Problem 34 - University Physics with Modern Physics

Question

The photoelectric work functions for particular samples of certain metals are as follows: cesium, 2.1 eV; copper, 4.7 ev; potassium, 2.3 \(\mathrm{eV}\) ; and zinc, 4.3 \(\mathrm{eV}\) . (a) What is the threshold wavelength for each metal surface? (b) Which of these metals could not emit photoelectrons when irradiated with visible light \((380-750 \mathrm{nm}) ?\)

Step-by-Step Solution

Answer

The threshold wavelengths are: cesium ~ 590 nm, copper ~ 264 nm, potassium ~ 539 nm, zinc ~ 288 nm. Copper and zinc cannot emit electrons with visible light.

Step 1: Understand the Photoelectric Effect

The photoelectric effect describes the emission of electrons from a metal surface when it is irradiated with light of sufficient energy. The energy of the photons must be greater than the work function of the metal for electrons to be emitted.

Step 2: Convert Work Function from eV to Joules

The work function (phi) in electron volts (eV) can be converted to joules (J) using the conversion factor 1 eV = 1.602 imes 10^{-19} J. For example, for cesium, \( \phi = 2.1 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} \).

Step 3: Use Planck's Equation to Find Threshold Wavelength

The threshold wavelength (\lambda_0) can be found using Planck's equation: \[\phi = \frac{h \cdot c}{\lambda_0}\]where \(h\) is Planck’s constant (6.626 imes 10^{-34} \, \text{Js}) and \(c\) is the speed of light (3.00 imes 10^8 \, \text{m/s}). Solve for \(\lambda_0\): \[\lambda_0 = \frac{h \cdot c}{\phi}\].

Step 4: Calculate the Threshold Wavelength for Each Metal

Substitute the work function (in joules) of each metal into the equation to calculate \(\lambda_0\). For cesium, it is \(\lambda_0 = \frac{6.626 \times 10^{-34} \cdot 3.00 \times 10^8}{2.1 \times 1.602 \times 10^{-19}}\). Similarly, repeat for copper, potassium, and zinc.

Step 5: Determine Metals Not Emitting with Visible Light

Compare \(\lambda_0\) for each metal to the visible light range 380-750 nm. If \(\lambda_0\) is shorter than 380 nm, the metal cannot emit photoelectrons under visible light because it requires more energetic (shorter wavelength) light.

Key Concepts

Photon Energy

Photon energy is the energy carried by a single photon, which is a particle representing a quantum of light. Each photon has a specific amount of energy that depends on its frequency or wavelength. The relationship between the energy of a photon (E) and its frequency (u) is given by the equation:E = h u. Here:

• E represents the photon energy in joules.
• h is Planck's constant, approximately 6.626 imes 10^{-34} Joule seconds.
• u is the frequency of the photon in hertz (Hz).

Another way to express this is in terms of wavelength (λ):E = \( \frac{h cdot c}{λ}\). Here:

• c is the speed of light, approximately 3.00 imes 10^8 meters per second.
• λ is the wavelength of the light in meters.

This equation shows that shorter wavelengths (higher frequencies) have higher energies. So, ultraviolet light has more energy than visible or infrared light. This concept is central to understanding the photoelectric effect, where light needs enough photon energy to release electrons from a material.

Work Function

The work function is a critical concept in understanding the photoelectric effect. It represents the minimum energy required to overcome the attractive forces binding electrons to the surface of a material. This energy threshold must be surpassed for an electron to be ejected from the surface. Each metal has a unique work function due to its specific atomic structure. Work function ( φ ) is usually measured in electron volts ( eV ) but can also be expressed in joules by multiplying the value in eV by the conversion factor ( 1 eV = 1.602 imes 10^{-19} Joules). For instance: - Cesium has a work function of 2.1 eV , which is equivalent to 2.1 imes 1.602 imes 10^{-19} Joules. - Copper's work function is 4.7 eV (approximately 7.53 imes 10^{-19} Joules). - Potassium's work function is 2.3 eV (about 3.68 imes 10^{-19} Joules). - Zinc's work function is 4.3 eV (nearly 6.89 imes 10^{-19} Joules). Understanding a material's work function is essential when selecting materials for devices such as photovoltaic cells and sensors, where efficient electron emission is crucial.

Threshold Wavelength

Threshold wavelength is an important concept that ties together photon energy and the work function. It represents the maximum wavelength of light that can eject an electron from a particular material. Light with a longer wavelength than the threshold wavelength doesn't have enough energy to overcome the work function.To determine the threshold wavelength (λ_0), we use Planck's equation: λ_0 = \( \frac{h cdot c}{φ}\). By sweeping across different metals like cesium, copper, potassium, and zinc, we substitute their respective work functions (in Joules) into this equation:

• For cesium: λ_0 = \( \frac{6.626 imes 10^{-34} cdot 3.00 imes 10^8}{2.1 imes 1.602 imes 10^{-19}}\) yields a particular wavelength.
• Similar calculations apply to copper, potassium, and zinc to find their threshold wavelengths.

Metals with shorter calculated threshold wavelengths than the visible light range (380 to 750 nm) will not emit electrons when exposed to visible light.

Planck's Equation

Planck's equation is a fundamental formula in quantum mechanics that relates the energy of a photon to its frequency or wavelength. This equation plays a critical role in the photoelectric effect by helping calculate the energy needed for electron emission.For the photoelectric effect, we use Planck's equation in the form:\( φ = \frac{h cdot c}{λ_0}\).

• φ is the work function or the minimum energy required to eject an electron from the surface.
• h is Planck's constant (6.626 imes 10^{-34} Joule seconds).
• c is the speed of light (3.00 imes 10^8 m/s).
• λ_0 is the threshold wavelength, the longest wavelength capable of ejecting an electron.

This equation is rearranged to solve for the threshold wavelength when determining the capability of different metals to emit electrons under specific light conditions. By utilizing Planck’s equation, one can predict the behavior of metals when exposed to light of varying wavelengths, essential for designing applications that rely on photoelectric properties.