) is the wavelength of the photon, which can vary depending on the type of electromagnetic radiation.

Understanding photon energy is vital for solving problems where we need to determine how much energy is required to cause an effect such as emitting electrons in a material.

Work Function

The work function is the minimum energy needed to remove an electron from the surface of a solid. It's a critical component in the study of the photoelectric effect because it sets the threshold energy level that photons must exceed to cause electron emission.
In essence, it's akin to a barrier that holds the electron in place until enough energy is supplied.For example, in the given exercise, the work function is calculated to be \(8.17 \times 10^{-19} \mathrm{J}\) for a gold surface. This value varies by material, due to the differences in atomic structure and bonding strength. Metallic surfaces usually have lower work functions than nonmetals, making metals more prone to participate in the photoelectric process.To calculate wavelengths capable of overcoming the work function, this value is used in conjunction with photon energy equations to determine the necessary wavelength of incoming light. Understanding work function is essential to measure and manipulate the electronic properties of materials in technology and experimental physics.

Wavelength Calculation

Calculating the wavelength of light involved in the photoelectric effect means determining how long the wave must be to provide sufficient energy to dislodge an electron. **Wavelength and energy are inversely related**, meaning **shorter wavelengths** have **higher energy**, suitable for ejecting electrons from a material surface.The formula to calculate the maximum wavelength (\(\lambda\)) capable of causing an electron emission is:\[ \lambda = \frac{hc}{E} \]Using values from the example, Planck's constant \(h = 6.626 \times 10^{-34} \mathrm{Js}\), the speed of light \(c = 3.00 \times 10^8 \mathrm{m/s}\), and the work function \(E = 8.17 \times 10^{-19} \mathrm{J}\), the calculation yields:\[ \lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{8.17 \times 10^{-19}} \≈ 2.43 \times 10^{-7} \mathrm{m} \]Converting from meters to nanometers (1 m = \(10^9\) nm), results in a wavelength of 243 nm.This final value tells us the required wavelength of light that can facilitate the removal of an electron at the energy level indicated by the work function. Understanding wavelength calculations allows for precise tuning in practical applications like lasers and photovoltaic cells.