Problem 32 - University Physics with Modern Physics

Question

A 2.50 -W beam of light of wavelength 124 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 \(\mathrm{cV}\) . Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts of this metal? (b) How many photoelectrons are ejected eachsecond from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

Step-by-Step Solution

Answer

(a) 5.84 eV (b) 1.56 × 10^18 electrons/s (c) 7.8 × 10^17 electrons/s (d) 7.81 × 10^17 electrons/s.

Step 1: Understanding the Given Data

We are given a power of the beam as 2.50 W, a wavelength of 124 nm, and the maximum kinetic energy of the ejected electrons as 4.16 eV. We need to find the work function of the metal, the number of ejected photoelectrons per second, and analyze changes with adjustments in light power and wavelength.

Step 2: Convert Wavelength to Energy

First, convert the given wavelength into energy using Planck's equation: \[ E = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \ J\cdot s \) and \( c = 3 \times 10^8 \ m/s \). Convert \(\lambda = 124 \ nm = 124 \times 10^{-9} \ m\).

Step 3: Calculate Photon Energy

Calculate the photon energy \(E\):\[ E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{124 \times 10^{-9}} = 1.60 \times 10^{-18} \ J \]Convert to electron volts (1 eV = \(1.602 \times 10^{-19} \ J\)): \[ E \approx \frac{1.60 \times 10^{-18}}{1.602 \times 10^{-19}} \approx 10 \ eV \]

Step 4: Determine Work Function (a)

Using the photoelectric equation: \[ KE_{max} = E - \phi \]where \( KE_{max} = 4.16 \ eV \). Therefore, \[ \phi = E - KE_{max} = 10 \ eV - 4.16 \ eV = 5.84 \ eV \]The work function is 5.84 eV.

Step 5: Calculate Number of Photons (b)

Determine the number of photons per second:The power \( P = 2.50 \ W = \frac{J}{s} \).So, \( \text{number of photons per second} = \frac{P}{E} \ = \frac{2.50}{1.60 \times 10^{-18}} = 1.56 \times 10^{18} \text{photons/s}. \)

Step 6: Calculate Ejected Photoelectrons (b)

Since each photon ejects one electron, the number of photoelectrons is the same as the number of photons, which is \(1.56 \times 10^{18} \text{electrons/s} \).

Step 7: Effect of Reducing Power by Half (c)

If power is halved, \( P_{new} = 1.25 \ W \). Number of photons = \( \frac{1.25}{1.60 \times 10^{-18}} = 7.8 \times 10^{17} \text{photons/s} \).This equals the number of photoelectrons, \(7.8 \times 10^{17} \text{electrons/s} \).

Step 8: Effect of Reducing Wavelength by Half (d)

A half wavelength \(\lambda = 62 \ nm \) increases energy:\[ E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{62 \times 10^{-9}} = 3.2 \times 10^{-18} \ J \]Convert to eV: \(E \approx 20 \ eV \).Number of photons = \( \frac{2.50}{3.2 \times 10^{-18}} \approx 7.81 \times 10^{17} \text{photons/s} \), same as photoelectrons.

Key Concepts

Work Function

In the realm of the photoelectric effect, the work function is a crucial concept. The work function, denoted as \( \phi \), is the minimum energy required to eject an electron from the surface of a metal. Essentially, it's the energy barrier that the incoming photon must overcome to release an electron. Different metals have different work functions, depending on the nature of their atomic structure and surface properties.
To calculate the work function, we use the photoelectric equation:

• \( KE_{\text{max}} = E - \phi \)

where \( KE_{\text{max}} \) is the kinetic energy of the ejected electron and \( E \) is the energy of the incident photon.
Understanding the work function helps in selecting the appropriate material for applications like solar cells and photodetectors, where efficient electron ejection is needed.

Photon Energy

Photon energy is the energy carried by a single photon. It's a fundamental component of quantum mechanics and is essential to understanding the photoelectric effect. Photons are elementary particles of light, and their energy can be calculated using Planck's equation:

• \( E = \frac{hc}{\lambda} \)

where \( h \) is Planck’s constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the light.
When light hits a metal surface, the photon energy determines whether electrons are ejected. If the photon energy is greater than the work function of the metal, electrons will be released. Otherwise, they remain bound to the metal surface. Knowing how to calculate photon energy is key to predicting electron emission and is crucial in experiments that employ the photoelectric effect.

Kinetic Energy of Electrons

Once an electron is ejected from a metal surface due to the photoelectric effect, it possesses kinetic energy, which is a measure of its motion. This kinetic energy is what ultimately gets measured in experiments involving the photoelectric effect.
The maximum kinetic energy \( KE_{\text{max}} \) of the ejected electrons can be calculated using the following relationship:

• \( KE_{\text{max}} = E - \phi \)

Here, \( E \) is the energy of the incoming photon and \( \phi \) is the work function of the metal. Understanding this relationship helps in identifying the energy that electrons will have once they escape the metal surface.
The ability to calculate and measure this kinetic energy is vital for various applications, including the development of photoemissive materials and devices.

Wavelength and Frequency Relationship

Wavelength and frequency share a fundamental relationship in the context of waves, including light. This relationship is expressed by the equation:

• \( c = \lambda \cdot f \)

where \( c \) is the speed of light, \( \lambda \) is the wavelength, and \( f \) is the frequency of the wave. This equation tells us that the wavelength and frequency are inversely proportional; meaning, as one increases, the other decreases.
In terms of the photoelectric effect, understanding this relationship is pivotal because the energy of a photon depends on its frequency, as given by:

• \( E = hf \)

Thus, knowing the wavelength allows us to find the frequency and subsequently calculate the photon’s energy. Exploring this concept helps in comprehending why shorter wavelengths (or higher frequencies) are more capable of ejecting electrons from metal surfaces.