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?

Short answer

The switch will not work; the light wavelength is too long to release an electron.

Step-by-step solution

Determine Energy Required

The energy required to remove an electron from the metal is provided as \(6.7 \times 10^{-19} \ \mathrm{J}\). This is the threshold energy, also known as the work function, \(\phi\).

Convert Wavelength to Energy

To find out if the light can remove an electron, convert the light's wavelength to energy using the equation: \[ E = \frac{hc}{\lambda} \] where \(h = 6.626 \times 10^{-34} \ \mathrm{J \cdot s}\) is Planck's constant, \(c = 3 \times 10^8 \ \mathrm{m/s}\) is the speed of light, and \(\lambda = 540 \ \mathrm{nm} = 540 \times 10^{-9} \ \mathrm{m}\) is the wavelength. Let's calculate this energy.

Calculation

Substitute the given values into the energy equation: \[ E = \frac{(6.626 \times 10^{-34} \ \mathrm{J \cdot s})(3 \times 10^8 \ \mathrm{m/s})}{540 \times 10^{-9} \ \mathrm{m}} \] \[ E = \frac{1.9878 \times 10^{-25} \ \mathrm{J \cdot m}}{540 \times 10^{-9} \ \mathrm{m}} \] \[ E = 3.681 \times 10^{-19} \ \mathrm{J} \]

Compare Energy with Work Function

Compare the energy of the light with the work function of the metal. The energy we calculated is \(3.681 \times 10^{-19} \ \mathrm{J}\), which is less than the required \(6.7 \times 10^{-19} \ \mathrm{J}\).

Conclusion

Since the energy of the incoming photons (\(3.681 \times 10^{-19} \ \mathrm{J}\)) is less than the work function of the metal (\(6.7 \times 10^{-19} \ \mathrm{J}\)), the light with a wavelength of 540 nm is insufficient to eject an electron. Consequently, the switch will not work.

Key concepts

Work Function

In the context of the photoelectric effect, the work function is a critical concept. It is the minimum energy needed to remove an electron from the surface of a material. This energy is specific to each material and is measured in joules (J) per atom. The work function is denoted by the symbol \( \phi \).
For instance, in the given problem, the metal used requires \(6.7 \times 10^{-19} \mathrm{J} \) to remove an electron. This value indicates the threshold energy or work function that must be surpassed by the incoming photons for electron emission to occur.
Understanding the work function allows engineers and scientists to determine the suitability of a particular material in photoelectric applications. If the photon energy does not meet or exceed this value, electrons will not be emitted, and devices relying on this principle, like the switch in the exercise, won't operate effectively.

Photon Energy

Photon energy refers to the amount of energy carried by a single photon, and it is fundamental in understanding the photoelectric effect. Photons are the basic units of light and electromagnetic radiation.
To calculate the energy of a photon, the equation \( E = \frac{hc}{\lambda} \) is used, where \( E \) is the energy, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \mathrm{J \cdot s}) \), \( c \) is the speed of light \( (3 \times 10^8 \mathrm{m/s}) \), and \( \lambda \) is the wavelength of the light in meters.
When solving the exercise, photons with a wavelength of 540 nm were considered. The energy calculated was \(3.681 \times 10^{-19} \mathrm{J}\). This energy reflects the photon’s potency in potentially ejecting electrons from the metal's surface if it meets the required conditions.

Wavelength and Frequency

Wavelength and frequency are important characteristics of electromagnetic waves like light, which impact their energy. The wavelength \( (\lambda) \) is the distance between consecutive crests of a wave and is usually measured in meters (m), while frequency \( (f) \) denotes how many waves pass a point in one second, measured in Hertz (Hz).
These two properties are inversely related through the equation \( c = \lambda f \), where \( c \) is the speed of light \((3 \times 10^8 \mathrm{m/s})\). Essentially, as wavelength increases, frequency decreases and vice versa. This relationship is crucial because it affects photon energy, as represented in the equation \( E = hf = \frac{hc}{\lambda} \).
In the exercise, the wavelength of 540 nm (or \( 540 \times 10^{-9} \mathrm{m} \) ) resulted in a photon energy calculation that was not sufficient to overcome the metal's work function. This illustrates how longer wavelengths, meaning lower frequencies and energies, might not meet the energy requirement for the photoelectric effect to occur.

Threshold Energy

Threshold energy is the minimum energy needed for an electron to be emitted from a material's surface in the photoelectric effect, and it is synonymous with the work function \((\phi)\).
This concept plays a pivotal role in determining whether an incoming photon can eject an electron. If the photon's energy is equal to or surpasses the threshold energy, electron emission will occur. Conversely, if the photon's energy is below this threshold, no electrons will be emitted.
In the context of the given problem, the threshold energy, or the work function of the metal, is \(6.7 \times 10^{-19} \mathrm{J}\). As calculated, the energy of the photons with a wavelength of 540 nm was \(3.681 \times 10^{-19} \mathrm{J}\), which is below the threshold needed. Thus, the photons lack the necessary energy to induce electron emission from the metal's surface, preventing the designed switch from operating as intended.