Problem 12 - Chemistry and Chemical Reactivity

Question

You are an engineer designing a switch that works by the photoelectric effect. The metal you wish to use in your device requires \(6.7 \times 10^{-19} \mathrm{J} /\) atom to remove an electron. Will the switch work if the light falling on the metal has a wavelength of 540 nm or greater? Why or why not?

Step-by-Step Solution

Answer

No, the photon energy is less than the work function, so the switch won't work.

Step 1: Understand the problem

We need to determine if light with a wavelength of 540 nm or greater can eject an electron from the surface of a metal, given the work function of the metal is \(6.7 \times 10^{-19} \mathrm{J}\).

Step 2: Calculate Frequency of Light

The frequency \( f \) of light can be calculated using the formula: \( c = \lambda \cdot f \), where \( c \) is the speed of light \(3 \times 10^8 \text{ m/s}\) and \( \lambda \) is the wavelength. For \( \lambda = 540 \text{ nm} = 540 \times 10^{-9}\text{ m}\), \( f = \frac{c}{\lambda} = \frac{3 \times 10^8}{540 \times 10^{-9}}\).

Step 3: Calculate Energy of Photons

Using Planck's equation, the energy \( E \) of a photon can be calculated as: \( E = h \cdot f \), where \( h \) is Planck's constant \(6.63 \times 10^{-34} \mathrm{Js}\). Substitute the frequency from Step 2 into this equation.

Step 4: Compare Photon Energy to Work Function

The energy calculated from Step 3 must be compared to the work function provided, \(6.7 \times 10^{-19} \mathrm{J}\), to determine if it is enough to remove an electron. If \( E \) is greater than or equal to the work function, the electron can be ejected.

Key Concepts

Understanding the Work Function

The work function is a fundamental concept when dealing with the photoelectric effect. It defines the minimum energy needed to remove an electron from the surface of a metal. Each metal has a unique work function, expressed in joules (J).

For this exercise, the work function of the metal in question is given as \(6.7 \times 10^{-19} \text{ J}\). This means that any photon aiming to eject an electron from this metal must have energy equal to or greater than this value.

In practical terms, if the photon's energy is less than the work function, it will not be able to overcome the attractive forces holding the electron in the metal. This energy, therefore, acts as a barrier that controls the photoelectric emission. If the energy is higher or at least equal, it triggers the ejection of an electron, allowing the switch in this scenario to potentially work.

Calculating Photon Energy

Once you understand the work function, the next step is calculating the energy of the photons that will interact with the metal. Photons carry energy that is calculated using Planck’s equation, \(E = h \cdot f\), where:

• \(E\) is the energy of the photon in joules.
• \(h\) is Planck's constant \((6.63 \times 10^{-34} \text{ Js})\).
• \(f\) is the frequency of the photon.

Determining the frequency requires going through a wavelength calculation, which we will discuss next. Once you have the frequency, multiplying it by Planck’s constant gives you the photon's energy. Remember, the photon energy must meet or exceed that of the work function to liberate an electron from the metal surface effectively.

Wavelength and Frequency Calculation

The wavelength of light, given as 540 nm in this case, needs to be converted to meters for consistent unit usage, resulting in \(540 \times 10^{-9} \text{ m}\). Using the formula \(c = \lambda \cdot f\), where \(c\) is the speed of light \((3 \times 10^8 \text{ m/s})\), we can find the frequency \(f\).

The equation rearranges to \(f = \frac{c}{\lambda}\). Substituting in our values, we get the frequency. This is a crucial step as frequency directly relates to energy.

Higher frequencies mean higher energy photons, and thus, increasing the likelihood of surpassing the metal's work function. Once calculated, this frequency is then used to find the photon's energy, which determines the feasibility of your switch design, according to the principles of the photoelectric effect.